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21873
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\section{Structure Induced by Commutative Operation is Commutative}
Tags: Abstract Algebra, Commutativity, Pointwise Operations, Mappings, Mapping Theory
\begin{theorem}
Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Let $\circ$ be a commutative operation.
Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also commutative.
\end{theorem}
\begin{proof}
Let $\struct {T, \circ}$ be a commutative algebraic structure.
Let $f, g \in T^S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \oplus g} } x
| r = \map f x \circ \map g x
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \map g x \circ \map f x
| c = $\circ$ is commutative
}}
{{eqn | r = \map {\paren {g \oplus f} } x
| c = {{Defof|Pointwise Operation}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21874
|
\section{Structure Induced by Commutative Ring Operations is Commutative Ring}
Tags: Mapping Theory, Ring Theory, Rings of Mappings
\begin{theorem}
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $S$ be a set.
Let $\struct {R^S, +', \circ'}$ be the structure on $R^S$ induced by $+'$ and $\circ'$.
Then $\struct {R^S, +', \circ'}$ is a commutative ring.
\end{theorem}
\begin{proof}
By Structure Induced by Ring Operations is Ring then $\struct {R^S, +', \circ'}$ is a ring.
From Structure Induced by Commutative Operation is Commutative, so is the pointwise operation $\circ$ induces on $R^S$.
The result follows by definition of commutative ring.
{{qed}}
Category:Rings of Mappings
\end{proof}
|
21875
|
\section{Structure Induced by Group Operation is Group}
Tags: Mapping Theory, Group Theory
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $S$ be a set.
Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\circ$.
Then $\struct {G^S, \oplus}$ is a group.
\end{theorem}
\begin{proof}
Taking the group axioms in turn:
\end{proof}
|
21876
|
\section{Structure Induced by Permutation on Algebra Loop is not necessarily Algebra Loop}
Tags: Operations Induced by Permutations, Algebra Loops
\begin{theorem}
Let $\struct {S, \circ}$ be an algebra loop.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\struct {S, \circ_\sigma}$ be the structure induced by $\sigma$ on $\circ$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\struct {S, \circ_\sigma}$ is not necessarily also an algebra loop.
\end{theorem}
\begin{proof}
Consider the Cayley table of the algebra loop on $S = \set {e, a, b}$:
:$\begin{array}{r|rrr}
\circ & e & a & b \\
\hline
e & e & a & b
\\
a & a & b & e
\\
b & b & e & a
\\
\end{array}$
Consider the permutation on $S$:
Let $\sigma$ denote the permutation on $S$ defined as:
{{begin-eqn}}
{{eqn | l = \map \sigma e
| r = a
}}
{{eqn | l = \map \sigma a
| r = e
}}
{{eqn | l = \map \sigma b
| r = b
}}
{{end-eqn}}
Then the Cayley table of the structure induced by $\sigma$ on $\circ$ is seen to be:
:$\begin{array}{r|rrr}
\circ & e & a & b \\
\hline
e & a & e & b
\\
a & e & b & a
\\
b & b & a & e
\\
\end{array}$
It is apparent by inspection that this is the Cayley table of a quasigroup.
However, there is no identity element.
Hence by definition $\struct {S, \circ_\sigma}$ is not an algebra loop.
{{qed}}
\end{proof}
|
21877
|
\section{Structure Induced by Permutation on Commutative Quasigroup is Commutative Quasigroup}
Tags: Operations Induced by Permutations, Commutativity, Quasigroups, Quasigroup
\begin{theorem}
Let $\struct {S, \circ}$ be a quasigroup such that $\circ$ is a commutative operation.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\struct {S, \circ_\sigma}$ be the structure induced by $\sigma$ on $\circ$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\struct {S, \circ_\sigma}$ is also a quasigroup such that $\circ_\sigma$ is a commutative operation.
\end{theorem}
\begin{proof}
From Structure Induced by Permutation on Quasigroup is Quasigroup we have that $\struct {S, \circ_\sigma}$ is a quasigroup.
Then we see that:
{{begin-eqn}}
{{eqn | q = \forall a, b \in S
| l = a \circ_\sigma b
| r = \map \sigma {a \circ b}
| c = {{Defof|Operation Induced by Permutation}}
}}
{{eqn | r = \map \sigma {b \circ a}
| c = {{Defof|Commutative Operation}}
}}
{{eqn | r = b \circ_\sigma a
| c = {{Defof|Operation Induced by Permutation}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21878
|
\section{Structure Induced by Permutation on Quasigroup is Quasigroup}
Tags: Operations Induced by Permutations, Semigroups, Quasigroups
\begin{theorem}
Let $\struct {S, \circ}$ be a quasigroup.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\struct {S, \circ_\sigma}$ be the structure induced by $\sigma$ on $\circ$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\struct {S, \circ_\sigma}$ is also a quasigroup.
\end{theorem}
\begin{proof}
By definition of quasigroup:
:$\forall a, b \in S: \exists ! x \in S: x \circ a = b$
:$\forall a, b \in S: \exists ! y \in S: a \circ y = b$
Let $a, b \in S$.
As $\sigma$ is a permutation, it is by definition both surjective and injective.
We have that:
:$\exists ! x: x \circ a = b$
Thus:
{{begin-eqn}}
{{eqn | q = \exists ! x \in S
| l = \map \sigma {x \circ a}
| r = b
| c =
}}
{{eqn | ll= \leadsto
| l = x \circ_\sigma a
| r = b
| c = {{Defof|Operation Induced by Permutation}}
}}
{{end-eqn}}
Similarly, we have that:
:$\exists ! x: a \circ x = b$
Thus:
{{begin-eqn}}
{{eqn | q = \exists ! x \in S
| l = \map \sigma {a \circ x}
| r = b
| c =
}}
{{eqn | ll= \leadsto
| l = a \circ_\sigma x
| r = b
| c = {{Defof|Operation Induced by Permutation}}
}}
{{end-eqn}}
The result follows.
{{qed}}
\end{proof}
|
21879
|
\section{Structure Induced by Permutation on Semigroup is not necessarily Semigroup}
Tags: Operations Induced by Permutations, Semigroups
\begin{theorem}
Let $\struct {S, \circ}$ be a semigroup.
Let $\sigma: S \to S$ be a permutation on $S$.
Let $\struct {S, \circ_\sigma}$ be the structure induced by $\sigma$ on $\circ$:
:$\forall x, y \in S: x \circ_\sigma y := \map \sigma {x \circ y}$
Then $\struct {S, \circ_\sigma}$ is not necessarily itself a semigroup.
\end{theorem}
\begin{proof}
From Operation Induced by Permutation on Magma is Closed we have that $\struct {S, \circ_\sigma}$ is a closed structure.
Hence {{SemigroupAxiom|0}} holds.
However, we have that Operation Induced by Permutation on Semigroup is not necessarily Associative.
Hence {{SemigroupAxiom|1}} does not necessarily hold for $\struct {S, \circ_\sigma}$
Hence $\struct {S, \circ_\sigma}$ is not necessarily a semigroup.
{{qed}}
\end{proof}
|
21880
|
\section{Structure Induced by Ring Operations is Ring}
Tags: Rings of Mappings, Ring Theory, Mappings, Examples of Rings, Rings, Mapping Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring.
Let $S$ be a set.
Then $\struct {R^S, +', \circ'}$ is a ring, where $+'$ and $\circ'$ are the pointwise operations induced on $R^S$ by $+$ and $\circ$.
\end{theorem}
\begin{proof}
As $R$ is a ring, both $+$ and $\circ$ are closed on $R$ by definition.
From Closure of Pointwise Operation on Algebraic Structure, it follows that both $+'$ and $\circ'$ are closed on $R^S$:
:$\forall f, g \in R^S: f +' g \in R^S$
:$\forall f, g \in R^S: f \circ' g \in R^S$
By Structure Induced by Abelian Group Operation is Abelian Group, $\struct {R^S, +'}$ is an abelian group.
By Structure Induced by Associative Operation is Associative, $\struct {R^S, \circ'}$ is a semigroup.
From Pointwise Operation on Distributive Structure is Distributive, $\circ'$ is distributive over $+'$.
The result follows by definition of ring.
{{qed}}
\end{proof}
|
21881
|
\section{Structure Induced by Ring with Unity Operations is Ring with Unity}
Tags: Mapping Theory, Ring Theory, Rings of Mappings, Rings with Unity
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring with unity whose unity is $1_R$.
Let $S$ be a set.
Let $\struct {R^S, +', \circ'}$ be the structure on $R^S$ induced by $+'$ and $\circ'$.
Then $\struct {R^S, +', \circ'}$ is a ring with unity whose unity is $f_{1_R}: S \to R$, defined by:
:$\forall s \in S: \map {f_{1_R} } s = 1_R$
\end{theorem}
\begin{proof}
By Structure Induced by Ring Operations is Ring then $\struct {R^S, +', \circ'}$ is a ring.
We have from Induced Structure Identity that the constant mapping $f_{1_R}: S \to R$ defined as:
:$\forall x \in S: \map {f_{1_R} } x = 1_R$
is the identity for $\struct {R^S, \circ'}$.
The result follows by definition of ring with unity and unity of ring.
{{qed}}
Category:Rings of Mappings
Category:Rings with Unity
\end{proof}
|
21882
|
\section{Structure Induced on Set of Self-Maps on Entropic Structure is Entropic}
Tags: Entropic Structures, Pointwise Operations
\begin{theorem}
Let $\struct {S, \odot}$ be a magma.
Let $\struct {S, \odot}$ be an entropic structure.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let $\struct {S^S, \oplus}$ denote the algebraic structure on $S^S$ induced by $\odot$.
Then $\struct {S^S, \oplus}$ is an entropic structure.
\end{theorem}
\begin{proof}
Recall the definition of algebraic structure on $S^S$ induced by $\odot$:
Let $f: S \to S$ and $g: S \to S$ be self-maps on $S$, and thus elements of $S^S$.
The pointwise operation on $S^S$ induced by $\odot$ is defined as:
:$\forall x \in S: \map {\paren {f \oplus g} } x = \map f x \odot \map g x$
Let $f, g, p, q \in S^S$ be arbitrary.
Let $x \in S$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {\paren {f \oplus g} \oplus \paren {p \oplus q} } } x
| r = \paren {\map f x \odot \map g x} \odot \paren {\map p x \odot \map q x}
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \paren {\map f x \odot \map p x} \odot \paren {\map g x \odot \map q x}
| c = {{Defof|Entropic Structure}}
}}
{{eqn | r = \map {\paren {\paren {f \oplus p} \oplus \paren {g \oplus q} } } x
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | ll= \leadsto
| l = \paren {f \oplus g} \oplus \paren {p \oplus q}
| r = \paren {f \oplus p} \oplus \paren {g \oplus q}
| c = Equality of Mappings
}}
{{end-eqn}}
Hence the result by definition of entropic structure.
{{qed}}
\end{proof}
|
21883
|
\section{Structure of Cardinality 3+ where Every Permutation is Automorphism is Idempotent}
Tags: Idempotence
\begin{theorem}
Let $S$ be a set whose cardinality is at least $3$.
Let $\struct {S, \circ}$ be an algebraic structure on $S$ such that every permutation on $S$ is an automorphism on $\struct {S, \circ}$.
Then $\circ$ is an idempotent operation.
\end{theorem}
\begin{proof}
{{AimForCont}} $\circ$ is not idempotent.
Then there exists $x \in S$ such that:
:$\exists y \in S: x \circ x = y$
where $x \ne y$.
Because there are at least $3$ distinct elements of $S$ {{hypothesis}}:
:$\exists z \in S: z \ne x, z \ne y$
Let $f: S \to S$ be a permutation on $S$ such that:
:$\map f x = x$
:$\map f y = z$
We have:
{{begin-eqn}}
{{eqn | l = z
| r = \map f y
| c = Definition of $f$
}}
{{eqn | r = \map f {x \circ x}
| c = {{hypothesis}}: $x \circ x = y$
}}
{{eqn | r = \map f x \circ \map f x
| c = {{hypothesis}}: $f$ is an automorphism
}}
{{eqn | r = x \circ x
| c = Definition of $f$
}}
{{eqn | r = y
| c = {{hypothesis}}: $x \circ x = y$
}}
{{end-eqn}}
This contradicts our assertion that $x$ and $z$ are distinct.
From Proof by Contradiction it follows that our assumption that $\circ$ is not idempotent must have been false.
Hence $\circ$ is idempotent.
{{qed}}
\end{proof}
|
21884
|
\section{Structure of Inverse Completion of Commutative Semigroup}
Tags: Commutative Semigroups, Inverse Completions
\begin{theorem}
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then:
:$T = S \circ' C^{-1}$
where:
:$C^{-1}$ is the inverse of $C$ in $T$
:$S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$.
\end{theorem}
\begin{proof}
Let $a \in C$.
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = x
| r = x \circ \paren {a \circ' a^{-1} }
| c = {{Defof|Invertible Element}}
}}
{{eqn | ll= \leadsto
| l = x
| r = \paren {x \circ a} \circ' a^{-1}
| c = {{Defof|Associative Operation}}
}}
{{eqn | ll= \leadsto
| l = x
| o = \in
| r = S \circ' C^{-1}
| c = {{Defof|Subset Product}}
}}
{{eqn | ll= \leadsto
| l = S
| o = \subseteq
| r = S \circ' C^{-1}
| c = {{Defof|Subset}}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = C
| c =
}}
{{eqn | ll= \leadsto
| l = y^{-1}
| o = \in
| r = C^{-1}
| c = {{Defof|Inverse of Subset of Monoid}}
}}
{{eqn | ll= \leadsto
| l = y^{-1}
| r = a \circ' a^{-1} \circ' y^{-1}
| c = {{Defof|Invertible Element}}
}}
{{eqn | ll= \leadsto
| l = y^{-1}
| r = a \circ' \paren {y \circ' a}^{-1}
| c = Inverse of Product
}}
{{eqn | ll= \leadsto
| l = y^{-1}
| r = a \circ' \paren {y \circ a}^{-1}
| c = {{Defof|Extension of Operation}}
}}
{{eqn | ll= \leadsto
| l = y^{-1}
| o = \in
| r = S \circ' C^{-1}
| c = {{Defof|Subset Product}}
}}
{{eqn | ll= \leadsto
| l = C^{-1}
| o = \subseteq
| r = S \circ' C^{-1}
| c = {{Defof|Subset}}
}}
{{end-eqn}}
Thus, as:
:$C^{-1} \subseteq S \circ' C^{-1}$
and:
:$S \subseteq S \circ' C^{-1}$
by Union is Smallest Superset it follows that:
:$S \cup C^{-1} \subseteq S \circ' C^{-1}$
From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup:
:$S \circ' C^{-1}$ is a commutative semigroup.
So $S \circ' C^{-1}$ is a semigroup which contains $S \cup C^{-1}$.
By definition of generator of semigroup, it follows that:
:$\gen {S \cup C^{-1} } \subseteq S \circ' C^{-1}$
Let $z = x \circ' y^{-1} \in S \circ' C^{-1}$.
Then by definition:
:$x \in S$
and:
:$y^{-1} \in C^{-1}$
and so by definition of generator of semigroup:
:$x \circ' y^{-1} \in \gen {S \cup C^{-1} }$
Thus:
$S \circ' C^{-1} \subseteq \gen {S \cup C^{-1} }$
By definition of set equality:
:$S \circ' C^{-1} = \gen {S \cup C^{-1} }$
and so by definition of the inverse completion:
:$T = S \circ' C^{-1}$
{{qed}}
\end{proof}
|
21885
|
\section{Structure of Recurring Decimal}
Tags: Number Theory
\begin{theorem}
Let $\dfrac 1 m$, when expressed as a decimal expansion, recur with a period of $p$ digits with no nonperiodic part.
Let $\dfrac 1 n$, when expressed as a decimal expansion, terminate after $q$ digits.
Then $\dfrac 1 {m n}$ has a nonperiodic part of $q$ digits, and a recurring part of $p$ digits.
\end{theorem}
\begin{proof}
Let $b \in \N_{>1}$ be the base we are working on.
Note that $b^p \times \dfrac 1 m$ is the result of shifting the decimal point of $\dfrac 1 m$ by $p$ digits.
Hence $b^p \times \dfrac 1 m - \dfrac 1 m$ is an integer, and $\paren {b^i - 1} \dfrac 1 m$ is not an integer for integers $0 < i < p$.
Therefore $m \divides b^p - 1$.
Also note that $b^q \times \dfrac 1 n$ is the result of shifting the decimal point of $\dfrac 1 n$ by $q$ digits.
Hence $b^q \times \dfrac 1 n$ is an integer, and $b^j \times \dfrac 1 n$ is not an integer for integers $0 < j < q$.
Therefore $n \divides b^q$ and $n \nmid b^{q - 1}$.
Write $m x = b^p - 1$ and $n y = b^q$.
Then $\dfrac 1 {m n} = \dfrac {x y} {\paren {b^p - 1} b^q}$.
By Division Theorem:
:$\exists! r, s \in \Z: x y = s \paren {b^p - 1} + r, 0 \le r < b^p - 1$
Then we would have:
:$\dfrac 1 {m n} = \dfrac {s \paren {b^p - 1} + r} {\paren {b^p - 1} b^q} = \dfrac s {b^q} + \dfrac r {\paren {b^p - 1} b^q}$
which is a fraction with a nonperiodic part of $q$ digits equal to $s$, followed by a recurring part of $p$ digits equal to $r$.
To show that the nonperiodic part and recurring part of $\dfrac 1 {m n}$ cannot be shorter, we must show:
:$r \not \equiv s \pmod b$: or else the nonperiodic part could be shortened by at least $1$ digit
:$\dfrac r {b^p - 1} \paren {b^i - 1}$ is not an integer for integers $0 < i < p$: or else the recurring part could be shortened to $i$ digits
To show the first condition, suppose the contrary.
Then:
{{begin-eqn}}
{{eqn | l = x y
| r = s \paren {b^p - 1} + r
}}
{{eqn | o = \equiv
| r = s \paren {b^p - 1} + s
| rr= \pmod b
}}
{{eqn | o = \equiv
| r = s b^p
| rr= \pmod b
}}
{{eqn | o = \equiv
| r = 0
| rr= \pmod b
}}
{{end-eqn}}
From GCD with Remainder:
:$\gcd \set {b, b^p - 1} = 1$
Since $x \divides b^p - 1$, by Divisor of One of Coprime Numbers is Coprime to Other:
:$\gcd \set {b, x} = 1$
By Euclid's Lemma:
:$b \divides y$
From $n y = b^q$:
:$n \times \dfrac y b = b^{q - 1}$
so $n \divides b^{q - 1}$, which contradicts the properties of $n$.
To show the second condition, suppose the contrary.
Then:
{{begin-eqn}}
{{eqn | l = \paren {b^i - 1} \dfrac 1 m
| r = \dfrac {n \paren {b^i - 1} } {m n}
}}
{{eqn | r = n \paren {b^i - 1} \paren {\dfrac s {b^q} + \dfrac r {\paren {b^p - 1} b^q} }
}}
{{eqn | r = \dfrac n {b^q} \paren {s \paren {b^i - 1} + \dfrac {r \paren {b^i - 1} } {b^p - 1} }
}}
{{end-eqn}}
Both fractions are integers, so $\paren {b^i - 1} \dfrac 1 m$ is also an integer, which contradicts the properties of $\dfrac 1 m$.
Hence the result.
{{qed}}
\end{proof}
|
21886
|
\section{Structure of Simple Algebraic Field Extension}
Tags: Fields, Field Extensions
\begin{theorem}
Let $F / K$ be a field extension.
Let $\alpha \in F$ be algebraic over $K$.
Let $\mu_\alpha$ be the minimal polynomial of $\alpha$ over $K$.
Let $K \sqbrk \alpha$ (resp. $\map K \alpha$) be the subring (resp. subfield) of $F$ generated by $K \cup \set \alpha$.
Then:
:$K \sqbrk \alpha = \map K \alpha \simeq K \sqbrk X / \gen {\mu_\alpha}$
where $\gen {\mu_\alpha}$ is the ideal of the ring of polynomial functions generated by $\mu_\alpha$.
{{explain|Link to Generator needs refining}}
Moreover:
:$n := \index {\map K \alpha} K = \deg \mu_\alpha$
and:
:$1, \alpha, \dotsc, \alpha^{n - 1}$ is a basis of $\map K \alpha$ over $K$.
\end{theorem}
\begin{proof}
Define $\phi: K \sqbrk X \to K \sqbrk \alpha$ by:
:$\map \phi f = \map f \alpha$
We have
:$\map \phi f = 0 \iff \map f \alpha = 0 \iff \mu_\alpha \divides f$
where the last equivalence is proved in Minimal Polynomial.
{{explain|Presumably $\mu_\alpha \vert f$ means divisibility -- clarification needed.}}
{{refactor|The above needs to be extracted into a different page, as Minimal Polynomial is to be refactored: merged, probably, into somewhere else.}}
Thus:
:$\ker \phi = \set {f \in K \sqbrk X: \mu_\alpha \divides f} =: \gen {\mu_\alpha}$
By the corollary to Field Adjoined Set $\phi$ is surjective.
So by the First Isomorphism Theorem,
:$K \sqbrk X / \gen {\mu_\alpha} \simeq K \sqbrk \alpha$
By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal:
:$\gen {\mu_\alpha}$ is maximal.
So $K \sqbrk \alpha$ is a field by Maximal Ideal iff Quotient Ring is Field.
Also $K \sqbrk \alpha$ is the smallest ring containing $K \cup \set \alpha$.
So because a field is a ring it is also the smallest field containing $K \cup \set \alpha$.
This shows that $K \sqbrk \alpha = \map K \alpha$.
By Field Adjoined Set:
:$K \sqbrk \alpha = \set {\map f \alpha: f \in K \sqbrk X}$
where $K \sqbrk X$ is the ring of polynomial functions over $K$.
From Division Theorem for Polynomial Forms over Field, for each $f \in K \sqbrk X$ there are $q, r \in K \sqbrk X$ such that:
:$f = q \mu_\alpha + r$
and:
:$\deg r < d =: \deg \mu_\alpha$
Therefore, since $\map {\mu_\alpha} \alpha = 0$, we have
{{begin-eqn}}
{{eqn | l = K \sqbrk \alpha
| r = \set {\map r \alpha: r \in K \sqbrk X, \ \deg r < \deg \mu_\alpha}
}}
{{eqn | r = \set {a_0 + a_1 \alpha + \dotsb + a_{d - 1} \alpha^{d - 1}: \alpha_0, \dotsc, \alpha_d \in K}
}}
{{end-eqn}}
Therefore $1, \dotsc, \alpha^{d - 1}$ span $K \sqbrk \alpha / K$ as a vector space.
Moreover if $s \in K \sqbrk X$, then $\map s \alpha = 0$ and $\deg s < \deg \mu_\alpha$ implies that $s = 0$.
Thus $1, \dotsc, \alpha^{d - 1}$ are linearly independent.
Therefore no non-zero $K$-combination of $1, \dotsc, \alpha^{d - 1}$ is zero.
Therefore $\set {1, \dotsc, \alpha^{d - 1} }$ is a basis of $K \sqbrk \alpha / K$.
The degree of $K \sqbrk \alpha / K$ is by definition the number of elements of a basis for $K \sqbrk \alpha / K$.
Therefore:
:$d = \deg \mu_\alpha = \index {K \sqbrk \alpha} K$
{{qed}}
Category:Field Extensions
\end{proof}
|
21887
|
\section{Structure under Left Operation is Semigroup}
Tags: Left Operation, Examples of Semigroups
\begin{theorem}
Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\gets$ is the left operation.
Then $\struct {S, \gets}$ is a semigroup.
\end{theorem}
\begin{proof}
We need to verify the semigroup axioms:
{{:Axiom:Semigroup Axioms}}
By the nature of the right operation, $\struct {S, \to}$ is closed:
:$\forall x, y \in S: x \gets y = y \in S$
whatever $S$ may be.
Hence {{SemigroupAxiom|0}} holds.
From Right Operation is Associative, $\gets$ is associative.
Hence {{SemigroupAxiom|1}} holds.
So $\struct {S, \gets}$ is a semigroup.
{{qed}}
\end{proof}
|
21888
|
\section{Structure under Right Operation is Semigroup}
Tags: Right Operation, Examples of Semigroups
\begin{theorem}
Let $\struct {S, \to}$ be an algebraic structure in which the operation $\to$ is the right operation.
Then $\struct {S, \to}$ is a semigroup.
\end{theorem}
\begin{proof}
We need to verify the semigroup axioms:
{{:Axiom:Semigroup Axioms}}
By the nature of the right operation, $\struct {S, \to}$ is closed:
:$\forall x, y \in S: x \to y = y \in S$
whatever $S$ may be.
Hence {{SemigroupAxiom|0}} holds.
From Right Operation is Associative, $\to$ is associative.
Hence {{SemigroupAxiom|1}} holds.
So $\struct {S, \to}$ is a semigroup.
{{qed}}
\end{proof}
|
21889
|
\section{Structure with Element both Identity and Zero has One Element}
Tags: Zero Elements, Identity Elements
\begin{theorem}
Let $\struct {S, \circ}$ be an algebraic structure.
Let $z \in S$ such that $z$ is both an identity element and a zero element.
Then:
:$S = \set z$
\end{theorem}
\begin{proof}
Let $x \in S$.
Then
{{begin-eqn}}
{{eqn | l = x
| r = x \circ z
| c = {{Defof|Identity Element}}
}}
{{eqn | r = z
| c = {{Defof|Zero Element}}
}}
{{end-eqn}}
and so there is no other element of $S$ but $z$.
{{qed}}
\end{proof}
|
21890
|
\section{Sturm-Liouville Problem}
Tags:
\begin{theorem}
Let:
:$P \in C^\infty \map P x > 0$
:$Q \in C^0$
:$-\paren {P y'}' + Q y = \lambda y$
:$\map y a = \map y b = 0$
Then the Sturm-Liouville problem has an infinite sequence of eigenvalues $\set {\lambda^{\paren n} }$, and to each $\lambda^{\paren n}$ corresponds an eigenfunction $y^{\paren n}$, unique up to a constant factor.
\end{theorem}
\begin{proof}
:$J \sqbrk y = \ds \int_a^b \paren {P y'^2 + Q y^2} \rd x$
:$\ds \int_a^b y^2 \rd x = 1$
:$\ds \int_a^b \paren {P y'^2 + Q y^2} \rd x > \int_a^b Q y^2 \rd x \ge M \int_a^b y^2 \rd x = M$
:$M = \min \limits_{a \mathop \le x \mathop \le b} \map Q x$
Assume $a = 0$, $b = \pi$.
Choose $\set {\map {\phi_n} x} = \set {\sin n x}$
For $k \ne l$ we have:
:$\ds \int_0^\pi \sin k x \sin l x \rd x = 0$
:$\ds \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2 \rd x = \dfrac \pi 2 \sum_{k \mathop = 1}^n \alpha_k^2 = 1$
:$\ds \map {J_n} {\alpha_1, \ldots, \alpha_n} = \int_o^\pi \paren {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin k x}'^2 + \map Q {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2} \rd x$
{{explain|The scope of the $'$ and ${}^2$ in $\ds \map P {\sum_{k \mathop {{=}} 1}^n \alpha_k \sin k x}'^2$ needs to be clarified in the above}}
:$\ds \map {y_n^{\paren 1} } x = \sum_{k \mathop = 1}^n \alpha_k^{\paren 1} \sin k x$
:$\set {\lambda_n^{\paren 1} }$
:$\set {y_n^{\paren 1} }$
:$\map {J_n} {\alpha_1, \ldots, \alpha_n} = \map {J_{n + 1} } {\alpha_1, \ldots, \alpha_n, 0}$
:$\lambda_{n + 1}^{\paren 1} \le \lambda_n^{\paren 1}$
:$\lambda^{\paren 1} = \lim_{n \mathop \to \infty} \lambda_n^{\paren 1}$
:$\ds \lambda_n^{\paren 1} = \int_0^\pi \paren {P y_n'^2 + Q y_n^2} \rd x$
:$\ds \int_0^\pi \paren {P y_n'^2 + Q y_n^2} \rd x \le M$
:$\ds \int_0^\pi Py_n'^2 \rd x \le M + \size {\int_0^\pi Q y_n^2 \rd x} \le M + \max \limits_{a \mathop \le x \mathop \le b } \size {\map Q x} = M_1$
:$\ds \int_0^\pi \map {y_n'^2} x \rd x \le \dfrac {M_1} {\min \limits_{a \mathop \le x \mathop \le b} \map P x} = M_2$
:$\map {y_n} 0 = 0$
:$\ds \size {\map {y_n} x}^2 = \size {\int_0^x \map {y_n'} \zeta \rd \zeta}^2 \le \int_0^x \map {y_n'^2} \zeta \rd \zeta \int_0^x \rd \zeta \le M_2 \pi$
:$\ds \size {\map {y_n} {x_2} - \map {y_n} {x_1} }^2 = \size {\int_{x_1}^{x_2} \map {y_n'} x \rd x}^2 \le \int_{x_1}^{x_2} y_n'^2 \rd x \size {\int_{x_1}^{x_2} \rd x}^2 \le M_2 \size {x_2 - x_1}$
:$\map {y^{\paren 1} } x = \lim_{m \mathop \to \infty} \map {y_{n_m} } x$
:$\int_0^\pi \paren {-\paren {P h'}' + Q_1 h} y \rd x = 0$
:$\map h x \in \map {\DD_2} {0, \pi}$
:$\map h 0 = \map h \pi = \map {h'} 0 = \map {h'} \pi = 0$
:$\map y x \in \map {\DD_2} {0, \pi}$
:$-\paren {P y'}' + Q_1 y = 0$
:$\ds \int_0^\pi \paren {-\paren {P y'}' + Q_1 y} y \rd x = -\int_0^\pi P h'' y \rd x - \int_0^\pi P' h' y \rd x + \int_0^\pi Q_1 h y \rd x = - \int_0^\pi \paren {- P y + \int_0^x P' y \rd \zeta + \int_0^x \paren {\int_0^\zeta Q_1 y \rd t} \rd \zeta} \rd x = 0$
:$-\paren {P y}' + P' y + \ds \int_0^x Q_1 y \rd \zeta = c_1$
:$-P y' + \ds \int_0^x Q_1 y \rd \zeta = c_1$
:$-\paren {P y'}' + Q_1 y = 0$
:$-\paren {P {y^{\paren 1} }'}' + Q y^{\paren 1} = \lambda^{\paren 1} y^{\paren 1}$
:$\dfrac \partial {\partial \alpha_r} \paren {\map {J_n} {\alpha_1, \ldots, \alpha_n} - \lambda_n^{\paren 1} \ds \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2 \rd x} = 0$
{{ProofWanted}}
\end{proof}
|
21891
|
\section{Sturm-Liouville Problem/Unit Weight Function}
Tags: Calculus of Variations
\begin{theorem}
Let $P, Q: \R \to \R$ be real mappings such that $P$ is smooth and positive, while $Q$ is continuous:
:$\map P x \in C^\infty$
:$\map P x > 0$
:$\map Q x \in C^0$
Let the Sturm-Liouville equation, with $\map w x = 1$, be of the form:
:$-\paren {P y'}' + Q y = \lambda y$
where $\lambda \in \R$.
Let it satisfy the following boundary conditions:
:$\map y a = \map y b = 0$
Then all solutions of the Sturm-Liouville equation, together with their eigenvalues, form infinite sequences $\sequence {y^{\paren n} }$ and $\sequence {\lambda^{\paren n} }$.
Furthermore, each $\lambda^{\paren n}$ corresponds to an eigenfunction $y^{\paren n}$, unique up to a constant factor.
\end{theorem}
\begin{proof}
Suppose, $y^{\paren r}$ and $\lambda^{\paren r}$ are known.
The next eigenfunction $y^{\paren {r + 1} }$ and the corresponding eigenvalue $\lambda^{\paren {r + 1} }$ can be found by minimising
:$\ds J \sqbrk y = \int_0^\pi \paren {P y'^2 + Q y^2} \rd x $
where boundary and subsidiary conditions are supplied with orthogonality conditions:
:$\forall m \in \N : {1 \le m \le r} : \ds \int_0^\pi \map {y^{\paren m} } t \map {y^{\paren {r + 1} } } t \rd t = 0$
The new solution of the form:
:$\ds \map {y_n^{\paren {r + 1} } } t = \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \sin {k t}$
is now also orthogonal to mappings:
:$\ds \map {y_n^{\paren m} } t = \sum_{k \mathop = 1}^n \alpha_k^{\paren m} \sin {k t}$
This results into:
:$\ds \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \int_0^\pi \sin {k t} \paren {\sum_{l \mathop = 1}^n \alpha_l^{\paren m} \sin {l t} } \rd t = \frac \pi 2 \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \alpha_k^{\paren m} = 0$
These equations describe $r$ distinct $\paren {n - 1}$-dimensional hyperplanes, passing through the origin of coordinates in $n$ dimensions.
These hyperplanes intersect the sphere $\sigma_n$, resulting in an $\paren {n - r}$-dimensional sphere $\hat \sigma_{n - r}$.
By definition, it is a compact set.
By Continuous Function on Compact Subspace of Euclidean Space is Bounded, $\map {J_n} {\boldsymbol \alpha}$ has a minimum on $\hat {\sigma}_{n - r}$.
Denote it as $\lambda_n^{\paren {r + 1} }$.
By Ritz Method implies Not Worse Approximation with Increased Number of Functions:
:$\lambda_{n + 1}^{\paren {r + 1} } \le \lambda_n^{\paren {r + 1} }$
This, together with $J$ being bounded from below, implies:
:$\ds \lambda^{\paren {r + 1} } = \lim_{n \mathop \to \infty} \lambda_n^{\paren {r + 1} }$
Additional constraints may or may not affect the new minimum:
:$\lambda^{\paren r} \le \lambda^{\paren {r + 1} }$
Let:
:$\ds \map {y_n^{\paren {r + 1} } } t = \sum_{k \mathop = 1}^n \alpha_k^{\paren {r + 1} } \sin {k t}$
$y^{\paren {r + 1} }$ satisfies Sturm-Liouville equation together with boundary, subsidiary and orthogonality conditions.
By Lemma 7, which is not affected by additional constraints, $\sequence {y_n^{\paren {r + 1} } }$ uniformly converges to $y^{\paren {r + 1} }$.
Thus, $y^{\paren {r + 1} }$ is an eigenfunction of Sturm-Liouville equation with an eigenvalue $\lambda^{\paren {r + 1} }$.
Orthogonal mappings are linearly independent.
Each eigenvalue corresponds only to one eigenfunction, unique up to a constant factor.
Thus:
:$\lambda^{\paren r} < \lambda^{\paren {r + 1} }$
{{qed}}
\end{proof}
|
21892
|
\section{Sturm-Liouville Problem/Unit Weight Function/Lemma}
Tags: Calculus of Variations
\begin{theorem}
Let $\map \alpha x: \R \to \R$ such that $\map \alpha x \in C^2 \closedint a b$.
Suppose:
:$\ds \forall \map h x \in C^2 \closedint a b: \map h a = \map h b = \map {h'} a = \map {h'} b = 0: \int_a^b \map \alpha x \, \map {h''} x \rd x = 0$
Then:
:$\forall x \in \closedint a b: \map \alpha x = c_0 + c_1 x$
where $ c_0, c_1 \in \R $ are constants.
\end{theorem}
\begin{proof}
Let $ c_0, c_1$ be defined by the conditions:
:$\ds \int_a^b \paren {\map \alpha x - c_0 - c_1 x} \rd x = 0$
:$\ds \int_a^b \rd x \int_a^x \paren {\map \alpha \xi - c_0 - c_1 \xi} \rd \xi = 0$
Suppose:
:$\ds \map h x = \int_a^x \xi \int_a^\xi \paren {\map \alpha t - c_0 - c_1 t} \rd t$
This form satisfies conditions on $h$ in the theorem.
Then:
{{begin-eqn}}
{{eqn | l = \int_a^b \paren {\map \alpha x - c_0 - c_1 x} \map {h''} x \rd x
| r = \int_a^b \map \alpha x \map {h''} x \rd x - c_0 \paren {\map {h'} b - \map {h'} a} - c_1 \int_a^b x \map {h''} x \rd x
}}
{{eqn | r = -c_1 \paren {b \map {h'} b - a \map {h'} a} - c_1 \paren {\map h b - \map h a}
}}
{{eqn | r = 0
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = \int_a^b \paren {\map \alpha x - c_0 - c_1 x} \map {h''} x \rd x
| r = \int_a^b \paren {\map \alpha x - c_0 - c_1 x}^2 \rd x
}}
{{eqn | r = 0
}}
{{end-eqn}}
Therefore:
:$\map \alpha x - c_0 - c_1 x = 0$
or:
:$\map \alpha x = c_0 + c_1 x$
{{qed}}
Category:Calculus of Variations
\end{proof}
|
21893
|
\section{Sub-Basis for Real Number Line}
Tags: Real Numbers, Topology, Topological Bases
\begin{theorem}
Let the real number line $\R$ be considered as a topology under the usual (Euclidean) topology.
Then:
:$\BB := \set {\openint \gets a, \openint b \to: a, b \in \R}$ is a sub-basis for $\R$.
\end{theorem}
\begin{proof}
Let $\openint c d$ be an open real interval.
Then by definition:
:$\openint c d = \openint \gets d \cap \openint c \to$
and so $\openint c d$ is the intersection of two elements of $\BB$.
From Open Sets in Real Number Line, any open set of $\R$ is the union of countably many open real intervals.
The result follows by definition of sub-basis.
{{qed}}
\end{proof}
|
21894
|
\section{Sub-Basis for Topological Subspace}
Tags: Topological Subspaces, Topological Bases
\begin{theorem}
Let $\struct {X, \tau}$ be a topological space.
Let $K$ be a sub-basis for $\tau$.
Let $\struct {S, \tau'}$ be a subspace of $\struct {X, \tau}$.
Let $K' = \set {U \cap S: U \in K}$.
That is, $K'$ consists of the $\tau'$-open sets in $S$ corresponding to elements of $K$.
Then $K'$ is a sub-basis for $\tau'$.
\end{theorem}
\begin{proof}
Let $B$ be the basis for $\tau$ generated by $K$.
By Basis for Topological Subspace, $B$ generates a basis, $B'$, for $\tau'$.
We will show that $K'$ generates $B'$.
Let $V' \in B'$.
Then for some $V \in B$:
:$V' = S \cap V$
By the definition of a sub-basis, there is a finite subset $K_V$ of $K$ such that:
:$\ds V = \bigcap K_V$
Let $K_V' = \set {S \cap U: U \in K_V}$.
Then:
:$\ds V' = S \cap V = S \cap \bigcap K_V$
so:
:$\ds V' = \bigcap_{P \mathop \in K_V} \paren {S \cap P} = \bigcap K_V'$
Thus $B'$ is generated by $K'$.
{{qed}}
Category:Topological Bases
Category:Topological Subspaces
\end{proof}
|
21895
|
\section{Subalgebra of Algebraic Field Extension is Field}
Tags: Integral Extensions, Integral Ring Extensions
\begin{theorem}
Let $E / F$ be an algebraic field extension.
Let $A \subseteq E$ be a unital subalgebra over $F$.
Then $A$ is a field.
\end{theorem}
\begin{proof}
By Integral Ring Extension is Integral over Intermediate Ring, $E$ is integral over $A$.
Let $a \in A$ be nonzero.
Because $E$ is a field, $a$ is a unit of $E$.
By Ring Element is Unit iff Unit in Integral Extension, $a$ is a unit of $A$.
Thus $A$ is a field.
{{qed}}
\end{proof}
|
21896
|
\section{Subclass of Set is Set}
Tags: Gödel-Bernays Class Theory
\begin{theorem}
Let $A$ be a set.
Let $\map \phi x$ be a condition in which $x$ is taken to be a set.
Then there exists a set that consists of all of the elements of $A$ that satisfies this condition.
In ZF, this result is known as the Axiom of Specification.
\end{theorem}
\begin{proof}
{{NotZFC}}
By the axiom of class comprehension, let $B$ be the class defined as:
{{begin-eqn}}
{{eqn | l = B
| r = \set {x: x \in A \land \map \phi x}
}}
{{eqn | r = \set {x \in A: \map \phi x}
| c = Set-Builder Notation
}}
{{end-eqn}}
{{AimForCont}} that $B$ is not a set.
Then $B$ must be a proper class.
It is easily seen that $B \subseteq A$.
So by the Axiom of Powers:
:$B \in \powerset A$
where $\powerset A$ is denotes the power set of $A$.
But by Proper Class is not Element of Class, this is a contradiction.
Therefore by contradiction it follows that $B$ is a set.
{{qed}}
\end{proof}
|
21897
|
\section{Subcover is Refinement of Cover}
Tags: Covers
\begin{theorem}
Let $S$ be a set.
Let $\CC$ be a cover for $S$.
Let $\VV$ be a subcover of $\CC$.
Then $\VV$ is a refinement of $\CC$.
\end{theorem}
\begin{proof}
From definition of subcover:
:$\VV \subseteq \CC$
That is, every element of $\VV$ is an element of $\CC$.
From definition of subset, every element of $\VV$ is the subset of some element of $\CC$.
This is precisely the definition of refinement.
{{qed}}
Category:Covers
\end{proof}
|
21898
|
\section{Subdomain Test}
Tags: Rings, Integral Domains, Subrings
\begin{theorem}
Let $S$ be a subset of an integral domain $\struct {R, +, \circ}$.
Then $\struct {S, + {\restriction_S}, \circ {\restriction_S} }$ is a subdomain of $\struct {R, +, \circ}$ {{iff}} these conditions hold:
:$(1): \quad \struct {S, + {\restriction_S}, \circ {\restriction_S} }$ is a subring of $\struct {R, +, \circ}$
:$(2): \quad$ The unity of $R$ is also in $S$, that is $1_R = 1_S$.
\end{theorem}
\begin{proof}
By Idempotent Elements of Ring with No Proper Zero Divisors, it follows that the unity of a subdomain is the unity of the integral domain it's a subdomain of.
{{qed}}
\end{proof}
|
21899
|
\section{Subfield of Subfield is Subfield}
Tags: Subfields
\begin{theorem}
Let $R$ be a ring with unity.
Let $K_1, K_2$ be fields, such that:
: $K_1$ is a subfield of $R$
: $K_2$ is a subfield of $K_1$
Then $K_2$ is a subfield of $R$.
\end{theorem}
\begin{proof}
Let $K_1$ be a subfield of $R$ and $K_2$ be a subfield of $K_1$.
Then by definition:
:$K_1 \subseteq R$
:$K_2 \subseteq K_1$
From Subset Relation is Transitive it follows that $K_2 \subseteq R$
So by definition $K_2$ is a subfield of $R$.
{{qed}}
Category:Subfields
\end{proof}
|
21900
|
\section{Subgroup Action is Group Action}
Tags: Subgroup Action, Examples of Group Actions
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $\struct {H, \circ}$ be a subgroup of $G$.
Let $*: H \times G \to G$ be the subgroup action defined for all $h \in H, g \in G$ as:
:$\forall h \in H, g \in G: h * g := h \circ g$
Then $*$ is a group action.
\end{theorem}
\begin{proof}
Let $g \in G$.
First we note that since $G$ is closed, and $h \circ g \in G$, it follows that $h * g \in G$.
Next we note:
:$e * g = e \circ g = g$
and so {{GroupActionAxiom|2}} is satisfied.
Now let $h_1, h_2 \in G$.
We have:
{{begin-eqn}}
{{eqn | l = \paren {h_1 \circ h_2} * g
| r = \paren {h_1 \circ h_2} \circ g
| c = Definition of $*$
}}
{{eqn | r = h_1 \circ \paren {h_2 \circ g}
| c = {{GroupAxiom|1}}
}}
{{eqn | r = h_1 * \paren {h_2 * g}
| c = Definition of $*$
}}
{{end-eqn}}
and so {{GroupActionAxiom|1}} is satisfied.
Hence the result.
{{qed}}
\end{proof}
|
21901
|
\section{Subgroup Containing all Squares of Group Elements is Normal}
Tags: Normal Subgroups
\begin{theorem}
Let $G$ be a group.
Let $H$ be a subgroup of $G$ with the property that:
:$\forall x \in G: x^2 \in H$
Then $H$ is normal in $G$.
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = \paren {x h}^2 h^{-1} \paren {x^{-1} }^2
| r = x h x h h^{-1} x^{-1} x^{-1}
| c = {{GroupAxiom|1}}
}}
{{eqn | r = x h x x^{-1} x^{-1}
| c = {{GroupAxiom|3}}
}}
{{eqn | r = x h x^{-1}
| c = {{GroupAxiom|3}}
}}
{{end-eqn}}
Because $\paren {x h}^2$ and $\paren {x^{-1} }^2$ are in the form $x^2$ for $x \in G$, they are both elements of $H$.
Thus:
:$x h x^{-1} \in H$
and so $H$ is normal in $G$ by definition.
{{qed}}
\end{proof}
|
21902
|
\section{Subgroup Generated by Commuting Elements is Abelian}
Tags: Abelian groups, Abelian Groups
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$ such that:
:$\forall x, y \in S: x \circ y = y \circ x$
Then the subgroup generated by $S$ is abelian.
\end{theorem}
\begin{proof}
Let $H = \gen S$ denote the subgroup generated by $S$.
Let $a, b \in H$.
Then:
:$a = s_1$
:$b = s_2$
for some words $s_1, s_2$ of the set of words $\map W S$ of $S$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ b
| r = s_1 \circ s_2
| c =
}}
{{eqn | r = s_2 \circ s_1
| c = as all elements of $S$ commute with each other
}}
{{eqn | r = b \circ a
| c =
}}
{{end-eqn}}
Hence the result, by definition of abelian group.
{{qed}}
\end{proof}
|
21903
|
\section{Subgroup Generated by Infinite Order Element is Infinite}
Tags: Infinite Groups, Generated Subgroups, Order of Group Elements
\begin{theorem}
Let $G$ be a group.
Let $a \in G$ be of infinite order.
Let $\gen a$ be the subgroup generated by $a$.
Then $\gen a$ is of infinite order.
\end{theorem}
\begin{proof}
{{AimForCont}} $\gen a$ is of finite order.
We have that $a \in \gen a$ by definition.
From Element of Finite Group is of Finite Order it follows that $a$ is of finite order.
From this contradiction it follows that $\gen a$ must be of infinite order after all.
{{qed}}
\end{proof}
|
21904
|
\section{Subgroup Generated by One Element is Cyclic}
Tags: Generated Subgroups
\begin{theorem}
Let $G$ be a group.
Let $a \in G$.
Then $\gen a$, the subgroup generated by $a$, is cyclic:
\end{theorem}
\begin{proof}
By Subgroup Generated by One Element is Set of Powers:
:$\gen a = \set {a^n : n \in \Z}$
The result follows by definition of cyclic group.
{{qed}}
\end{proof}
|
21905
|
\section{Subgroup Generated by One Element is Set of Powers}
Tags: Generated Subgroups, Subgroups
\begin{theorem}
Let $G$ be a group.
Let $a \in G$.
Then the subgroup generated by $a$ is the set of powers:
:$\gen a = \set {a^n : n \in \Z}$
\end{theorem}
\begin{proof}
By definition, the subgroup generated by $a$ is the intersection of all subgroups containing $a$.
By Powers of Element form Subgroup, the set $H = \set {a^n : n \in \Z}$ is a subgroup.
Thus $\gen a \subseteq H$.
By Power of Element in Subgroup, $H \subseteq \gen a$.
By definition of set equality, $\gen a = H$.
{{qed}}
\end{proof}
|
21906
|
\section{Subgroup Generated by Subgroup}
Tags: Generated Subgroups
\begin{theorem}
Let $G$ be a group.
Let $H \le G$ be a subgroup of $G$.
Then:
:$H = \gen H$
where $\gen H$ denotes the subgroup generated by $H$.
\end{theorem}
\begin{proof}
By definition of generated subgroup, $\gen H$ is the smallest subgroup of $H$ containing $H$.
Hence the result.
{{qed}}
\end{proof}
|
21907
|
\section{Subgroup Subset of Subgroup Product}
Tags: Subset Products, Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Then:
:$H \subseteq H \circ K \supseteq K$
where $H \circ K$ denotes the subset product of $H$ and $K$.
\end{theorem}
\begin{proof}
By definition of subset product:
:$H \circ K = \set {h \circ k: h \in H, k \in K}$
So:
{{begin-eqn}}
{{eqn | l=x
| o=\in
| r=H
| c=
}}
{{eqn | ll=\leadsto
| l=x
| r=x \circ e
| c={{Defof|Identity Element}}
}}
{{eqn | o=\in
| r=H \circ K
| c=Identity of Subgroup
}}
{{eqn | ll=\leadsto
| l=H
| o=\subseteq
| r=H \circ K
| c={{Defof|Subset}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l=y
| o=\in
| r=K
| c=
}}
{{eqn | ll=\leadsto
| l=y
| r=e \circ y
| c={{Defof|Identity Element}}
}}
{{eqn | o=\in
| r=H \circ K
| c=Identity of Subgroup
}}
{{eqn | ll=\leadsto
| l=K
| o=\subseteq
| r=H \circ K
| c={{Defof|Subset}}
}}
{{end-eqn}}
{{qed}}
Category:Subset Products
Category:Subgroups
\end{proof}
|
21908
|
\section{Subgroup is Closed iff Quotient is Hausdorff}
Tags: Hausdorff Spaces, Quotient Spaces, Topological Groups
\begin{theorem}
Let $G$ be a topological group.
Let $H \le G$ be a subgroup.
Let $G / H$ be their quotient.
Then the following are equivalent:
:$H$ is closed in $G$
:$G / H$ is Hausdorff
\end{theorem}
\begin{proof}
{{ProofWanted|use Group Acts by Homeomorphisms Implies Projection on Quotient Space is Open and Open Projection and Closed Graph Implies Quotient is Hausdorff}}
Category:Topological Groups
Category:Hausdorff Spaces
\end{proof}
|
21909
|
\section{Subgroup is Normal Subgroup of Normalizer}
Tags: Normal Subgroups, Normalizers, Subgroups
\begin{theorem}
Let $G$ be a group.
A subgroup $H \le G$ is a normal subgroup of its normalizer:
:$H \le G \implies H \lhd \map {N_G} H$
\end{theorem}
\begin{proof}
From Subgroup is Subgroup of Normalizer we have that $H \le \map {N_G} H$.
It remains to show that $H$ is normal in $\map {N_G} H$.
Let $a \in H$ and $b \in \map {N_G} H$.
By the definition of normalizer:
:$b a b^{-1} \in H$
Thus $H$ is normal in $\map {N_G} H$.
{{qed}}
\end{proof}
|
21910
|
\section{Subgroup is Normal iff Contains Conjugate Elements}
Tags: Conjugacy, Normal Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ {{iff}}:
:$\forall g \in G: \paren {n \in N \iff g \circ n \circ g^{-1} \in N}$
:$\forall g \in G: \paren {n \in N \iff g^{-1} \circ n \circ g \in N}$
\end{theorem}
\begin{proof}
By definition, a subgroup is normal in $G$ {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
\end{proof}
|
21911
|
\section{Subgroup is Subset of Conjugate iff Normal}
Tags: Conjugacy, Normal Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) {{iff}}:
:$\forall g \in G: N \subseteq g \circ N \circ g^{-1}$
:$\forall g \in G: N \subseteq g^{-1} \circ N \circ g$
\end{theorem}
\begin{proof}
By definition, a subgroup is normal in $G$ {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: N \subseteq g \circ N \circ g^{-1} } \iff \paren {\forall g \in G: N \subseteq g^{-1} \circ N \circ g}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
\end{proof}
|
21912
|
\section{Subgroup is Superset of Conjugate iff Normal}
Tags: Conjugacy, Normal Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) {{iff}}:
:$\forall g \in G: g \circ N \circ g^{-1} \subseteq N$
:$\forall g \in G: g^{-1} \circ N \circ g \subseteq N$
\end{theorem}
\begin{proof}
By definition, a subgroup is normal in $G$ {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} \subseteq N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g \subseteq N}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
\end{proof}
|
21913
|
\section{Subgroup of Abelian Group is Abelian}
Tags: Abelian Groups, Subgroups, Group Theory
\begin{theorem}
A subgroup of an abelian group is itself abelian.
\end{theorem}
\begin{proof}
Follows directly from Restriction of Commutative Operation is Commutative.
{{qed}}
\end{proof}
|
21914
|
\section{Subgroup of Abelian Group is Normal}
Tags: Abelian Groups, Normal Subgroups
\begin{theorem}
Every subgroup of an abelian group is normal.
\end{theorem}
\begin{proof}
Let $G$ be an abelian group.
Let $H \le G$ be a subgroup of $G$.
Then for all $a \in G$:
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = H^a
| c = where $H^a$ is the conjugate of $H$ by $a$
}}
{{eqn | ll= \leadstoandfrom
| l = a y a^{-1}
| o = \in
| r = H
| c = {{Defof|Conjugate of Group Subset}}
}}
{{eqn | ll= \leadstoandfrom
| l = y
| o = \in
| r = H
| c = because $a y a^{-1} = y$ as $G$ is abelian
}}
{{end-eqn}}
{{Qed}}
\end{proof}
|
21915
|
\section{Subgroup of Additive Group Modulo m is Ideal of Ring}
Tags: Subgroups, Additive Group of Integers Modulo m, Additive Groups of Integers Modulo m, Modulo Addition, Ideal Theory, Modulo Arithmetic
\begin{theorem}
Let $m \in \Z: m > 1$.
Let $\struct {\Z_m, +_m}$ be the additive group of integers modulo $m$.
Then every subgroup of $\struct {\Z_m, +_m}$ is an ideal of the ring of integers modulo $m$ $\struct {\Z_m, +_m, \times_m}$.
\end{theorem}
\begin{proof}
Let $H$ be a subgroup of $\struct {\Z_m, +_m}$
Suppose:
: $(1): \quad h + \ideal m \in H$, where $\ideal m$ is a principal ideal of $\struct {\Z_m, +_m, \times_m}$
and
: $(2): \quad n \in \N_{>0}$.
Then by definition of multiplication on integers and Homomorphism of Powers as applied to integers:
{{begin-eqn}}
{{eqn | l = \paren {n + \ideal m} \times \paren {h + \ideal m}
| r = \map {q_m} n \times \map {q_m} h
| c = where $q_m$ is the quotient mapping
}}
{{eqn | r = \map {q_m} {n \times h}
| c =
}}
{{eqn | r = \map {q_m} {n \cdot h}
| c =
}}
{{eqn | r = n \cdot \map {q_m} h
| c =
}}
{{end-eqn}}
But:
:$n \cdot \map {q_m} h \in \gen {\map {q_m} h}$
where $\gen {\map {q_m} h}$ is the group generated by $\map {q_m} h$.
Hence by Epimorphism from Integers to Cyclic Group, $n \cdot \map {q_m} h \in H$.
The result follows.
{{qed}}
\end{proof}
|
21916
|
\section{Subgroup of Additive Group of Integers Generated by Two Integers}
Tags: Additive Group of Integer Multiples, Additive Groups of Integer Multiples
\begin{theorem}
Let $m, n \in \Z_{> 0}$ be (strictly) positive integers.
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $\gen {m, n}$ be the subgroup of $\struct {\Z, +}$ generated by $m$ and $n$.
Then:
:$\gen {m, n} = \struct {\gcd \set {m, n} \Z, +}$
That is, the additive groups of integer multiples of $\gcd \set {m, n}$, where $\gcd \set {m, n}$ is the greatest common divisor of $m$ and $n$.
\end{theorem}
\begin{proof}
By definition:
:$\gen {m, n} = \set {x \in \Z: \gcd \set {m, n} \divides x}$
{{Handwaving|Sorry, I would make the effort, but it's tedious.}}
Hence the result.
{{qed}}
\end{proof}
|
21917
|
\section{Subgroup of Cyclic Group is Cyclic}
Tags: Cyclic Groups, Subgroups, Subgroup of Cyclic Group is Cyclic, Group Theory
\begin{theorem}
Let $G$ be a cyclic group.
Let $H$ be a subgroup of $G$.
Then $H$ is cyclic.
\end{theorem}
\begin{proof}
Let <math>G</math> be a cyclic group generated by <math>a</math>.
Let <math>H</math> be a subgroup of <math>G</math>.
If <math>H = \left\{{e}\right\}</math>, then <math>H</math> is a cyclic subgroup generated by <math>e</math>.
If <math>H \ne \left\{{e}\right\}</math>, then <math>a^n \in H</math> for some <math>n \in \Z</math> (since every element in <math>G</math> has the form <math>a^n</math> and <math>H</math> is a subgroup of <math>G</math>).
Let <math>m</math> be the smallest positive integer such that <math>a^m \in H</math>.
Consider an arbitrary element <math>b</math> of <math>H</math>.
Since <math>H</math> is a subgroup of <math>G</math>, <math>b = a^n</math> for some <math>n</math>.
Find integers <math>q</math> and <math>r</math> such that <math>n = mq + r</math> with <math>0 \leq r < m</math> by the Division Algorithm.
It follows that <math>a^n = a^{mq + r} = \left({a^m}\right)^qa^r</math>
and hence that <math>a^r = \left({a^m}\right)^{-q} a^n</math>.
Since <math>a^m \in H</math> so is its inverse <math>\left({a^m}\right)^{-1}</math>, and all powers of its inverse by closure.
Now <math>a^n</math> and <math>\left({a^m}\right)^{-q}</math> are both in <math>H</math>, thus so is their product <math>a^r</math> by closure.
However, <math>m</math> was the smallest positive integer such that <math>a^m \in H</math> and <math>0 \leq r < m</math>, so <math>r = 0</math>.
Therefore <math>n = q m</math> and <math>b = a^n = (a^m)^q</math>.
We conclude that any arbitrary element <math>b = a^n</math> of <math>H</math> is generated by <math>a^m</math> so <math>H = \left \langle {a^m}\right \rangle </math> is cyclic.
{{qed}}
\end{proof}
|
21918
|
\section{Subgroup of Direct Product is not necessarily Direct Product of Subgroups}
Tags: Group Direct Products
\begin{theorem}
Let $G$ and $H$ be groups.
Let $G \times H$ denote the direct product of $G$ and $H$.
Let $K$ be a subgroup of $G \times H$.
Then it is not necessarily the case that $K$ is of the form:
:$G' \times H'$
where:
:$G'$ is a subgroup of $G$
:$H'$ is a subgroup of $H$.
\end{theorem}
\begin{proof}
Let $G = H = C_2$, the cyclic group of order $2$.
Let $G = \gen x$ and $H = \gen y$, so that:
:$G = \set {e_G, x}$
:$H = \set {e_H, y}$
where $e_G$ and $e_H$ are the identity elements of $G$ and $H$ respectively.
Consider the element $\tuple {x, y} \in G \times H$.
We have that:
:$\gen {\tuple {x, y} } =\set {\tuple {e_G, e_H}, \tuple {x, y} }$
but this is not the direct product of a subgroup of $G$ with a subgroup of $H$.
{{qed}}
\end{proof}
|
21919
|
\section{Subgroup of Elements whose Order Divides Integer}
Tags: Abelian Groups
\begin{theorem}
Let $A$ be an abelian group.
Let $k \in \Z$ and $B$ be a set of the form:
:$\left\{{x \in A : x^k = e}\right\}$
Then $B$ is a subgroup of $A$.
\end{theorem}
\begin{proof}
First note that the identity $e$ satisfies $e^k = e$ and so $B$ is non-empty.
Now assume that $a, b \in B$.
Then:
{{begin-eqn}}
{{eqn | l=\left({ab^{-1} }\right)^k
| r=a^k \left({b^{-1} }\right)^k
| c=Power of Product in Abelian Group
}}
{{eqn | r=a^k \left({b^k}\right)^{-1}
| c=Powers of Group Elements
}}
{{eqn | r=ee^{-1}
| c=as $a^k = b^k = e \in B$
}}
{{eqn | r=e
| c=Identity is Self-Inverse
}}
{{end-eqn}}
Hence, by the One-Step Subgroup Test, $B$ is a subgroup of $A$.
{{qed}}
Category:Abelian Groups
\end{proof}
|
21920
|
\section{Subgroup of Finite Cyclic Group is Determined by Order}
Tags: Cyclic Group, Cyclic Groups
\begin{theorem}
Let $G = \gen g$ be a cyclic group whose order is $n$ and whose identity is $e$.
Let $d \divides n$, where $\divides$ denotes divisibility.
Then there exists exactly one subgroup $G_d = \gen {g^{n / d} }$ of $G$ with $d$ elements.
\end{theorem}
\begin{proof}
Let $G$ be generated by $g$, such that $\order g = n$.
From Number of Powers of Cyclic Group Element, $g^{n/d}$ has $d$ distinct powers.
Thus $\gen {g^{n / d} }$ has $d$ elements.
Now suppose $H$ is another subgroup of $G$ of order $d$.
Then by Subgroup of Cyclic Group is Cyclic, $H$ is cyclic.
Let $H = \gen y$ where $y \in G$.
Thus $\order y = d$.
Thus $\exists r \in \Z: y = g^r$.
Since $\order y = d$, it follows that $y^d = g^{r d} = e$.
From Equal Powers of Finite Order Element:
: $n \divides r d$
Thus:
{{begin-eqn}}
{{eqn | q = \exists k \in \N
| l = k n
| r = r d
| c =
}}
{{eqn | r = k \paren {\dfrac n d} d
| c =
}}
{{eqn | ll= \leadsto
| l = r
| r = k \paren {\dfrac n d}
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac n d
| o = \divides
| r = r
| c =
}}
{{end-eqn}}
Thus $y$ is a power of $g^{n / d}$.
Hence $H$ is a subgroup of $\gen {g^{n / d} }$.
Since both $H$ and $\gen {g^{n / d} }$ have order $d$, they must be equal.
{{qed}}
\end{proof}
|
21921
|
\section{Subgroup of Index 2 contains all Squares of Group Elements}
Tags: Subgroups
\begin{theorem}
Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index is $2$.
Then:
:$\forall x \in G: x^2 \in H$
\end{theorem}
\begin{proof}
By Subgroup of Index 2 is Normal, $H$ is normal in $G$.
Hence the quotient group $G / H$ exists.
Then we have:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = \paren {x^2} H
| r = \paren {x H}^2
| c =
}}
{{eqn | r = H
| c = as $G / H$ is of order $2$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21922
|
\section{Subgroup of Index 2 is Normal}
Tags: Normal Subgroups
\begin{theorem}
A subgroup of index $2$ is always normal.
\end{theorem}
\begin{proof}
Suppose $H \le G$ such that $\index G H = 2$.
Thus $H$ has two left cosets (and two right cosets) in $G$.
If $g \in H$, then $g H = H = H g$.
If $g \notin H$, then $g H = G \setminus H$ as there are only two cosets and the cosets partition $G$.
For the same reason, $g \notin H \implies H g = G \setminus H$.
That is, $g H = H g$.
The result follows from the definition of normal subgroup.
{{qed}}
\end{proof}
|
21923
|
\section{Subgroup of Index 3 does not necessarily contain all Cubes of Group Elements}
Tags: Subgroups
\begin{theorem}
Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index is $3$.
Then it is not necessarily the case that:
:$\forall x \in G: x^3 \in H$
\end{theorem}
\begin{proof}
Proof by Counterexample:
Consider $S_3$, the symmetric group on $3$ letters.
From Subgroups of Symmetric Group on 3 Letters, the subgroups of $S_3$ are:
subsets of $S_3$ which form subgroups of $S_3$ are:
{{begin-eqn}}
{{eqn | o =
| r = S_3
}}
{{eqn | o =
| r = \set e
}}
{{eqn | o =
| r = \set {e, \tuple {123}, \tuple {132} }
}}
{{eqn | o =
| r = \set {e, \tuple {12} }
}}
{{eqn | o =
| r = \set {e, \tuple {13} }
}}
{{eqn | o =
| r = \set {e, \tuple {23} }
}}
{{end-eqn}}
One such subgroup of $G$ whose index is $3$ is $\set {e, \tuple {12} }$
But $\set {e, \tuple {12} }$ does not contain $\tuple {123}$ or $\tuple {132}$, both of which are of order $3$.
{{qed}}
\end{proof}
|
21924
|
\section{Subgroup of Infinite Cyclic Group is Infinite Cyclic Group}
Tags: Cyclic Groups, Subgroups
\begin{theorem}
Let $G = \gen a$ be an infinite cyclic group generated by $a$, whose identity is $e$.
Let $g \in G, g \ne e: \exists k \in \Z, k \ne 0: g = a^k$.
Let $H = \gen g$.
Then $H \le G$ and $H \cong G$.
Thus, all non-trivial subgroups of an infinite cyclic group are themselves infinite cyclic groups.
A subgroup of $G = \gen a$ is denoted as follows:
:$n G := \gen {a^n}$
This notation is usually used in the context of $\struct {\Z, +}$, where $n \Z$ is (informally) understood as '''the set of integer multiples of $n$'''.
\end{theorem}
\begin{proof}
The fact that $H \le G$ follows from the definition of subgroup generator.
By Infinite Cyclic Group is Isomorphic to Integers:
:$G \cong \struct {\Z, +}$
Now we show that $H$ is of infinite order.
Suppose $\exists h \in H, h \ne e: \exists r \in \Z, r > 0: h^r = e$.
But:
:$h \in H \implies \exists s \in \Z, s > 0: h = g^s$
where $g = a^k$.
Thus:
:$e = h^r = \paren {g^s}^r = \paren {\paren {a^k}^s}^r = a^{k s r}$
and thus $a$ is of finite order.
This would mean that $G$ was also of finite order.
So $H$ must be of infinite order.
From Subgroup of Cyclic Group is Cyclic, as $G$ is cyclic, then $H$ must also be cyclic.
From Infinite Cyclic Group is Isomorphic to Integers:
:$H \cong \struct {\Z, +}$
Therefore, as $G \cong \struct {\Z, +}$:
:$H \cong G$
{{qed}}
\end{proof}
|
21925
|
\section{Subgroup of Integers is Ideal}
Tags: Subgroups, Ideals, Ideal Theory, Integers, Subrings
\begin{theorem}
Let $\struct {\Z, +}$ be the additive group of integers.
Every subgroup of $\struct {\Z, +}$ is an ideal of the ring $\struct {\Z, +, \times}$.
\end{theorem}
\begin{proof}
Let $H$ be a subgroup of $\struct {\Z, +}$.
Let $n \in \Z, h \in H$.
Then from the definition of cyclic group and Negative Index Law for Monoids:
:$n h = n \cdot h \in \gen h \subseteq H$
The result follows.
{{Qed}}
\end{proof}
|
21926
|
\section{Subgroup of Order 1 is Trivial}
Tags: Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Then $\struct {G, \circ}$ has exactly $1$ subgroup of order $1$: the trivial subgroup.
\end{theorem}
\begin{proof}
From Trivial Subgroup is Subgroup, $\struct {\set e, \circ}$ is a subgroup of $\struct {G, \circ}$.
Suppose $\struct {\set g, \circ}$ is a subgroup of $\struct {G, \circ}$.
From Group is not Empty, $e \in \set g$.
Thus it follows trivially that $\struct {\set g, \circ} = \struct {\set e, \circ}$.
That is, $\struct {\set e, \circ}$ is the only subgroup of $\struct {G, \circ}$ of order $1$.
{{qed}}
\end{proof}
|
21927
|
\section{Subgroup of Order p in Group of Order 2p is Normal}
Tags: Groups of Order 2 p
\begin{theorem}
Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$.
Let $a \in G$ be of order $p$.
Let $K = \gen a$ be the subgroup of $G$ generated by $a$.
Then $K$ is normal in $G$.
\end{theorem}
\begin{proof}
The result Non-Abelian Order 2p Group has Order p Element demonstrates that such an element $a$ exists in $G$.
By definition of generator of cyclic group, $K$ is the cyclic group $C_p$ of order $p$.
By Lagrange's Theorem, the index of $K$ is:
:$\index G K = \dfrac {\order G} {\order K} = \dfrac {2 p} p = 2$
The result follows from Subgroup of Index 2 is Normal.
{{qed}}
\end{proof}
|
21928
|
\section{Subgroup of Order p in Group of Order 2p is Normal/Corollary}
Tags: Groups of Order 2 p
\begin{theorem}
Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$ whose identity element is $e$.
Let $a \in G$ be of order $p$.
Let $K = \gen a$ be the subgroup of $G$ generated by $a$.
Let $G$ be non-abelian.
Every element of $G \setminus K$ is of order $2$, and:
:$\forall b \in G \setminus K: b a b^{-1} = a^{-1}$
\end{theorem}
\begin{proof}
By Lagrange's Theorem, the elements of $G \setminus K$ can be of order $1$, $2$, $p$ or $2 p$.
$1$ is not possible because Identity is Only Group Element of Order 1.
Then we have that $G$ is non-abelian.
Hence from Cyclic Group is Abelian, $G$ is not cyclic.
Thus $\order b \ne 2 p$.
It remains to investigate $2$ and $p$.
Let $b \in G \setminus K$.
By Subgroup of Index 2 contains all Squares of Group Elements:
:$b^2 \in K$
{{AimForCont}} $b$ has order $p$.
Then by Intersection of Subgroups of Prime Order:
:$K \cap \gen b = e$
which contradicts $b^2 \in K$.
Thus by Proof by Contradiction $\order b \ne p$
Hence it must follow that $\order b = 2$.
{{qed|lemma}}
We have that:
:$b a \in G \setminus K$
Thus:
:$\paren {b a}^2 = e$
{{begin-eqn}}
{{eqn | l = b a
| o = \in
| r = G \setminus K
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {b a}^2
| r = e
| c =
}}
{{eqn | ll= \leadsto
| l = b a b
| r = \paren {b a}^2 a^{-1}
| c =
}}
{{eqn | r = a^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = b a b^{-1}
| r = a^{-1}
| c = as $b = b^{-1}$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21929
|
\section{Subgroup of Real Numbers is Discrete or Dense}
Tags: Group Theory, Real Numbers, Topological Groups
\begin{theorem}
Let $G$ be a subgroup of the additive group of real numbers.
Then one of the following holds:
:$G$ is dense in $\R$.
:$G$ is discrete and there exists $a \in \R$ such that $G = a \Z$, that is, $G$ is cyclic.
\end{theorem}
\begin{proof}
If $G$ is trivial, then $G$ is discrete and cyclic.
Let $G$ be non-trivial.
Because $x \in G \iff -x \in G$, $G$ has a strictly positive element.
Thus $G \cap \R_{>0}$ is non-empty.
We have that $G \cap \R_{>0}$ is bounded below by $0$.
Hence, by the Continuum Property, $G \cap \R_{>0}$ admits an infimum.
So, let $a = \map \inf {G \cap \R_{>0} }$.
\end{proof}
|
21930
|
\section{Subgroup of Solvable Group is Solvable}
Tags: Subgroups, Solvable Groups, Subgroup of Solvable Group is Solvable
\begin{theorem}
Let $G$ be a solvable group.
Let $H$ be a subgroup of $G$.
Then $H$ is solvable.
\end{theorem}
\begin{proof}
{{tidy|much work to be done}}
{{MissingLinks}}
{{refactor|proof per page, and we also go back to the original proof and reinstate it}}
We will provide two proofs here, for the two equivalent definitions of solvable groups.
Firstly we, know that a group is solvable if and only if its derived series
$D(G)=[G,G] \ , \ D^i(G)= [D^{i-1}(G),D^{i-1}(G)$
becomes trivial after finite iteration. Meaning
$D^j(G) = \{1\}$
for some finite j. Now it is trivial that
$D(H) \leq D(G)$
since H is smaller than G. Further since $D^i(H)$ is dominated by $D^i(G)$ it too has to become trivial after a finite amount of steps.
{{qed}}
A different proof using the more common definition of solvability:
Let $H \leq G$ and $G$ be solvable with normal series
$1 = G_0 \lhd G_1 \lhd \dots \lhd G_m = G$
such that $G_{i+1}/G_i$ is abelian for all i.
Define $N_i= G_i \cap H$. These $N_i$ will form a normal series with abelian factors.
Normality:
Let $x \in N_i$ and $y \in N{i+1}$ then $yxy^{-1} \in N$, since N is a group, and $yxy^{-1} \in G_i$, since $G_i$ is normal in $G_{i+1}$. Hence $N_i$ is stable under conjugation and therefore normal.
Abelian Factors:
We have:
:$\dfrac{N_{i+1}}{N_i} = \dfrac{N_{i+1}}{N_{i+1}\cap G_i} \cong \dfrac{N_{i+1}G_i}{G_i} \leq \dfrac{G_{i+1}}{G_i}$
so the quotient $\dfrac{N_{i+1}}{N_i}$ is isomorphic to a subgroup of the abelian group $\dfrac{G_{i+1}}{G_i}$, due to the Second Isomorphism Theorem for Groups, hence its abelian, proving the theorem.
{{qed}}
{{Proofread|grammar's all over the place for a start}}
\end{proof}
|
21931
|
\section{Subgroup of Subgroup with Prime Index}
Tags: Index of Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Let $K$ be a subgroup of $H$.
Let:
:$\index G K = p$
where:
:$p$ denotes a prime number
:$\index G K$ denotes the index of $K$ in $G$.
Then either:
:$H = K$
or:
:$H = G$
\end{theorem}
\begin{proof}
From the Tower Law for Subgroups:
:$\index G K = \index G H \index H K$
As $\index G K = p$ is prime, either $\index G H = p$ or $\index H K = p$.
Thus either $\index G H = 1$ or $\index H K = 1$.
The result follows from Index is One iff Subgroup equals Group.
{{qed}}
\end{proof}
|
21932
|
\section{Subgroup of Subgroup with Prime Index/Corollary}
Tags: Index of Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Let $K \subsetneq H$.
Let:
:$\index G K = p$
where:
:$p$ denotes a prime number
:$\index G K$ denotes the index of $K$ in $G$.
Then:
:$H = G$
\end{theorem}
\begin{proof}
As $K \subsetneq H$ and $K$ is a subgroups of $G$, it follows that $K$ is a proper subgroup of $H$.
That is, $K \ne H$
Hence from Subgroup of Subgroup with Prime Index:
:$H = G$
{{qed}}
\end{proof}
|
21933
|
\section{Subgroup of Symmetric Group that Fixes n}
Tags: Symmetric Groups
\begin{theorem}
Let $S_n$ denote the symmetric group on $n$ letters.
Let $H$ denote the subgroup of $S_n$ which consists of all $\pi \in S_n$ such that:
:$\map \pi n = n$
Then:
:$H = S_{n - 1}$
and the index of $H$ in $S_n$ is given by:
:$\index {S_n} H = n$
\end{theorem}
\begin{proof}
We have that $S_{n - 1}$ is the symmetric group on $n - 1$ letters.
Let $\pi \in S_{n - 1}$.
Then $\pi$ is a permutation on $n - 1$ letters.
Hence $\pi$ is also a permutation on $n$ letters which fixes $n$.
So $S_{n - 1} \subseteq H$.
Now let $\pi \in H$.
Then $\pi$ is a permutation on $n$ letters which fixes $n$.
That is, $\pi$ is a permutation on $n - 1$ letters.
Thus $\pi \in S_{n - 1}$.
So we have that $H = S_{n - 1}$.
We also have that $S_{n - 1}$ is a group, and:
:$\forall \rho \in S_{n - 1}: \rho \in S_n$
So $S_{n - 1}$ is a subset of $S_n$ which is a group.
Hence $S_{n - 1}$ is a subgroup of $S_n$ by definition.
Then we have:
{{begin-eqn}}
{{eqn | l = \index {S_n} {S_{n - 1} }
| r = \dfrac {\order {S_n} } {\order {S_{n - 1} } }
| c = {{Defof|Index of Subgroup}}
}}
{{eqn | r = \dfrac {n!} {\paren {n - 1}!}
| c = Order of Symmetric Group
}}
{{eqn | r = \dfrac {n \paren {n - 1}!} {\paren {n - 1}!}
| c = {{Defof|Factorial}}
}}
{{eqn | r = n
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21934
|
\section{Subgroups of Additive Group of Integers}
Tags: Subgroups, Additive Groups of Integer Multiples, Additive Group of Integers, Subgroups of Additive Group of Integers, Group Theory, Integers, Additive Group of Integer Multiples
\begin{theorem}
Let $\struct {\Z, +}$ be the additive group of integers.
Let $n \Z$ be the additive group of integer multiples of $n$.
Every non-trivial subgroup of $\struct {\Z, +}$ has the form $n \Z$.
\end{theorem}
\begin{proof}
First we note that, from Integer Multiples under Addition form Infinite Cyclic Group, $\struct {n \Z, +}$ is an infinite cyclic group.
From Cyclic Group is Abelian, it follows that $\struct {n \Z, +}$ is an infinite abelian group.
Let $H$ be a non-trivial subgroup of $\struct {\Z, +}$.
Because $H$ is non-trivial:
:$\exists m \in \Z: m \in H: m \ne 0$
Because $H$ is itself a group:
:$-m \in H$
So either $m$ or $-m$ is positive and therefore in $\Z_{>0}$.
Thus:
:$H \cap \Z_{>0} \ne \O$
From the Well-Ordering Principle, $H \cap \Z_{>0}$ has a smallest element, which we can call $n$.
It follows from Subgroup of Infinite Cyclic Group is Infinite Cyclic Group that:
: $\forall a \in \Z: a n \in H$
Thus:
: $n \Z \subseteq H$
{{AimForCont}}:
:$\exists m \in \Z: m \in H \setminus n \Z$
Then $m \ne 0$, and also $-m \in H \setminus n \Z$.
Assume $m > 0$, otherwise we consider $-m$.
By the Division Theorem:
:$m = q n + r$
If $r = 0$, then $m = q n \in n \Z$, so $0 \le r < n$.
Now this means $r = m - q n \in H$ and $0 \le r < n$.
This would mean $n$ was not the smallest element of $H \cap \Z$.
Hence, by Proof by Contradiction, there can be no such $m \in H \setminus n \Z$.
Thus:
:$H \setminus n \Z = \O$
Thus from Set Difference with Superset is Empty Set:
:$H \subseteq n \Z$
Thus we have $n \Z \subseteq H$ and $H \subseteq n \Z$.
Hence:
:$H = n \Z$
{{qed}}
\end{proof}
|
21935
|
\section{Subgroups of Additive Group of Integers Modulo m}
Tags: Additive Groups of Integers Modulo m, Additive Group of Integers Modulo m
\begin{theorem}
Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$.
The subgroups of $\struct {\Z_m, +_m}$ are the additive groups of integers modulo $k$ where:
:$k \divides m$
\end{theorem}
\begin{proof}
From Integers Modulo m under Addition form Cyclic Group, $\struct {\Z_m, +_m}$ is cyclic.
Let $H$ be a subgroup of $\struct {\Z_m, +_m}$
From Subgroup of Cyclic Group is Cyclic, $H$ is of the form $\struct {\Z_k, +_k}$ for some $k \in \Z$.
From Lagrange's Theorem, it follows that $k \divides m$.
Hence the result.
{{qed}}
\end{proof}
|
21936
|
\section{Subgroups of Cartesian Product of Additive Group of Integers}
Tags: Additive Group of Integers, Additive Group of Integer Multiples, Additive Groups of Integer Multiples
\begin{theorem}
Let $\struct {\Z, +}$ denote the additive group of integers.
Let $m, n \in \Z_{> 0}$ be (strictly) positive integers.
Let $\struct {\Z \times \Z, +}$ denote the Cartesian product of $\struct {\Z, +}$ with itself.
The subgroups of $\struct {\Z \times \Z, +}$ are not all of the form:
:$\struct {m \Z, +} \times \struct {n \Z, +}$
where $\struct {m \Z, +}$ denotes the additive group of integer multiples of $m$.
\end{theorem}
\begin{proof}
Consider the map $\phi: \struct {m \Z, +} \times \struct {n \Z, +} \mapsto \struct {\Z, +} \times \struct {\Z, +}$ defined by:
:$\forall c, d \in \Z: \map \phi {m c, n d} = \tuple {c, d}$
which is a group isomorphism.
{{explain|Prove the above statement}}
Hence, $\struct {m \Z, +} \times \struct {n \Z, +}$ is a free abelian group of rank $2$.
Therefore, any subgroup generated by a singleton, for example, $\set {\tuple {x, 0}: x \in \Z}$ is a subgroup not in the form $\struct {m \Z, +} \times \struct {n \Z, +}$.
{{refactor|Implement "singly generated" as a definition page in its own right and then link to it, hence keep this page clean}}
{{qed}}
\end{proof}
|
21937
|
\section{Subring Generated by Unity of Ring with Unity}
Tags: Ideal Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let the mapping $g: \Z \to R$ be defined as $\forall n \in \Z: \map g n = n 1_R$, where $n 1_R$ the $n$th power of $1_R$.
Let $\ideal x$ be the principal ideal of $\struct {R, +, \circ}$ generated by $x$.
Then $g$ is an epimorphism from $\Z$ onto the subring $S$ of $R$ generated by $1_R$.
If $R$ has no proper zero divisors, then $g$ is the only nonzero homomorphism from $\Z$ into $R$.
The kernel of $g$ is either:
:$(1): \quad \ideal {0_R}$, in which case $g$ is an isomorphism from $\Z$ onto $S$
or:
:$(2): \quad \ideal p$ for some prime $p$, in which case $S$ is isomorphic to the field $\Z_p$.
\end{theorem}
\begin{proof}
By the Index Law for Sum of Indices and Powers of Ring Elements, we have $\paren {n 1_R} \paren {m 1_R} = n \paren {m 1_R} = \paren {n m} 1_R$.
Thus $g$ is an epimorphism from $\Z$ onto $S$.
{{AimForCont}} $R$ has no proper zero divisors.
By Kernel of Ring Epimorphism is Ideal, the kernel of $g$ is an ideal of $\Z$.
By Ring of Integers is Principal Ideal Domain, the kernel of $g$ is $\ideal p$ for some $p \in \Z_{>0}$.
By Kernel of Ring Epimorphism is Ideal (don't think this is the correct reference - check it), $S$ is isomorphic to $\Z_p$ and also has no proper zero divisors.
So from Integral Domain of Prime Order is Field either $p = 0$ or $p$ is prime.
Now we need to show that $g$ is unique.
Let $h$ be a non-zero (ring) homomorphism from $\Z$ into $R$.
As $\map h 1 = \map h {1^2} = \paren {\map h 1}^2$, either $\map h 1 = 1_R$ or $\map h 1 = 0_R$ by Idempotent Elements of Ring with No Proper Zero Divisors.
But, by Homomorphism of Powers: Integers, $\forall n \in \Z: \map h n = \map h {n 1} = n \map h 1$
So if $\map h 1 = 0_R$, then $\forall n \in \Z: \map h n = n 0_R = 0_R$.
Hence $h$ would be a zero homomorphism, which contradicts our stipulation that it is not.
So $\map h 1 = 1_R$, and thus $\forall n \in \Z: \map h n = n 1 = \map g n$.
{{qed}}
{{Proofread}}
\end{proof}
|
21938
|
\section{Subring Module is Module}
Tags: Subring Module is Module, Module Theory, Subrings, Modules
\begin{theorem}
Let $\struct {R, +, \times}$ be a ring.
Let $\struct {S, +_S, \times_S}$ be a subring of $R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $\circ_S$ be the restriction of $\circ$ to $S \times G$.
Let $\struct {G, +_G, \circ_S}_S$ be subring module induced by $S$.
Then $\struct {G, +_G, \circ_S}_S$ is an $S$-module.
\end{theorem}
\begin{proof}
We have that:
:$\forall a, b \in S: a +_S b = a + b$
:$\forall a, b \in S: a \times_S b = a \times b$
:$\forall a \in S: \forall x \in G = a \circ_S x = a \circ x$
as $+_S$, $\times_S$ and $\circ_S$ are restrictions.
Let us verify the module axioms.
\end{proof}
|
21939
|
\section{Subring Module is Module/Special Case}
Tags: Subring Module is Module, Module Theory, Subrings
\begin{theorem}
Let $S$ be a subring of the ring $\struct {R, +, \circ}$.
Let $\circ_S$ be the restriction of $\circ$ to $S \times R$.
Then $\struct {R, +, \circ_S}_S$ is an $S$-module.
\end{theorem}
\begin{proof}
From Ring is Module over Itself, it follows that:
:$\struct {R, +, \circ}_R$ is an $R$-module.
The result follows directly from Subring Module is Module.
{{qed}}
\end{proof}
|
21940
|
\section{Subring Module is Module/Special Case/Unitary Module}
Tags: Subring Module is Module
\begin{theorem}
Let $S$ be a subring of the ring $\struct {R, +, \circ}$.
Let $\circ_S$ be the restriction of $\circ$ to $S \times R$.
Let $\struct {R, +, \circ}$ be a ring with unity such that $1_R$ is that unity.
Let $1_R \in S$.
Then $\struct {R, +, \circ_S}_S$ is a unitary $S$-module.
\end{theorem}
\begin{proof}
From Subring Module is Module: Special Case, we have that $\struct {R, +, \circ_S}_S$ is an $S$-module.
Then {{hypothesis}} $1_R$ is the unity of $\struct {S, +, \circ_S}$.
Thus $\struct {S, +, \circ_S}$ is also a ring with unity.
It follows from Ring with Unity is Module over Itself that $\struct {R, +, \circ_S}_S$ is a unitary module.
{{qed}}
\end{proof}
|
21941
|
\section{Subring Module is Module/Unitary}
Tags: Subring Module is Module
\begin{theorem}
Let $\struct {R, +, \times}$ be a ring.
Let $\struct {S, +_S, \times_S}$ be a subring of $R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $\circ_S$ be the restriction of $\circ$ to $S \times G$.
Let $\struct {R, +, \times}$ be a ring with unity.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Let $1_R \in S$.
Then $\struct{G, +_G, \circ_S}_S$ is also unitary.
\end{theorem}
\begin{proof}
From Subring Module is Module, we have that $\struct {G, +_G, \circ_S}_S$ is an $S$-module.
It remains to be demonstrated that $\struct{G, +_G, \circ_S}_S$ is unitary.
To show this, we must prove that:
:$\forall x \in G: 1_R \circ_S x = x$
Since $1_R \in S$ by assumption, the product $1_R \circ_S x$ is defined.
We now have:
{{begin-eqn}}
{{eqn | l = 1_R \circ_S x
| r = 1_R \circ x
}}
{{eqn | r = x
| c = {{Module-axiom|4}} on $\struct {G, +_G, \circ}_R$
}}
{{end-eqn}}
and the proof is complete.
{{qed}}
\end{proof}
|
21942
|
\section{Subring is not necessarily Ideal}
Tags: Ideal Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +_S, \circ_S}$ be a subring of $R$.
Then it is not necessarily the case that $S$ is also an ideal of $R$.
\end{theorem}
\begin{proof}
Consider the field of real numbers $\struct {\R, +, \times}$.
We have that a field is by definition a ring, hence so is $\struct {\R, +, \times}$.
From Rational Numbers form Subfield of Real Numbers and Integers form Subdomain of Rationals, it follows that the integers $\struct {\Z, +, \times}$ are a subring of $\struct {\R, +, \times}$.
Consider $1 \in \Z$, and consider $\dfrac 1 2 \in \R$.
We have that $1 \times \dfrac 1 2 = \dfrac 1 2 \notin \Z$.
From this counterexample it is seen that $\Z$ is not an ideal of $R$.
Hence the result, again by Proof by Counterexample.
{{qed}}
\end{proof}
|
21943
|
\section{Subring of Integers is Ideal}
Tags: Integers, Subrings, Ideal Theory
\begin{theorem}
Let $\struct {\Z, +}$ be the additive group of integers.
Every subring of $\struct {\Z, +, \times}$ is an ideal of the ring $\struct {\Z, +, \times}$.
\end{theorem}
\begin{proof}
Follows directly from:
:Subrings of Integers are Sets of Integer Multiples
and:
:Subgroup of Integers is Ideal.
{{qed}}
\end{proof}
|
21944
|
\section{Subring of Non-Archimedean Division Ring}
Tags: Normed Division Rings, Definitions: Norm Theory, Definitions: Division Rings
\begin{theorem}
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with non-archimedean norm $\norm {\, \cdot \,}$.
Let $\struct {S, \norm {\, \cdot \,}_S }$ be a normed division subring of $R$.
Then:
:$\norm {\, \cdot \,}_S$ is a non-archimedean norm.
\end{theorem}
\begin{proof}
$\forall x, y \in S$:
{{begin-eqn}}
{{eqn | l = \norm {x + y}_S
| r = \norm {x + y}
| c = Definition of $\norm {\,\cdot\,}_S$
}}
{{eqn | o = \le
| r = \max \set {\norm x, \norm y}
| c = $(\text N 4)$: Ultrametric Inequality
}}
{{eqn | r = \max \set {\norm x_S, \norm y_S}
| c = Definition of $\norm {\, \cdot \,}_S$
}}
{{end-eqn}}
{{qed}}
Category:Normed Division Rings
\end{proof}
|
21945
|
\section{Subring of Polynomials over Integral Domain Contains that Domain}
Tags: Subrings, Polynomial Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $x \in R$.
Let $D \sqbrk x$ denote the ring of polynomials in $x$ over $D$.
Then $D \sqbrk x$ contains $D$ as a subring and $x$ as an element.
\end{theorem}
\begin{proof}
We have that $\ds \sum_{k \mathop = 0}^m a_k \circ x^k$ is a polynomial for all $m \in \Z_{\ge 0}$.
Set $m = 0$:
:$\ds \sum_{k \mathop = 0}^0 a_k \circ x^k = a_k \circ x^0 = a_k \circ 1_D = a_k$
Thus:
:$\ds \forall a_k \in D: \sum_{k \mathop = 0}^0 a_k \circ x^k \in D$
It follows directly that $D$ is a subring of $D \sqbrk x$ by applying the Subring Test on elements of $D$.
{{qed}}
\end{proof}
|
21946
|
\section{Subrings of Integers are Sets of Integer Multiples}
Tags: Subrings of Integers are Sets of Integer Multiples, Integers, Subrings
\begin{theorem}
Let $\struct {\Z, +, \times}$ be the integral domain of integers.
The subrings of $\struct {\Z, +, \times}$ are the rings of integer multiples:
:$\struct {n \Z, +, \times}$
where $n \in \Z: n \ge 0$.
There are no other subrings of $\struct {\Z, +, \times}$ but these.
\end{theorem}
\begin{proof}
From Integer Multiples form Commutative Ring, it is clear that $\struct {n \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$ when $n \ge 1$.
We also note that when $n = 0$, we have:
:$\struct {n \Z, +, \times} = \struct {0, +, \times}$
which is the null ring.
When $n = 1$, we have:
:$\struct {n \Z, +, \times} = \struct {\Z, +, \times}$
From Null Ring and Ring Itself Subrings, these extreme cases are also subrings of $\struct {\Z, +, \times}$.
From Subgroups of Additive Group of Integers, the only additive subgroups of $\struct {\Z, +, \times}$ are $\struct {n \Z, +}$.
So there can be no subrings of $\struct {\Z, +, \times}$ which do not have $\struct {n \Z, +}$ as their additive group.
Hence the result.
{{qed}}
\end{proof}
|
21947
|
\section{Subrings of Integers are Sets of Integer Multiples/Examples/Even Integers}
Tags: Subrings of Integers are Sets of Integer Multiples
\begin{theorem}
Let $2 \Z$ be the set of even integers.
Then $\struct {2 \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$.
\end{theorem}
\begin{proof}
From Subrings of Integers are Sets of Integer Multiples, a ring of the form $\struct {n \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$ when $n \ge 1$.
$\struct {2 \Z, +, \times}$ is such an example.
{{qed}}
\end{proof}
|
21948
|
\section{Subsemigroup Closure Test}
Tags: Semigroups, Subsemigroups, Abstract Algebra
\begin{theorem}
To show that an algebraic structure $\struct {T, \circ}$ is a subsemigroup of a semigroup $\struct {S, \circ}$, we need to show only that:
:$(1): \quad T \subseteq S$
:$(2): \quad \circ$ is a closed operation in $T$.
\end{theorem}
\begin{proof}
From Restriction of Associative Operation is Associative, if $\circ$ is associative on $\struct {S, \circ}$, then it will also be associative on $\struct {T, \circ}$.
Thus we do not need to check for associativity in $\struct {T, \circ}$, as that has been inherited from its extension $\struct {S, \circ}$.
So, once we have established that $T \subseteq S$, all we need to do is to check for $\circ$ to be a closed operation.
{{Qed}}
\end{proof}
|
21949
|
\section{Subsemigroup of Cancellable Mappings is Subgroup of Invertible Mappings}
Tags: Examples of Subgroups, Cancellability, Examples of Subsemigroups, Inverse Mappings
\begin{theorem}
Let $S$ be a set.
Let $S^S$ denote the set of mappings from $S$ to itself.
Let $\CC \subseteq S^S$ denote the set of cancellable mappings on $S$.
Let $\MM \subseteq S^S$ denote the set of invertible mappings on $S$.
Then:
:the subsemigroup $\struct {\CC, \circ}$ of $\struct {S^S, \circ}$ coincides with the subgroup $\struct {\MM, \circ}$ of $\struct {S^S, \circ}$
where $\circ$ denotes composition of mappings.
\end{theorem}
\begin{proof}
From Set of Invertible Mappings forms Symmetric Group, we have that $\struct {\MM, \circ}$ is a group.
Hence, by definition, $\struct {\MM, \circ}$ is a subgroup of $\struct {S^S, \circ}$.
Recall from Bijection iff Left and Right Inverse that a mapping is invertible {{iff}} it is a bijection.
By definition, a cancellable mapping is a mapping both left cancellable and right cancellable.
From Injection iff Left Cancellable and Surjection iff Right Cancellable, a cancellable mapping is both an injection and a surjection.
That is, a mapping is cancellable mapping {{iff}} it is a bijection.
That is, $\struct {\CC, \circ}$ is exactly the same as $\struct {\MM, \circ}$.
{{qed}}
\end{proof}
|
21950
|
\section{Subsemigroup of Monoid is not necessarily Monoid}
Tags: Subsemigroups, Monoids, Identity Elements
\begin{theorem}
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $\struct {T, \circ}$ be a subsemigroup of $\struct {S, \circ}$
Then it is not necessarily the case that $\struct {T, \circ}$ has an identity.
\end{theorem}
\begin{proof}
Consider the set of integers under multiplication $\struct {\Z, \times}$.
From Integers under Multiplication form Monoid, $\struct {\Z, \times}$ is a monoid.
Let $n \in \Z$ such that $n > 1$.
Let $n \Z$ be the set of integer multiples of $n$:
:$\set {x \in \Z: n \divides x}$
where $\divides$ denotes divisibility.
From Integer Multiples under Multiplication form Semigroup, $\struct {n \Z, \times}$ is a semigroup which has no identity.
By construction, $n \Z$ is a subset of $\Z$.
Hence $\struct {n \Z, \times}$ is a subsemigroup of $\struct {\Z, \times}$ which has no identity.
{{qed}}
\end{proof}
|
21951
|
\section{Subsemigroup of Ordered Semigroup is Ordered}
Tags: Ordered Semigroups
\begin{theorem}
Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $\struct {T, \circ_T}$ be a subsemigroup of $\struct {S, \circ}$.
Then the ordered structure $\struct {T, \circ_T, \preceq_T}$ is also an ordered semigroup.
In the above:
:$\circ_T$ denotes the operation induced on $T$ by $\circ$
:$\preceq_T$ denotes the restriction of $\preceq$ to $T \times T$.
\end{theorem}
\begin{proof}
It is necessary to ascertain that $\struct {T, \circ {\restriction_T} }$ fulfils the ordered semigroup axioms:
{{:Axiom:Ordered Semigroup Axioms}}
In this context, we see that $\text {OS} 0$ and $\text {OS} 1$ are fulfilled a fortiori by dint of $\struct {T, \circ {\restriction_T} }$ being a subsemigroup of $\struct {S, \circ}$.
We have that $\struct {S, \circ, \preceq}$ is an ordered semigroup.
From Restriction of Ordering is Ordering, we have that $\preceq_T$ is an ordering.
Hence:
{{begin-eqn}}
{{eqn | q = \forall a, b \in S
| l = a \preceq b
| o = \implies
| r = \paren {a \circ c} \preceq \paren {b \circ c}
| c = Ordered Semigroup Axiom $\text {OS} 2$ on $\struct {S, \circ, \preceq}$
}}
{{eqn | ll= \leadsto
| q = \forall a, b \in T
| l = a \preceq_T b
| o = \implies
| r = \paren {a \circ_T c} \preceq_T \paren {b \circ_T c}
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | q = \forall a, b \in S
| l = a \preceq b
| o = \implies
| r = \paren {c \circ a} \preceq \paren {c \circ b}
| c = Ordered Semigroup Axiom $\text {OS} 2$ on $\struct {S, \circ, \preceq}$
}}
{{eqn | ll= \leadsto
| q = \forall a, b \in T
| l = a \preceq_T b
| o = \implies
| r = \paren {c \circ_T a} \preceq_T \paren {c \circ_T b}
| c =
}}
{{end-eqn}}
Hence $\preceq_T$ fulfils Ordered Semigroup Axiom $\text {OS} 2$ on $\struct {T, \circ_T, \preceq_T}$.
{{qed}}
\end{proof}
|
21952
|
\section{Subsequence Characterisation of Compact Linear Transformations}
Tags: Compact Linear Transformations
\begin{theorem}
Let $\mathbb F \in \set {\R, \C}$.
Let $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$ be normed vector spaces over $\mathbb F$.
Let $A : X \to Y$ be a linear transformation.
Then $A$ is compact {{iff}}:
:all bounded sequences $\sequence {x_n}$ in $X$ have a subsequence $\sequence {x_{n_j} }$ such that the sequence $\sequence {A x_{n_j} }$ converges.
\end{theorem}
\begin{proof}
Let $\operatorname {ball} X$ be the closed unit ball of $X$.
\end{proof}
|
21953
|
\section{Subsequence is Equivalent to Cauchy Sequence}
Tags: Cauchy Sequences, Normed Division Rings
\begin{theorem}
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Let $\sequence {x_{m_n} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\ds \lim_{n \mathop \to \infty} {x_n - x_{m_n} } = 0$
\end{theorem}
\begin{proof}
From Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence:
:$\sequence {x_{m_n} }$ is a Cauchy sequence
Let $\epsilon > 0$.
By definition of a Cauchy sequence:
:$\exists N: \forall n, m > N: \norm {x_n - x_m } < \epsilon$
From Index of Subsequence not Less than its Index:
$\forall n \in \N : m_n \ge n$
Thus:
:$\exists N: \forall n > N: \norm {x_n - x_{m_n} } < \epsilon$
By definition of convergence:
:$\ds \lim_{n \mathop \to \infty} {x_n - x_{m_n} } = 0$
{{qed}}
Category:Cauchy Sequences
Category:Normed Division Rings
\end{proof}
|
21954
|
\section{Subsequence of Cauchy Sequence in Normed Division Ring is Cauchy Sequence}
Tags: Cauchy Sequences, Normed Division Rings
\begin{theorem}
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero: $0$.
Let $\sequence {x_n}$ be a Cauchy sequence in $R$.
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\sequence {x_{n_r} }$ is a Cauchy sequence in $R$.
\end{theorem}
\begin{proof}
Let $\epsilon > 0$.
Since $\sequence {x_n}$ is a Cauchy sequence then:
:$\exists N: \forall n,m > N: \norm {x_n - x_m } < \epsilon$
Now let $R = N$.
Then from Strictly Increasing Sequence of Natural Numbers:
:$\forall r, s > R: n_r \ge r$ and $n_s \ge s$
Thus $n_r, n_s > N$ and so:
:$\norm {x_{n_r} - x_{n_s} } < \epsilon$
The result follows.
{{qed}}
\end{proof}
|
21955
|
\section{Subsequence of Real Sequence Diverging to Negative Infinity Diverges to Negative Infinity}
Tags: Divergent Sequences, Limits of Sequences
\begin{theorem}
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to -\infty$.
Let $\sequence {x_{n_j} }_{j \mathop \in \N}$ be a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
Then:
:$x_{n_j} \to -\infty$
\end{theorem}
\begin{proof}
Let $M > 0$ be a real number.
From the definition of divergence to $-\infty$, there exists $N \in \N$ such that:
:$x_n < -M$
for each $n \ge N$.
From Strictly Increasing Sequence of Natural Numbers, we have:
:$n_j \ge j$ for each $j$.
So, we have:
:$x_{n_j} < -M$
for each $j \ge N$.
Since $M$ was arbitrary, we have:
:$x_{n_j} \to -\infty$
by the definition of divergence to $-\infty$.
{{qed}}
Category:Divergent Sequences
Category:Limits of Sequences
\end{proof}
|
21956
|
\section{Subsequence of Real Sequence Diverging to Positive Infinity Diverges to Positive Infinity}
Tags: Divergent Sequences, Limits of Sequences
\begin{theorem}
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to +\infty$.
Let $\sequence {x_{n_j} }_{j \mathop \in \N}$ be a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
Then:
:$x_{n_j} \to +\infty$
\end{theorem}
\begin{proof}
Let $M > 0$ be a real number.
From the definition of divergence to $+\infty$, there exists $N \in \N$ such that:
:$x_n > M$
for each $n \ge N$.
From Strictly Increasing Sequence of Natural Numbers, we have:
:$n_j \ge j$ for each $j$.
So, we have:
:$x_{n_j} > M$
for each $j \ge N$.
Since $M$ was arbitrary, we have:
:$x_{n_j} \to \infty$
by the definition of divergence to $+\infty$.
{{qed}}
Category:Divergent Sequences
Category:Limits of Sequences
\end{proof}
|
21957
|
\section{Subsequence of Sequence in Metric Space with Limit}
Tags: Metric Spaces, Limits of Sequences
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $M$.
Let $x$ be a limit point of $S = \set {x_n: n \in \N}$, the set of members of $\sequence {x_n}$.
Then $\sequence {x_n}$ has a subsequence which converges to $x$.
\end{theorem}
\begin{proof}
By Finite Subset of Metric Space has no Limit Points, $S$ is infinite (or it has no limit points).
We may assume that $x_n$ never equals $x$.
Otherwise we delete all instances of $x$ from $\sequence {x_n}$ and create a new sequence $x_m$ which is a subsequence of $x_n$ which ''does'' never equal $x$.
Then $S \setminus \set x$ is still infinite, and it still has $x$ as a limit point.
So, since $x$ is a limit point of $S$, there is an integer $\map n 1$, say, such that $x_{\map n 1} \in \map {B_1} x$, where $\map {B_1} x$ is the open $1$-ball of $x$.
Suppose we choose the integers $\map n 1 < \map n 2 < \cdots < \map n k$ so that $x_{\map n i} \in \map {B_{1/i} } x$ for $i = 1, 2, \ldots, k$.
Now we put $\epsilon = \min \set {\dfrac 1 {k+1}, \map d {x_1, x}, \map d {x_2, x}, \map d {x_{\map n k}, x} }$.
Since $x_n \ne x$ for any $n$, it follows that $\epsilon > 0$.
Since $x$ is a limit point of $S$, there exists an integer $\map n {k+1}$ such that $x_{\map n {k+1} } \in \map {B_\epsilon} x$.
Now $\epsilon$ has been chosen so as to force $\map n {k+1} > \map n k$, since $\map d {x_i, x} \ge \epsilon$ for all $i \le \map n k$.
Also, note that $x_{\map n {k+1} } \in \map {B_{\frac 1 {k+1}} } x$.
This completes the inductive step in constructing a subsequence.
This converges to $x$, because $\forall k_0 \in \N_{>0}$ and $\forall k \ge k_0$ we have $\map d {x_{\map n k}, x} < \dfrac 1 k \ge \dfrac 1 {k_0}$.
Hence the result.
{{qed}}
\end{proof}
|
21958
|
\section{Subsequence of Subsequence}
Tags: Subsequences
\begin{theorem}
Let $s$ be a set.
Let $\sequence {s_n}$ be a sequence in $S$.
Let $\sequence {s_m}$ be a subsequence of $\sequence {s_n}$.
Let $\sequence {s_k}$ be a subsequence of $\sequence {s_m}$.
Then $\sequence {s_k}$ is a subsequence of $\sequence {s_n}$.
\end{theorem}
\begin{proof}
By definition, there exists a strictly increasing sequence $\sequence {n_r}$ in $\N$ such that:
:$\forall m \in \N: s_m = s_{n_r}$
Similarly, there exists a strictly increasing sequence $\sequence {m_s}$ in $\N$ such that:
:$\forall k \in \N: s_k = s_{m_s}$
We have that:
:$\forall k \in \N: s_k \in \sequence {s_m}$
and that:
:$\forall m \in \N: s_m \in \sequence {s_n}$
hence:
:$\forall k \in \N: s_k \in \sequence {s_n}$
Because $\sequence {s_k}$ is a subsequence of $\sequence {s_m}$, we have that:
:$k = m_s \implies k + 1 = m_s + p$ for some $p \in \Z_{>0}$
and:
:$m_s = n_r \implies m_s + p = n_r + q$ for some $q \in \Z_{>0}$
Hence:
:$k = n_r \implies k + 1 = n_r + q$
and so $\sequence {s_k}$ is a subsequence of $\sequence {s_n}$.
{{qed}}
\end{proof}
|
21959
|
\section{Subset Equivalences}
Tags: Set Theory, Set Complement, Intersection, Empty Set, Subsets, Universe, Union, Subset, Set Difference
\begin{theorem}
These statements concerning subsets are equivalent:
* <math>S \subseteq T</math>
* <math>S \cup T = T</math>
* <math>S \cap T = S</math>
* <math>S \setminus T = \varnothing</math>
* <math>S \cap \complement \left({T}\right) = \varnothing</math>
* <math>\complement \left({S}\right) \cup T = \mathbb{U}</math>
* <math>\complement \left({T}\right) \subseteq \complement \left({S}\right)</math>
where:
* <math>S \subseteq T</math> denotes that <math>S</math> is a subset of <math>T</math>;
* <math>S \cup T</math> denotes the union of <math>S</math> and <math>T</math>;
* <math>S \cap T</math> denotes the intersection of <math>S</math> and <math>T</math>;
* <math>S \setminus T</math> denotes the set difference between <math>S</math> and <math>T</math>;
* <math>\varnothing</math> denotes the empty set;
* <math>\complement</math> denotes the complement of <math>S</math>;
* <math>\mathbb{U}</math> denotes the universal set.
\end{theorem}
\begin{proof}
* Let <math>S \cup T = T</math>.
Then from the definition of set equality, <math>S \cup T \subseteq T</math>.
Thus:
{{begin-equation}}
{{equation | l=<math>S</math>
| o=<math>\subseteq</math>
| r=<math>S \cup T</math>
| c=Subset of Union
}}
{{equation | ll=<math>\implies</math>
| l=<math>S</math>
| o=<math>\subseteq</math>
| r=<math>T</math>
| c=Subsets Transitive
}}
{{end-equation}}
* Now let <math>S \subseteq T</math>.
We have:
{{begin-equation}}
{{equation | l=<math>S</math>
| o=<math>\subseteq</math>
| r=<math>T</math>
| c=
}}
{{equation | ll=<math>\implies</math>
| l=<math>S \cup T</math>
| o=<math>\subseteq</math>
| r=<math>T \cup T</math>
| c=Set Union Preserves Subsets
}}
{{equation | ll=<math>\implies</math>
| l=<math>S \cup T</math>
| o=<math>\subseteq</math>
| r=<math>T</math>
| c=Union is Idempotent
}}
{{end-equation}}
Alternatively:
{{begin-equation}}
{{equation | l=<math>T</math>
| o=<math>\subseteq</math>
| r=<math>T</math>
| c=Subset of Itself
}}
{{equation | ll=<math>\implies</math>
| l=<math>S \subseteq T</math>
| o=<math>\and</math>
| r=<math>T \subseteq T</math>
| c=Rule of Conjunction
}}
{{equation | ll=<math>\implies</math>
| l=<math>S \cup T</math>
| o=<math>\subseteq</math>
| r=<math>T</math>
| c=Union Smallest
}}
{{end-equation}}
From Subset of Union, we have <math>T \subseteq S \cup T</math>.
So we have <math>T \subseteq S \cup T \and S \cup T \subseteq T</math>.
Hence by the definition of set equality, <math>S \cup T = T</math>.
Thus <math>S \subseteq T \implies S \cup T = T</math>.
Hence <math>S \subseteq T \iff S \cup T = T</math>, so <math>S \cup T = T</math> and <math>S \subseteq T</math> are equivalent.
{{Qed}}
----
* Let <math>S \cap T = S</math>.
Then by the definition of set equality, <math>S \subseteq S \cap T</math>.
Thus:
{{begin-equation}}
{{equation | l=<math>S \cap T</math>
| o=<math>\subseteq</math>
| r=<math>T</math>
| c=Intersection Subset
}}
{{equation | ll=<math>\implies</math>
| l=<math>S</math>
| o=<math>\subseteq</math>
| r=<math>T</math>
| c=Subsets Transitive
}}
{{end-equation}}
* Now let <math>S \subseteq T</math>.
We have:
{{begin-equation}}
{{equation | l=<math>S</math>
| o=<math>\subseteq</math>
| r=<math>T</math>
| c=
}}
{{equation | ll=<math>\implies</math>
| l=<math>S \cap S</math>
| o=<math>\subseteq</math>
| r=<math>T \cap S</math>
| c=Set Intersection Preserves Subsets
}}
{{equation | ll=<math>\implies</math>
| l=<math>S</math>
| o=<math>\subseteq</math>
| r=<math>S \cap T</math>
| c=Intersection is Idempotent and Intersection is Commutative
}}
{{end-equation}}
Alternatively:
{{begin-equation}}
{{equation | l=<math>S</math>
| o=<math>\subseteq</math>
| r=<math>S</math>
| c=Subset of Itself
}}
{{equation | ll=<math>\implies</math>
| l=<math>S \subseteq S</math>
| o=<math>\and</math>
| r=<math>S \subseteq T</math>
| c=Rule of Conjunction
}}
{{equation | ll=<math>\implies</math>
| l=<math>S</math>
| o=<math>\subseteq</math>
| r=<math>S \cap T</math>
| c=Intersection Largest
}}
{{end-equation}}
Then we have:
{{begin-equation}}
{{equation | l=<math>S \cap T</math>
| o=<math>\subseteq</math>
| r=<math>S</math>
| c=Intersection Subset
}}
{{equation | ll=<math>\implies</math>
| l=<math>S \cap T</math>
| o=<math>=</math>
| r=<math>S</math>
| c=Set Equality
}}
{{end-equation}}
So we have <math>S \cap T = S \implies S \subseteq T</math> and <math>S \subseteq T \implies S \cap T = S</math> and so:
:<math>S \subseteq T \iff S \cap T = S</math>.
{{Qed}}
----
{{begin-equation}}
{{equation | l=<math>S \setminus T = \varnothing</math>
| o=<math>\iff</math>
| r=<math>\neg \left({\exists x: x \in S \and x \notin T}\right)</math>
| c=Definition of Empty Set
}}
{{equation | o=<math>\iff</math>
| r=<math>\forall x: \neg \left({x \in S \and x \notin T}\right)</math>
| c=De Morgan's Laws (Predicate Logic)
}}
{{equation | o=<math>\iff</math>
| r=<math>\forall x: x \notin S \or x \in T</math>
| c=De Morgan's Laws
}}
{{equation | o=<math>\iff</math>
| r=<math>\forall x: x \in S \implies x \in T</math>
| c=Material Implication
}}
{{equation | o=<math>\iff</math>
| r=<math>S \subseteq T</math>
| c=Definition of Subset
}}
{{end-equation}}
Thus <math>S \subseteq T \iff S \setminus T = \varnothing</math>.
{{qed}}
----
{{begin-equation}}
{{equation | l=<math>S \subseteq T</math>
| o=<math>\iff</math>
| r=<math>S \setminus T = \varnothing</math>
| c=from above
}}
{{equation | o=<math>\iff</math>
| r=<math>S \cap \complement \left({T}\right) = \varnothing</math>
| c=Set Difference as Intersection with Complement
}}
{{end-equation}}
Thus <math>S \subseteq T \iff S \cap \complement \left({T}\right) = \varnothing</math>.
{{qed}}
----
{{begin-equation}}
{{equation | l=<math>S \subseteq T</math>
| o=<math>\iff</math>
| r=<math>S \cap \complement \left({T}\right) = \varnothing</math>
| c=from above
}}
{{equation | o=<math>\iff</math>
| r=<math>\complement \left({S \cap \complement \left({T}\right)}\right) = \mathbb{U}</math>
| c=Complement of Null
}}
{{equation | o=<math>\iff</math>
| r=<math>\complement \left({S}\right) \cup \complement \left({\complement \left({T}\right)}\right) = \mathbb{U}</math>
| c=De Morgan's Laws
}}
{{equation | o=<math>\iff</math>
| r=<math>\complement \left({S}\right) \cup T = \mathbb{U}</math>
| c=Complement of Complement
}}
{{end-equation}}
Thus <math>S \subseteq T \iff \complement \left({S}\right) \cup T = \mathbb{U}</math>
{{qed}}
----
{{begin-equation}}
{{equation | l=<math>S \subseteq T</math>
| o=<math>\iff</math>
| r=<math>\left({x \in S \implies x \in T}\right)</math>
| c=Definition of Subset
}}
{{equation | o=<math>\iff</math>
| r=<math>\left({x \notin T \implies x \notin S}\right)</math>
| c=Rule of Transposition
}}
{{equation | o=<math>\iff</math>
| r=<math>\left({x \in \complement \left({T}\right) \implies x \in \complement \left({S}\right)}\right)</math>
| c=Definition of Complement
}}
{{equation | o=<math>\iff</math>
| r=<math>\complement \left({T}\right) \subseteq \complement \left({S}\right)</math>
| c=Definition of Subset
}}
{{end-equation}}
Thus <math>S \subseteq T \iff \complement \left({T}\right) \subseteq \complement \left({S}\right)</math>.
{{qed}}
\end{proof}
|
21960
|
\section{Subset Product Action is Group Action}
Tags: Subset Products, Cosets, Group Actions, Subset Product Action
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\powerset G$ be the power set of $\struct {G, \circ}$.
For any $S \in \powerset G$ and for any $g \in G$, the subset product action:
:$\forall g \in G: \forall S \in \powerset G: g * S = g \circ S$
is a group action.
\end{theorem}
\begin{proof}
Let $g \in G$.
First we note that since $G$ is closed, and $g \circ S$ consists of products of elements of $G$, it follows that:
:$g * S \subseteq G$
Next we note:
:$e * S = e \circ S = \set {e \circ s: s \in S} = \set {s: s \in S} = S$
and so {{GroupActionAxiom|2}} is satisfied.
Now let $g, h \in G$.
We have:
{{begin-eqn}}
{{eqn | l = \paren {g \circ h} * S
| r = \paren {g \circ h} \circ S
| c =
}}
{{eqn | r = \set {\paren {g \circ h} \circ s: s \in S}
| c =
}}
{{eqn | r = \set {g \circ \paren {h \circ s}: s \in S}
| c =
}}
{{eqn | r = g * \set {h \circ s: s \in S}
| c =
}}
{{eqn | r = g * \paren {h \circ S}
| c =
}}
{{eqn | r = g * \paren {h * S}
| c =
}}
{{end-eqn}}
and so {{GroupActionAxiom|1}} is satisfied.
Hence the result.
{{qed}}
\end{proof}
|
21961
|
\section{Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup}
Tags: Inverse Completions
\begin{theorem}
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then:
:$S \circ' C^{-1}$ is a commutative semigroup
where $S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$ under $\circ'$ in $T$.
\end{theorem}
\begin{proof}
Note that by definition of inverse completion, $\struct {T, \circ'}$ is a semigroup.
Thus $\circ'$ is associative.
First it is demonstrated that $S \circ' C^{-1}$ is a semigroup.
Let $x, z \in S$.
Let $y, w \in C$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ' y^{-1} } \circ' \paren {z \circ' w^{-1} }
| r = x \circ' \paren {y^{-1} \circ' z} \circ' w^{-1}
| c = Associativity of $\circ'$
}}
{{eqn | r = x \circ' \paren {z \circ' y^{-1} } \circ' w^{-1}
| c = Commutation with Inverse in Monoid
}}
{{eqn | r = \paren {x \circ' z} \circ' \paren {y^{-1} \circ' w^{-1} }
| c = Associativity of $\circ'$
}}
{{eqn | r = \paren {x \circ' z} \circ' \paren {w \circ' y}^{-1}
| c = Inverse of Product in Monoid
}}
{{eqn | r = \paren {x \circ z} \circ' \paren {w \circ y}^{-1}
| c = $\circ'$ extends $\circ$
}}
{{end-eqn}}
Thus:
:$\paren {x \circ z} \circ' \paren {w \circ y}^{-1} \in S \circ' C^{-1}$
proving that $S \circ' C^{-1}$ is closed.
Therefore by Subsemigroup Closure Test:
:$S \circ' C^{-1}$ is a subsemigroup of $\struct {T, \circ'}$
and thus a semigroup.
{{qed|lemma}}
It remains to be shown that $\circ'$ is a commutative operation.
Let $\paren {x \circ' y^{-1} }$ and $\paren {z \circ' w^{-1} }$ be two arbitrary elements of $S \circ' C^{-1}$.
By Element Commutes with Product of Commuting Elements, $x, y, z, w$ all commute with each other under $\circ$.
As $\circ'$ is an extension of $\circ$, it follows that $x, y, z, w$ also all commute with each other under $\circ'$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ' y^{-1} } \circ' \paren {z \circ' w^{-1} }
| r = x \circ' \paren {y^{-1} \circ' z} \circ' w^{-1}
| c = Associativity of $\circ'$
}}
{{eqn | r = x \circ' \paren {z \circ' y^{-1} } \circ' w^{-1}
| c = Commutation with Inverse in Monoid
}}
{{eqn | r = \paren {x \circ' z} \circ' \paren {y^{-1} \circ' w^{-1} }
| c = Associativity of $\circ'$
}}
{{eqn | r = \paren {z \circ' x} \circ' \paren {w^{-1} \circ' y^{-1} }
| c = Commutation of Inverses in Monoid
}}
{{eqn | r = z \circ' \paren {x \circ' w^{-1} } \circ' y^{-1}
| c = Associativity of $\circ'$
}}
{{eqn | r = z \circ' \paren {w^{-1} \circ' x} \circ' y^{-1}
| c = Commutation with Inverse in Monoid
}}
{{eqn | r = \paren {z \circ' w^{-1} } \circ' \paren {x \circ' y^{-1} }
| c = Associativity of $\circ'$
}}
{{end-eqn}}
So $x \circ' y^{-1}$ commutes with $z \circ' w^{-1}$.
It follows by definition that $S \circ' C^{-1}$ is a commutative semigroup.
{{qed}}
\end{proof}
|
21962
|
\section{Subset Product is Subset of Generator}
Tags: Subset Products, Group Theory
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $X, Y \subseteq \struct {G, \circ}$.
Then $X \circ Y \subseteq \gen {X, Y}$ where:
:$X \circ Y$ is the Subset Product of $X$ and $Y$ in $G$.
:$\gen {X, Y}$ is the subgroup of $G$ generated by $X$ and $Y$.
\end{theorem}
\begin{proof}
It is clear from Set of Words Generates Group that $\map W {\hat X \cup \hat Y} = \gen {X, Y}$.
It is equally clear that $X \circ Y \subseteq \map W {\hat X \cup \hat Y}$.
{{qed}}
Category:Group Theory
Category:Subset Products
\end{proof}
|
21963
|
\section{Subset Product of Abelian Subgroups}
Tags: Abelian Groups, Subset Products
\begin{theorem}
Let $\left({G, \circ}\right)$ be an abelian group.
Let $H_1$ and $H_2$ be subgroups of $G$.
Then $H_1 \circ H_2$ is a subgroup of $G$.
\end{theorem}
\begin{proof}
From Subgroup of Abelian Group is Normal, $H_1$ and $H_2$ are normal.
The result follows from Subset Product with Normal Subgroup is Subgroup.
{{qed}}
Category:Abelian Groups
Category:Subset Products
\end{proof}
|
21964
|
\section{Subset Product of Normal Subgroups is Normal}
Tags: Normal Subgroups, Subset Products
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $N$ and $N'$ be normal subgroups of $G$.
Then $N N'$ is also a normal subgroup of $G$.
\end{theorem}
\begin{proof}
From Subset Product with Normal Subgroup is Subgroup, we already have that $N N'$ is a subgroup of $G$.
Let $n n' \in N N'$, so that $n \in N, n' \in N'$.
Let $g \in G$.
From Subgroup is Normal iff Contains Conjugate Elements:
:$g n g^{-1}\in N, g n' g^{-1}\in N'$
So:
:$\paren {g n g^{-1} } \paren {g n' g^{-1} } = g n n' g^{-1} \in N N'$
So $N N'$ is normal.
{{qed}}
\end{proof}
|
21965
|
\section{Subset Product of Normal Subgroups with Trivial Intersection}
Tags: Group Direct Products, Normal Subgroups
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H, K$ be normal subgroups of $G$.
Let $H \cap K = e$.
Then $H K$ is isomorphic to $H \times K$ where:
: $H K$ denotes the subset product of $H$ and $K$
: $H \times K$ denotes the direct product of $H$ and $K$.
\end{theorem}
\begin{proof}
Let $G' = H K$.
From Subset Product of Normal Subgroups is Normal, $G'$ is a normal subgroup of $G$.
That is $G'$ is itself a group.
So by the Internal Direct Product Theorem, $G'$ is the internal group direct product of $H$ and $K$.
The result follows by definition of the internal group direct product.
{{qed}}
\end{proof}
|
21966
|
\section{Subset Product with Identity}
Tags: Subset Products, Abstract Algebra
\begin{theorem}
Let $\struct {S, \circ}$ be a magma.
Let $\struct {S, \circ}$ have an identity element $e$.
Then $e \circ S = S \circ e = S$, where $\circ$ is understood to be the subset product with singleton.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = e \circ S
| r = \set e \circ S
| c = {{Defof|Subset Product with Singleton}}
}}
{{eqn | r = \set {x \circ y: x \in \set e, \, y \in S}
| c = {{Defof|Subset Product}}
}}
{{eqn | r = \set {e \circ y: y \in S}
| c =
}}
{{eqn | r = \set {y: y \in S}
| c = {{Defof|Identity Element}}
}}
{{eqn | r = S
| c =
}}
{{end-eqn}}
Thus:
:$e \circ S = S$
A similar argument shows that:
:$S \circ e = S$
{{qed}}
Category:Subset Products
\end{proof}
|
21967
|
\section{Subset Product with Normal Subgroup as Generator}
Tags: Normal Subgroups, Subset Products
\begin{theorem}
Let $G$ be a group whose identity is $e$.
Let:
:$H$ be a subgroup of $G$
:$N$ be a normal subgroup of $G$.
Then:
:$N \lhd \gen {N, H} = N H = H N \le G$
where:
:$\le$ denotes subgroup
:$\lhd$ denotes normal subgroup
:$\gen {N, H}$ denotes a subgroup generator
:$N H$ denotes subset product.
\end{theorem}
\begin{proof}
From Subset Product is Subset of Generator:
:$N H \subseteq \gen {N, H}$
From Subset Product with Normal Subgroup is Subgroup:
:$N H = H N \le G$
Then by the definition of a subgroup generator, $\gen {N, H}$ is the smallest subgroup containing $N H$ and so:
:$\gen {N, H} = N H = H N \le G$
From Normal Subgroup of Subset Product of Subgroups we have that:
:$N \lhd N H$
Hence the result.
{{qed}}
\end{proof}
|
21968
|
\section{Subset Product with Normal Subgroup is Subgroup}
Tags: Normal Subgroups, Subset Products
\begin{theorem}
Let $G$ be a group whose identity is $e$.
Let:
:$(1): \quad H$ be a subgroup of $G$
:$(2): \quad N$ be a normal subgroup of $G$.
Let $H N$ denote subset product.
Then $H N$ and $N H$ are both subgroups of $G$.
\end{theorem}
\begin{proof}
It is clear that $e \in N H$, so $N H \ne \O$.
Suppose $n_1, n_2 \in N$ and $h_1, h_2 \in H$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {n_1 h_1} \paren {n_2 h_2}
| r = n_1 \paren {h_1 n_2 h_1^{-1} h_1} h_2
}}
{{eqn | r = n_1 \paren {h_1 n_2 h_1^{-1} } \paren {h_1 h_2}
}}
{{end-eqn}}
Since $N$ is normal in $G$:
:$\exists n \in N: n = h_1 n_2 h_1^{-1}$
Thus:
{{begin-eqn}}
{{eqn | l = \paren {n_1 h_1} \paren {n_2 h_2}
| r = \paren {n_1 n} \paren {h_1 h_2}
}}
{{eqn | o = \in
| r = N H
}}
{{end-eqn}}
Also:
{{begin-eqn}}
{{eqn | l = \paren {n_1 h_1}^{-1}
| r = h_1^{-1} n_1^{-1}
}}
{{eqn | r = \paren {h_1^{-1} n_1^{-1} h_1} h_1^{-1}
}}
{{eqn | o = \in
| r = N H
}}
{{end-eqn}}
so from the Two-Step Subgroup Test, $N H$ is a subgroup of $G$.
The fact that $N H = H N$ follows from Subset Product of Subgroups.
{{qed}}
\end{proof}
|
21969
|
\section{Subset Product within Commutative Structure is Commutative}
Tags: Subset Products, Abstract Algebra
\begin{theorem}
Let $\struct {S, \circ}$ be a magma.
If $\circ$ is commutative, then the operation $\circ_\PP$ induced on the power set of $S$ is also commutative.
\end{theorem}
\begin{proof}
Let $\struct {S, \circ}$ be a magma in which $\circ$ is commutative.
Let $X, Y \in \powerset S$.
Then:
{{begin-eqn}}
{{eqn | l = X \circ_\PP Y
| r = \set {x \circ y: x \in X, y \in Y}
}}
{{eqn | l = Y \circ_\PP X
| r = \set {y \circ x: x \in X, y \in Y}
}}
{{end-eqn}}
from which it follows that $\circ_\PP$ is commutative on $\powerset S$.
{{qed}}
\end{proof}
|
21970
|
\section{Subset Product within Semigroup is Associative}
Tags: Semigroups, Abstract Algebra, Subset Products, Subset Product within Semigroup is Associative, Associativity
\begin{theorem}
Let $\struct {S, \circ}$ be a semigroup.
Then the operation $\circ_\PP$ induced on the power set of $S$ is also associative.
\end{theorem}
\begin{proof}
Let $X, Y, Z \in \powerset S$.
Then:
{{begin-eqn}}
{{eqn | l = X \circ_\PP \paren {Y \circ_\PP Z}
| r = \set {x \circ \paren {y \circ z}: x \in X, y \in Y, z \in Z}
| c =
}}
{{eqn | r = \set {\paren {x \circ y} \circ z: x \in X, y \in Y, z \in Z}
| c =
}}
{{eqn | r = \paren {X \circ_\PP Y} \circ_\PP Z
| c =
}}
{{end-eqn}}
demonstrating that $\circ_\PP$ is associative on $\powerset S$.
{{qed}}
\end{proof}
|
21971
|
\section{Subset Product within Semigroup is Associative/Corollary}
Tags: Subset Products, Subset Product within Semigroup is Associative
\begin{theorem}
Let $\left({S, \circ}\right)$ be a magma.
If $\circ$ is associative, then:
* $x \left({y S}\right) = \left({x y}\right) S$
* $x \left({S y}\right) = \left({x S}\right) y$
* $\left({S x}\right) y = S \left({x y}\right)$
\end{theorem}
\begin{proof}
From the definition of Subset Product with Singleton:
{{begin-eqn}}
{{eqn | l = x \paren {y S}
| r = \set x \paren {\set y S}
}}
{{eqn | l = x \paren {S y}
| r = \set x \paren {S \set y}
}}
{{eqn | l = \paren {S x} y
| r = \paren {S \set x} \set y
}}
{{end-eqn}}
The result then follows directly from Subset Product within Semigroup is Associative.
{{qed}}
\end{proof}
|
21972
|
\section{Subset Products of Normal Subgroup with Normal Subgroup of Subgroup}
Tags: Normal Subgroups, Subset Products
\begin{theorem}
Let $G$ be a group.
Let:
:$(1): \quad H$ be a subgroup of $G$
:$(2): \quad K$ be a normal subgroup of $H$
:$(3): \quad N$ be a normal subgroup of $G$
Then:
:$N K \lhd N H$
where:
: $N K$ and $N H$ denote subset product
: $\lhd$ denotes the relation of being a normal subgroup.
\end{theorem}
\begin{proof}
Consider arbitrary $x_n \in N, x_h \in H$.
Thus:
:$x_n x_h \in N H$
We aim to show that:
:$x_n x_h N K \paren {x_n x_h}^{-1} \subseteq N K$
thus demonstrating $N K \lhd N H$ by the Normal Subgroup Test.
We have:
{{begin-eqn}}
{{eqn | l = x_n x_h N K \paren {x_n x_h}^{-1}
| r = x_n x_h N K {x_h}^{-1} {x_n}^{-1}
| c = Inverse of Group Product
}}
{{eqn | r = x_n N x_h K {x_h}^{-1} {x_n}^{-1}
| c = as $N \lhd G$ and $H \le G$
}}
{{eqn | r = x_n N K {x_n}^{-1}
| c = as $K \lhd H$
}}
{{eqn | r = N K {x_n}^{-1}
}}
{{eqn | r = K N {x_n}^{-1}
| c = as $N \lhd G$ and $K \le G$
}}
{{eqn | r = K N
}}
{{eqn | r = N K
| c = as $N \lhd G$ and $K \le G$
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
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