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23673
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\begin{definition}[Definition:Abstract Space]
An '''abstract space''' is:
:a set of objects
together with:
:a set of axioms which define operations on and relations between those objects.
\end{definition}
|
23674
|
\begin{definition}[Definition:Abstraction]
'''Abstraction''' is the process of making a general statement from a collection of particular statements so as to summarise a property that applies to the subject of those particular statements.
\end{definition}
|
23675
|
\begin{definition}[Definition:Abstraction (Mathematical Logic)]
An '''abstraction''' is an operator that forms a class name or predicate from a given expression.
\end{definition}
|
23676
|
\begin{definition}[Definition:Absurd]
'''Absurd''' is a word used mainly in logic meaning either '''meaningless''', '''contradictory''' or '''internally inconsistent'''.
\end{definition}
|
23677
|
\begin{definition}[Definition:Abundance]
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $\map {\sigma_1} n$ be the divisor sum function of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then the '''abundance''' of $n$ is defined as $\map A n = \map {\sigma_1} n - 2 n$.
\end{definition}
|
23678
|
\begin{definition}[Definition:Abundancy Index]
Let $n$ be a positive integer.
Let $\map {\sigma_1} n$ be the divisor sum function of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then the '''abundancy index''' of $n$ is defined as $\dfrac {\map {\sigma_1} n} n$.
\end{definition}
|
23679
|
\begin{definition}[Definition:Abundant Number]
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $\map A n$ denote the abundance of $n$.
\end{definition}
|
23680
|
\begin{definition}[Definition:Abundant Number/Definition 1]
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $A \left({n}\right)$ denote the abundance of $n$.
$n$ is '''abundant''' {{iff}} $A \left({n}\right) > 0$.
\end{definition}
|
23681
|
\begin{definition}[Definition:Abundant Number/Definition 2]
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $\map {\sigma_1} n$ be the divisor sum function of $n$.
$n$ is '''abundant''' {{iff}} $\dfrac {\map {\sigma_1} n} n > 2$.
\end{definition}
|
23682
|
\begin{definition}[Definition:Abundant Number/Definition 3]
Let $n \in \Z_{\ge 0}$ be a positive integer.
$n$ is '''abundant''' {{iff}} it is smaller than its aliquot sum.
\end{definition}
|
23683
|
\begin{definition}[Definition:Abuse of Notation]
Mathematical notation can be considered in two ways:
* As an aid to mathematical understanding, no more and no less than a useful convention to encapsulate more or less complicated ideas in a completely unambiguous format;
* As the reason for mathematical effort, so as to encapsulate a truth as a documented piece of aesthetic beauty in its own right.
'''Abuse of notation''' is a technique of using a system of symbology in a way different from that for which it was originally defined.
Such abuse may make a train of thought more streamlined, as it is often possible to save considerable redefinition of one's terms.
However, such abuse is frequently considered to be incorrect, improper and (in the eyes of many mathematicians) illegal.
Philosophers of the various schools practising pragmatism tend to consider that if a notation has been adequately explained, then one should be allowed to use it in whatever way is most useful to communicate one's ideas.
On the other hand, such an attitude causes indignation, rage and fury among philosophers whose attraction to mathematics is purely aesthetic.
\end{definition}
|
23684
|
\begin{definition}[Definition:Académie Parisienne]
The '''Académie Parisienne''' was an informal academic society established in $1635$ by the Franciscan friar {{AuthorRef|Marin Mersenne}}, sponsored and encouraged by {{WP|Cardinal_Richelieu|Cardinal Richelieu}}.
It consisted of some $140$ correspondents from around Europe, and formed a hub of mathematical interaction through which the leading minds of the time could communicate with each other.
{{AuthorRef|Marin Mersenne|Mersenne}}'s copious correspondence held it all together.
It was to form the nucleus of the later more formal {{WP|French_Academy_of_Sciences|Académie des Sciences}}, which was chartered in $1666$.
\end{definition}
|
23685
|
\begin{definition}[Definition:Acceleration]
The '''acceleration''' $\mathbf a$ of a body $M$ is defined as the first derivative of the velocity $\mathbf v$ of $M$ relative to a given point of reference {{WRT|Differentiation}} time $t$:
:$\mathbf a = \dfrac {\d \mathbf v} {\d t}$
Colloquially, it is described as the ''rate of change of velocity''.
It is important to note that as velocity is a vector quantity, then it follows by definition of derivative of a vector that so is acceleration.
\end{definition}
|
23686
|
\begin{definition}[Definition:Acceleration/Dimension]
'''Acceleration''' has dimension $\mathsf {L T}^{-2}$.
Category:Definitions/Dimensions of Measurement
\end{definition}
|
23687
|
\begin{definition}[Definition:Acceleration (Continuum Mechanics)]
In the context of continuum mechanics, the '''acceleration''' of a point $P$ in a body $B$ is the material derivative of the velocity of $P$ evaluated at $P$.
\end{definition}
|
23688
|
\begin{definition}[Definition:Accumulation Point]
Let $\struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
\end{definition}
|
23689
|
\begin{definition}[Definition:Accumulation Point/Sequence]
Let $\struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $A$.
Let $x \in S$.
Then $x \in S$ is an '''accumulation point''' of $\sequence {x_n}$ {{iff}}:
:$\forall U \in \tau: x \in U \implies \set {n \in \N: x_n \in U}$ is infinite
\end{definition}
|
23690
|
\begin{definition}[Definition:Accumulation Point/Set]
Let $\struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
Let $x \in S$.
Then $x$ is an '''accumulation point''' of $A$ {{iff}}:
:$x \in \map \cl {A \setminus \set x}$
where $\cl$ denotes the (topological) closure of a set.
\end{definition}
|
23691
|
\begin{definition}[Definition:Accuracy]
'''Accuracy''' is a measure of how closely a reported number matches the true value of that number.
\end{definition}
|
23692
|
\begin{definition}[Definition:Accuracy/Decimal Places]
A representation $y$ in decimal notation of a number $x$ is '''accurate to $n$ decimal places''' {{iff}} the number $n$ of digits after the decimal point of $x$ match the first $n$ digits after the decimal point of $y$, ''without'' rounding.
\end{definition}
|
23693
|
\begin{definition}[Definition:Ackermann-Péter Function]
The '''Ackermann-Péter function''' $A: \Z_{\ge 0} \times \Z_{\ge 0} \to \Z_{> 0}$ is an integer-valued function defined on the set of ordered pairs of positive integers as:
:$\map A {x, y} = \begin{cases} y + 1 & : x = 0 \\
\map A {x - 1, 1} & : x > 0, y = 0 \\
\map A {x - 1, \map A {x, y - 1} } & : \text{otherwise} \end{cases}$
\end{definition}
|
23694
|
\begin{definition}[Definition:Acnode]
An '''acnode''' is a point of the locus of an equation describing a curve which is not actually on that curve.
\end{definition}
|
23695
|
\begin{definition}[Definition:Acoustics]
'''Acoustics''' is the branch of physics which studies the properties of sound.
\end{definition}
|
23696
|
\begin{definition}[Definition:Action Applied by System]
The '''action applied by a system''' from state $1$ to state $2$ is defined as the definite integral of the Lagrangian over time from state $1$ to state $2$:
:$\ds S_{12} = \int_{t_1}^{t_2} \LL \rd t$
where:
:$S_{12}$ is the '''action''' from $1$ to $2$
:$t$ is time
:$\LL$ is the Lagrangian.
\end{definition}
|
23697
|
\begin{definition}[Definition:Action of Physical System]
Let $S$ be an integral functional.
Suppose that equations of motion of a physical system are to be derived from $S$ by the Principle of Stationary Action.
Then $S$ is the '''action''' of a '''physical system'''.
\end{definition}
|
23698
|
\begin{definition}[Definition:Acute Angle]
An '''acute angle''' is an angle which has a measure between that of a right angle and that of a zero angle.
{{EuclidDefinition|book = I|def = 12|name = Acute Angle}}
\end{definition}
|
23699
|
\begin{definition}[Definition:Acyclic Graph]
An '''acyclic graph''' is a graph or digraph with no cycles.
An acyclic connected undirected graph is a tree.
An acyclic disconnected undirected graph is a forest.
Category:Definitions/Graph Theory
\end{definition}
|
23700
|
\begin{definition}[Definition:Acyclic Object]
Let $\mathbf A$ be an abelian category with enough injectives.
Let $\mathbf B$ be an abelian category.
Let $F : \mathbf A \to \mathbf B$ be a left exact functor.
Let $X$ be an object of $\mathbf A$.
Then $X$ is '''$F$-acyclic''' {{iff}} $\mathrm R^i \map F X = 0$ for all positive integers $i \in \Z_{i \mathop \ge 1}$.
In the above $\mathrm R^i F$ denotes the $i$-th right derived functor of $F$.
Category:Definitions/Homological Algebra
\end{definition}
|
23701
|
\begin{definition}[Definition:Acyclic Resolution]
Let $\mathbf A$ be an abelian category with enough injectives.
Let $\mathbf B$ be an arbitrary abelian category.
Let $F : \mathbf A \to \mathbf B$ be a left-exact functor.
Let $X$ be an object in $\mathbf A$.
An $F$-'''acyclic resolution''' of $X$ is a cochain complex $P := \family {d^i : I^i \to I^{i + 1} }_{i \mathop \in \Z}$ in $\mathbf A$, such that:
:$(1): \quad \forall i < 0 : I^i = 0$
:$(2): \quad I^i$ is $F$-acyclic for all $i \ge 0$
together with a morphism $\varepsilon : X \to I^0$, such that the cochain complex:
:$\begin{xy}
\xymatrix{
\dots \ar[r] & 0 \ar[r] & 0 \ar[r] & X \ar[r]^{\varepsilon} & I^0 \ar[r]^{d^0} & I^1 \ar[r]^{d^1} & I^2 \ar[r]^{d^2} & \dots
}
\end{xy}$
is exact.
\end{definition}
|
23702
|
\begin{definition}[Definition:Acyclic Sheaf]
Let $X$ be a topological space.
Let $\FF$ be an abelian sheaf on $X$.
Let $\map \Gamma {X, -} : \map {\mathbf {Ab} } X \to \mathbf {Ab}$ be the global sections functor on $X$.
Then $\FF$ is '''acyclic''' {{iff}} $\FF$ is $\map \Gamma {X, -}$-acyclic in the category of abelian sheaves $\map {\mathbf {Ab} } X$ on $X$.
\end{definition}
|
23703
|
\begin{definition}[Definition:Adding Machine]
An '''adding machine''' is a device for performing simple arithmetic.
Usually such machines were large and barely portable, being used at a desk.
They also did not provide a printout, so the answer (read from dials) would need to be manually transcribed.
Such machines are generally obsolete nowadays.
\end{definition}
|
23704
|
\section{Normal Space is Regular Space}
Tags: T3 Spaces, Regular Spaces, Normal Spaces
\begin{theorem}
Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a regular space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a normal space.
From Normal Space is $T_3$ Space, we have that $T$ is a $T_3$ space.
We also have by definition of normal space that $T$ is a $T_1$ (Fréchet) space.
From $T_1$ Space is $T_0$ Space we have that $T$ is a $T_0$ (Kolmogorov) space
So $T$ is both a $T_3$ space and a $T_0$ (Kolmogorov) space.
Hence $T$ is a regular space by definition.
{{qed}}
\end{proof}
|
23705
|
\section{Normal Space is T3 Space}
Tags: T3 Spaces, Regular Spaces, Separation Axioms, Normal Spaces
\begin{theorem}
Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a $T_3$ space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a normal space.
From the definition of normal space:
:$\struct {S, \tau}$ is a $T_4$ space
:$\struct {S, \tau}$ is a Fréchet ($T_1$) space.
Let $F$ be any closed set in $T$.
Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$.
As $T$ is a Fréchet ($T_1$) space it follows from Equivalence of Definitions of $T_1$ Space that $\set y$ is closed.
As $T = \struct {S, \tau}$ is a normal space, we have that:
:$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$
That is, for any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.
But $F$ and $\set y$ are disjoint closed sets.
So:
:$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$
which is precisely the definition of a $T_3$ space.
{{qed}}
\end{proof}
|
23706
|
\section{Normal Space is Tychonoff Space}
Tags: Tychonoff Spaces, Separation Axioms, Normal Spaces
\begin{theorem}
Let $\struct {S, \tau}$ be a normal space.
Then $\struct {S, \tau}$ is also a Tychonoff (completely regular) space.
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a normal space.
From the definition of normal space:
:$\struct {S, \tau}$ is a $T_4$ space
:$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.
Let $F$ be any closed set in $T$.
Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$.
As $T$ is a $T_1$ (Fréchet) space it follows from Equivalence of Definitions of $T_1$ Space that $\set y$ is closed.
As $T = \struct {S, \tau}$ is a $T_4$ space, we have that for any two disjoint closed sets $A, B \subseteq S$ there exists an Urysohn function for $A$ and $B$.
But $F$ and $\set y$ are disjoint closed sets.
So there exists an Urysohn function for $F$ and $\set y$.
This is the definition of a $T_{3 \frac 1 2}$ space.
Next we note that as $T$ is a $T_1$ (Fréchet) space, from $T_1$ Space is $T_0$ Space it follows that $T$ is a $T_0$ (Kolmogorov) space.
So $T$ is both a $T_{3 \frac 1 2}$ space and $T_0$ (Kolmogorov) space.
So, by definition, $T$ is a Tychonoff space.
{{qed}}
\end{proof}
|
23707
|
\section{Normal Subgroup Test}
Tags: Normal Subgroups, Group Theory
\begin{theorem}
Let $G$ be a group and $H \le G$.
Then $H$ is a normal subgroup of $G$ {{iff}}:
:$\forall x \in G: x H x^{-1} \subseteq H$.
\end{theorem}
\begin{proof}
Let $H$ be a subgroup of $G$.
Suppose $H$ is normal in $G$.
Then $\forall x \in G, a \in H: \exists b \in H: x a = b x$.
Thus, $x a x^{-1} = b \in H$ implying $x H x^{-1} \subseteq H$.
Conversely, suppose $\forall x \in G: x H x^{-1} \subseteq H$.
Then for $g \in G$, we have $g H g^{-1} \subseteq H$, which implies $g H \subseteq H g$.
Also, for $g^{-1} \in G$, we have $g^{-1} H (g^{-1})^{-1} = g^{-1} H g \subseteq H$ which implies $H g \subseteq g H$.
Therefore, $g H = H g$ meaning $H \lhd G$.
{{qed}}
Category:Normal Subgroups
\end{proof}
|
23708
|
\section{Normal Subgroup induced by Congruence Relation defines that Congruence}
Tags: Normal Subgroups, Congruence Relations
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\RR$ be a congruence relation for $\circ$.
Let $\eqclass e \RR$ be the equivalence class of $e$ under $\RR$.
Let $N = \eqclass e \RR$ be the normal subgroup induced by $\RR$.
Then $\RR$ is the equivalence relation $\RR_N$ defined by $N$.
\end{theorem}
\begin{proof}
Let $\RR_N$ be the equivalence defined by $N$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \RR
| r = y
| c =
}}
{{eqn | ll= \leadsto
| l = e
| o = \RR
| r = \paren {x^{-1} \circ y}
| c = $\RR$ is compatible with $\circ$
}}
{{eqn | ll= \leadsto
| l = \paren {e \circ e}
| o = \RR
| r = \paren {x^{-1} \circ y}
| c = Group properties
}}
{{eqn | ll= \leadsto
| l = x^{-1} \circ y
| o = \in
| r = N
| c = Definition of $N$
}}
{{end-eqn}}
But from Congruence Class Modulo Subgroup is Coset:
:$x \mathrel {\RR_N} y \iff x^{-1} \circ y \in N$
Thus:
:$\RR = \RR_N$
{{Qed}}
\end{proof}
|
23709
|
\section{Normal Subgroup is Kernel of Group Homomorphism}
Tags: Normal Subgroups, Group Homomorphisms, Kernels of Group Homomorphisms
\begin{theorem}
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Then there exists a group homomorphism of which $N$ is the kernel.
\end{theorem}
\begin{proof}
Let $G / N$ be the quotient group of $G$ by $N$.
Let $q_N: G \to G / N$ be the quotient epimorphism from $G$ to $G / N$:
:$\forall x \in G: \map {q_N} x = x N$
Then from Quotient Group Epimorphism is Epimorphism, $N$ is the kernel of $q_n$
Thus $q_N$ is that group homomorphism of which $N$ is the kernel.
{{qed}}
\end{proof}
|
23710
|
\section{Normal Subgroup of Group of Order 24}
Tags: Groups of Order 24
\begin{theorem}
Let $G$ be a group of order $24$.
Then $G$ has either:
:a normal subgroup of order $8$
or:
:a normal subgroup of order $4$.
\end{theorem}
\begin{proof}
We note that:
:$24 = 3 \times 2^3$
Hence a Sylow $2$-subgroup of $G$ is of order $8$.
Let $n_2$ denote the number of Sylow $2$-subgroups of $G$.
By the Fourth Sylow Theorem:
:$n_2 \equiv 1 \pmod 2$ (that is, $n_2$ is odd
and from the Fifth Sylow Theorem:
:$n_2 \divides 24$
where $\divides$ denotes divisibility.
It follows that $n_3 \in \set {1, 3}$.
Suppose $n_2 = 1$.
Let $P$ denote the unique Sylow $2$-subgroup of $G$.
From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
This is therefore the required normal subgroup of order $8$.
{{qed|lemma}}
Suppose $n_2 = 3$.
Let $S_1$, $S_2$ and $S_3$ denote the $3$ Sylow $2$-subgroups of $G$.
Each has order $8$, as noted earlier.
Consider the product $S_1 S_2$.
By Order of Subgroup Product:
:$\order {S_1 S_2} = \dfrac {2^3 2^3} {2^r}$
where:
:$\order {S_1 S_2}$ denotes the order of $S_1 S_2$
:$\order {S_1 \cap S_2} = 2^r$
As $S_1 S_2 \subseteq G$ and $\order G = 24$:
:$\order {S_1 S_2} \le 24$
Hence:
{{begin-eqn}}
{{eqn | l = 2^3 2^3
| o = \le
| r = \order {S_1 S_2} \times 2^r
| c =
}}
{{eqn | ll= \leadsto
| l = 64
| o = \le
| r = 24 \times 2^r
| c =
}}
{{eqn | ll= \leadsto
| l = r
| o = \ge
| r = 2
| c =
}}
{{end-eqn}}
As $S_1 \cap S_2$ is a proper subgroup of $S_1$, it can have no more than $2^2 = 4$ elements.
So if $G$ has $3$ Sylow $2$-subgroups, the intersection of any $2$ of them is of order $4$.
Let $T = S_1 \cap S_2$, so that $\order T = 4$.
We have that:
:$\index T {S_1} = 2$
where $\index T {S_1}$ denotes the index of $T$ in $S_1$.
From Subgroup of Index 2 is Normal, $T$ is normal in $S_1$.
Similarly, $T$ is normal in $S_2$.
Thus $S_1$ and $S_2$ are both subgroups of the normalizer $\map {N_G} T$ of $T$ in $G$.
Thus $H = \gen {S_1, S_2}$ is a subgroup of $\map {N_G} T$.
Hence $T$ is a normal subgroup of $H$.
As $H$ is a subgroup of $\map {N_G} T$, it contains $S_1 S_2$ as a subset.
But $S_1 S_2$ contains $\dfrac {2^6} {2^2} = 16$ elements.
The only subgroup of $G$ containing $16$ elements has to be $G$ itself.
So $H = G$ and so $\map {N_G} T = G$.
That is, $T$ is a normal subgroup of $G$.
This is therefore the required normal subgroup of order $4$.
{{qed}}
\end{proof}
|
23711
|
\section{Normal Subgroup of Order 25 in Group of Order 100}
Tags: Groups of Order 100
\begin{theorem}
Let $G$ be a group of order $100$.
Then $G$ has a normal subgroup of order $25$.
\end{theorem}
\begin{proof}
Let $r$ be the number of Sylow $p$-subgroup of order $5^2 = 25$
The First Sylow Theorem guarantees existence, so $r \ge 1$.
From the Fourth Sylow Theorem:
:$r \equiv 1 \pmod 5$
That is:
:$r \in \set {1, 6, 11, 16, \ldots}$
From the Fifth Sylow Theorem:
:$r \divides \dfrac {100} {25} = 4$
from which it follows that $r = 1$.
From Sylow p-Subgroup is Unique iff Normal, this unique Sylow $5$-subgroup must be normal.
{{qed}}
\end{proof}
|
23712
|
\section{Normal Subgroup of Subset Product of Subgroups}
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:
: $N \lhd N H$
where:
: $\lhd$ denotes normal subgroup
: $N H$ denotes subset product.
\end{theorem}
\begin{proof}
From Subset Product with Normal Subgroup is Subgroup:
:$N H = H N$ is a subgroup of $G$.
By definition of subset product all elements of $H N$ can be written in the form:
:$h n \in H N$
where $h \in H, n \in N$.
Let $h n \in H N$.
Let $n_1 \in N$.
From Inverse of Group Product:
:$\paren {h n} n_1 \paren {h n}^{-1} = h n n_1 n^{-1} h^{-1}$
We have that:
:$n n_1 n \in N$
:$h, h^{-1} \in G$.
Then, since $N$ is a normal subgroup of $G$:
:$\paren {h n} n_1 \paren {n^{-1} h^{-1} } \in N$
Thus:
: $N \lhd N H$
{{qed}}
\end{proof}
|
23713
|
\section{Normal Subgroup of Symmetric Group on More than 4 Letters is Alternating Group}
Tags: Examples of Normal Subgroups, Alternating Groups, Symmetric Groups
\begin{theorem}
Let $n \in \N$ be a natural number such that $n > 4$.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $A_n$ denote the alternating group on $n$ letters.
$A_n$ is the only proper non-trivial normal subgroup of $S_n$.
\end{theorem}
\begin{proof}
From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is seen to be normal in $S_n$.
It remains to be shown that $A_n$ is the only such normal subgroup of $S_n$.
{{AimForCont}} $N$ is a proper non-trivial normal subgroup of $S_n$ such that $N$ is a proper subset of $A_n$.
From Intersection with Normal Subgroup is Normal, we have that:
:$A_n \cap N$ is a normal subgroup of $A_n$.
As $N \subseteq A_n$, from Intersection with Subset is Subset we have that:
:$A_n \cap N = N$
That is:
:$N$ is a normal subgroup of $A_n$.
But from Alternating Group is Simple except on 4 Letters, $A_n$ has no proper non-trivial normal subgroup.
So $N$ cannot be a normal subgroup of $A_n$.
Hence by Proof by Contradiction it follows that $N$ cannot be a normal subgroup of $S_n$.
So $S_n$ has no proper non-trivial normal subgroup apart from $A_n$.
{{qed}}
\end{proof}
|
23714
|
\section{Normal Subgroup of p-Group of Order p is Subset of Center}
Tags: Normal Subgroups, Centers of Groups, P-Groups
\begin{theorem}
Let $p$ be a prime number.
Let $G$ be a $p$-group.
Let $N$ be a normal subgroup of $G$ of order $p$.
Then:
:$N \subseteq \map Z G$
where $\map Z G$ denotes the center of $G$.
\end{theorem}
\begin{proof}
From Intersection of Normal Subgroup with Center in p-Group:
:$\order {N \cap \map Z G} > 1$
From Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $N$.
It follows from Lagrange's Theorem that:
:$\order {N \cap \map Z G} = p$
and so:
:$N \cap \map Z G = N$
But from Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $\map Z G$
That is:
:$N$ is a subgroup of $\map Z G$
and the result follows.
{{qed}}
\end{proof}
|
23715
|
\section{Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup}
Tags: Normal Subgroups
\begin{theorem}
Let $G$ be a group.
Let $H$ and $K$ be normal subgroups of $G$.
Let $H \subseteq K$.
Then $H$ is a normal subgroup of $K$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | q = \forall g \in G
| l = g K
| r = K g
| c = {{Defof|Normal Subgroup}}
}}
{{eqn | q = \forall g \in G
| l = g H
| r = H g
| c = {{Defof|Normal Subgroup}}
}}
{{eqn | ll= \leadsto
| q = \forall g \in K
| l = g H
| r = H g
| c = as $K \subseteq G$
}}
{{end-eqn}}
Hence the result by definition of normal subgroup.
{{qed}}
\end{proof}
|
23716
|
\section{Normal Sylow p-Subgroups in Group of Order 12}
Tags: Groups of Order 12, Sylow p-Subgroups
\begin{theorem}
Let $G$ be of order $12$.
Then $G$ has either:
:a normal Sylow $2$-subgroup
or:
:a normal Sylow $3$-subgroup.
\end{theorem}
\begin{proof}
Note that a Sylow $2$-subgroup of $G$ is of order $4$.
From Sylow $3$-Subgroups in Group of Order 12, there are either $1$ or $4$ Sylow $3$-subgroups.
Suppose there is exactly $1$ Sylow $3$-subgroup $P$.
Then from Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
{{qed|lemma}}
Suppose there are $4$ Sylow $3$-subgroups $P_1$, $P_2$, $P_3$ and $P_4$.
Each intersection $P_i \cap P_j$ for $i, j \in \set {1, 2, 3, 4}, i \ne j$ is the trivial subgroup of $G$:
:$P_i \cap P_j = \set e$
Thus $G$ contains:
:The identity element $e$
:$8$ elements of order $3$, of which $2$ each are in $P_1$, $P_2$, $P_3$ and $P_4$
:$3$ more elements, which (along with $e$) must form the Sylow $2$-subgroup of order $4$.
This Sylow $2$-subgroup $Q$ must be unique.
Hence by Sylow $p$-Subgroup is Unique iff Normal, $Q$ is normal.
{{qed|lemma}}
Hence the result.
{{qed}}
\end{proof}
|
23717
|
\section{Normal p-Subgroup contained in All Sylow p-Subgroups}
Tags: Sylow P-Subgroups, Sylow p-Subgroups, P-Groups
\begin{theorem}
Let $G$ be a finite group.
Let $p$ be a prime number.
Let $H$ be a normal subgroup of $G$ which is a $p$-group.
Then $H$ is a subset of every Sylow $p$-subgroup of $G$.
\end{theorem}
\begin{proof}
Let $P$ be a Sylow $p$-subgroup of $G$.
By Second Sylow Theorem, $H$ is a subset of a conjugate of $P$.
Then:
:$\exists g \in G: H \subseteq g P g^{-1}$.
This implies:
{{begin-eqn}}
{{eqn | l = H
| r = g^{-1} H g
| c = Subgroup equals Conjugate iff Normal
}}
{{eqn | o = \subseteq
| r = g^{-1} \paren {g P g^{-1} } g
| c = Subset Relation is Compatible with Subset Product/Corollary 2
}}
{{eqn | r = P
}}
{{end-eqn}}
Since $P$ is arbitrary, $H$ is a subset of every Sylow $p$-subgroup of $G$.
{{qed}}
\end{proof}
|
23718
|
\section{Normal to Circle passes through Center}
Tags: Circles, Normals to Curves
\begin{theorem}
A normal $\NN$ to a circle $\CC$ passes through the center of $\CC$.
\end{theorem}
\begin{proof}
Let $\CC$ be positioned in a Cartesian plane with its center at the origin.
Let $\NN$ pass through the point $\tuple {x_1, y_1}$.
From Equation of Normal to Circle Centered at Origin, $\NN$ has the equation:
:$y_1 x - x_1 y = 0$
or:
:$y = \dfrac {y_1} {x_1} x$
From the Equation of Straight Line in Plane: Slope-Intercept Form, this is the equation of a straight line passing through the origin.
As the geometry of a circle is unchanged by a change of coordinate axes, the result follows for a general circle in whatever frame.
{{qed}}
\end{proof}
|
23719
|
\section{Normal to Curve is Tangent to Evolute}
Tags: Evolutes
\begin{theorem}
Let $C$ be a curve defined by a real function which is twice differentiable.
Let the curvature of $C$ be non-constant.
Let $P$ be a point on $C$.
Let $Q$ be the center of curvature of $C$ at $P$.
The normal to $C$ at $P$ is tangent to the evolute $E$ of $C$ at $Q$.
\end{theorem}
\begin{proof}
:400px
Let $P = \tuple {x, y}$ be a general point on $C$.
Let $Q = \tuple {X, Y}$ be the center of curvature of $C$ at $P$.
From the above diagram:
:$(1): \quad \begin{cases} x - X = \pm \rho \sin \psi \\
Y - y = \pm \rho \cos \psi \end{cases}$
where:
:$\rho$ is the radius of curvature of $C$ at $P$
:$\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis.
Whether the sign is plus or minus depends on whether the curve is convex or concave.
For simplicity, let it be assumed that the curvature $k$ at each point under consideration on $C$ is positive.
The case for $k < 0$ can then be treated similarly.
Thus we have $k > 0$ and so $(1)$ can be written:
:$(2): \quad \begin{cases} X = x - \rho \sin \psi \\
Y = y +\rho \cos \psi \end{cases}$
By definition of curvature:
:$k = \dfrac {\d \psi} {\d s}$
and:
:$\rho = \dfrac 1 k = \dfrac {\d s} {\d \psi}$
Hence:
{{begin-eqn}}
{{eqn | l = \rho \sin \psi
| r = \dfrac {\d s} {\d \psi} \dfrac {\d y} {\d s}
| c =
}}
{{eqn | n = 3
| r = \dfrac {\d y} {\d \psi}
| c =
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \rho \cos \psi
| r = \dfrac {\d s} {\d \psi} \dfrac {\d x} {\d s}
| c =
}}
{{eqn | n = 4
| r = \dfrac {\d x} {\d \psi}
| c =
}}
{{end-eqn}}
Differentiating $(2)$ {{WRT|Differentiation}} $\psi$:
{{begin-eqn}}
{{eqn | l = \dfrac {\d X} {\d \psi}
| r = \dfrac {\d x} {\d \psi} - \rho \cos \psi - \dfrac {\d \rho} {\d \psi} \sin \psi
| c =
}}
{{eqn | r = - \dfrac {\d \rho} {\d \psi} \sin \psi
| c = from $(3)$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \dfrac {\d Y} {\d \psi}
| r = \dfrac {\d y} {\d \psi} - \rho \sin \psi + \dfrac {\d \rho} {\d \psi} \cos \psi
| c =
}}
{{eqn | r = \dfrac {\d \rho} {\d \psi} \cos \psi
| c = from $(4)$
}}
{{end-eqn}}
By assumption, $\dfrac {\d \rho} {\d \psi} \ne 0$ on $C$.
Hence we have:
{{begin-eqn}}
{{eqn | l = \dfrac {\d Y} {\d X}
| r = -\dfrac {\cos \psi} {\sin \psi}
| c =
}}
{{eqn | r = -\dfrac 1 {\tan \psi}
| c =
}}
{{end-eqn}}
We have Slope of Normal is Minus Reciprocal of Tangent.
Thus the slope of the tangent to $E$ equals the slope of the normal to $C$.
The result follows.
Note the case when $\dfrac {\d \rho} {\d \psi} = 0$ on $C$.
In this case $C$ is a circle.
By Evolute of Circle is its Center its evolute is a single point and so has no tangent.
{{qed}}
\end{proof}
|
23720
|
\section{Normality Relation is not Transitive}
Tags: Normal Subgroups, Normality Relation is not Transitive
\begin{theorem}
Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $K$ be a normal subgroup of $N$.
Then it is not necessarily the case that $K$ is a normal subgroup of $G$.
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $A_4$ denote the alternating group on $4$ letters, whose Cayley table is given as:
{{:Alternating Group on 4 Letters/Cayley Table}}
From Alternating Group on 4 Letters: Normality of Subgroups:
:$K := \set {e, t, u, v}$ is a normal subgroup of $A_4$
:$T := \set {e, t}$ is not a normal subgroup of $A_4$.
But by Subgroup of Abelian Group is Normal:
:$T$ is a normal subgroup of $K$.
Thus we have:
:$T \lhd K$, $K \lhd A_4$
but:
:$T \not \lhd A_4$
{{qed}}
\end{proof}
|
23721
|
\section{Normalized URM Program}
Tags: URM Programs
\begin{theorem}
Let $P$ be a URM program.
Let $l = \map \lambda P$ be the number of basic instructions in $P$.
Let $u = \map \rho P$ be the number of registers used by $P$.
Then $P$ can be modified as follows:
:Every <tt>Jump</tt> of the form $\map J {m, n, q}$ where $q > l$ may be replaced by $\map J {m, n, l + 1}$
:If $u > 0$, a Clear Registers Program $\map Z {2, u}$ can be appended to the end of $P$ at lines $l + 1$ to $l + u - 1$.
The new program $P'$ that results from the above modifications produces exactly the same output as $P$ for each input.
Note now though that $\map \lambda {P'} = l + u - 1$.
Such a program as $P'$ is called a '''normalized URM program'''.
The point of doing this is so as to make programs easier to concatenate. Once the above have been done, each program has a well-defined exit line which can be used as the start line of the program that immediately follows it.
\end{theorem}
\begin{proof}
Each <tt>Jump</tt> of the form $\map J {m, n, q}$ where $q > l$ leads to a line which does not contain an instruction.
The line $\map J {m, n, l + 1}$ likewise contains no instructions, by definition.
Therefore, when jumping to $\map J {m, n, l + 1}$ the program behaves in exactly the same way: that is, it stops when the instruction $\map J {m, n, l + 1}$ causes the program to jump to line $l + 1$.
After the URM program has terminated, its output sits in $R_1$ by convention.
Once the URM program reaches line $l + 1$ it has by definition terminated, and because of the modifications to the <tt>Jump</tt>s as defined above, there is no other way that it ''can'' terminate.
Any further instructions that are added to a URM program that are placed at line $l + 1$ and those following will therefore be executed in order (as long as none of them are <tt>Jump</tt>s) and the program will then ''really'' terminate.
By adding the Clear Registers Program $\map Z {2, u}$ to the end of $P$, the only effect this will have on the operation of the program is to clear all the registers to $0$ except the one the output is in.
{{qed}}
Category:URM Programs
\end{proof}
|
23722
|
\section{Normalizer is Subgroup}
Tags: Normal Subgroups, Normalizers
\begin{theorem}
Let $G$ be a group.
The normalizer of a subset $S \subseteq G$ is a subgroup of $G$.
:$S \subseteq G \implies \map {N_G} S \le G$
\end{theorem}
\begin{proof}
Let $a, b \in \map {N_G} S$.
Then:
{{begin-eqn}}
{{eqn | l = S^{a b}
| r = \paren {S^b}^a
| c = Conjugate of Set by Group Product
}}
{{eqn | r = S^a
| c = {{Defof|Normal Subgroup}}
}}
{{eqn | r = S
| c = {{Defof|Normal Subgroup}}
}}
{{end-eqn}}
Therefore $a b \in \map {N_G} S$.
Now let $a \in \map {N_G} S$:
:$a \in \map {N_G} S \implies S^{a^{-1} } = \paren {S^a}^{a^{-1} } = S^{a^{-1} a} = S$
Therefore $a^{-1} \in \map {N_G} S$.
Thus, by the Two-Step Subgroup Test, $\map {N_G} S \le G$.
{{qed}}
\end{proof}
|
23723
|
\section{Normalizer of Center is Group}
Tags: Normal Subgroups, Normalizers, Centers of Groups
\begin{theorem}
Let $G$ be a group.
Let $\map Z G$ be the center of $G$.
Let $x \in G$.
Let $\map {N_G} x$ be the normalizer of $x$ in $G$.
Then:
:$\map Z G = \set {x \in G: \map {N_G} x = G}$
That is, the center of a group $G$ is the set of elements $x$ of $G$ such that the normalizer of $x$ is the whole of $G$.
Thus:
:$x \in \map Z G \iff \map {N_G} x = G$
and so:
:$\index G {\map {N_G} x} = 1$
where $\index G {\map {N_G} x}$ is the index of $\map {N_G} x$ in $G$.
\end{theorem}
\begin{proof}
$\map {N_G} x$ is the normalizer of the set $\set x$.
Thus:
{{begin-eqn}}
{{eqn | l = \map {N_G} x
| r = G
| c =
}}
{{eqn | ll= \leadstoandfrom
| q = \forall a \in G
| l = \set x^a
| r = \set x
| c = {{Defof|Normalizer}}
}}
{{eqn | ll= \leadstoandfrom
| q = \forall a \in G
| l = a x a^{-1}
| r = x
| c = {{Defof|Conjugate of Group Element}}
}}
{{eqn | ll= \leadstoandfrom
| q = \forall a \in G
| l = a x
| r = x a
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x
| o = \in
| r = \map Z G
| c = {{Defof|Center of Group}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23724
|
\section{Normalizer of Conjugate is Conjugate of Normalizer}
Tags: Conjugacy, Normal Subgroups, Normalizers
\begin{theorem}
Let $G$ be a group.
Let $a \mathop \in G$.
Let $S$ be a subset of $G$.
Let $S^a$ denote the conjugate of $S$ by $a$.
Let $\map {N_G} S$ denote the normalizer of $S$ in $G$.
Then:
:$\map {N_G} {S^a} = \paren {\map {N_G} S}^a$
That is, the normalizer of a conjugate is the conjugate of the normalizer:
\end{theorem}
\begin{proof}
From the definition of conjugate:
:$S^a = \set {y \in G: \exists x \in S: y = a x a^{-1} } = a S a^{-1}$
From the definition of normalizer:
:$\map {N_G} S = \set {x \in G: S^x = S}$
Thus:
{{begin-eqn}}
{{eqn | l = \map {N_G} {S^a}
| r = \set {x \in G: \paren {S^a}^x = S^a}
| c =
}}
{{eqn | r = \set {x \in G: x a S a^{-1} x^{-1} = a S a^{-1} }
| c =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {\map {N_G} S}^a
| r = \set {x \in G: S^x = S}^a
}}
{{eqn | r = \set {x \in G: x S x^{-1} = S}^a
}}
{{eqn | r = \set {y \in G: \exists z \in \set {x \in G: x S x^{-1} = S}: y = a z a^{-1} }
}}
{{end-eqn}}
Suppose that $x \in \map {N_G} S$.
It is to be shown that:
:$a x a^{-1} \in \map {N_G} {S^a} = \map {N_G} {a S a^{-1} }$
To this end, compute:
{{begin-eqn}}
{{eqn | l = \paren {a x a^{-1} } a S a^{-1} \paren {a x a^{-1} }^{-1}
| r = \paren {a x a^{-1} } a S a^{-1} \paren {a x^{-1} a^{-1} }
| c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = a x S x^{-1} a^{-1}
}}
{{eqn | r = a S a^{-1}
| c = as $x \in \map {N_G} S$
}}
{{end-eqn}}
Hence $a x a^{-1} \in \map {N_G} {S^a}$, and it follows that:
:$z \in \paren {\map {N_G} S}^a \implies z \in \map {N_G} {S^a}$
Conversely, let $x \in \map {N_G} {S^a}$.
That is, let $x \in G$ such that $x a S a^{-1} x^{-1} = a S a^{-1}$.
Now if we can show that $a^{-1} x a \in \map {N_G} S$, then:
:$x = a \left({a^{-1} x a}\right) a^{-1} \in \paren {\map {N_G} S}^a$
establishing the remaining inclusion.
Thus, we compute:
{{begin-eqn}}
{{eqn | l = \paren {a^{-1} x a} S \paren {a^{-1} x a}^{-1}
| r = a^{-1} x a S a^{-1} x^{-1} a
| c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = a^{-1} a S a^{-1} a
| c = as $x \in \map {N_G} {S^a}$
}}
{{eqn | r = S
}}
{{end-eqn}}
Combined with the above observation, this establishes that:
:$z \in \map {N_G} {S^a} \implies z \in \paren {\map {N_G} S}^a$
Hence the result, by definition of set equality.
{{qed}}
\end{proof}
|
23725
|
\section{Normalizer of Reflection in Dihedral Group}
Tags: Normalizers, Dihedral Groups
\begin{theorem}
Let $n \in \N$ be a natural number such that $n \ge 3$.
Let $D_n$ be the dihedral group of order $2 n$, given by:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
Let $\map {N_{D_n} } {\set \beta}$ denote the normalizer of the singleton containing the reflection element $\beta$.
Then:
:$\map {N_{D_n} } {\set \beta} = \begin{cases} \set {e, \beta} & : n \text { odd} \\ \set {e, \beta, \alpha^{n / 2}, \alpha^{n / 2} \beta} & : n \text { even} \end{cases}$
\end{theorem}
\begin{proof}
By definition, the normalizer of $\set \beta$ is:
:$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \set \beta g^{-1} = \set \beta}$
That is:
:$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \beta = \beta g}$
First let $g = \beta^k$ for $k \in \set {0, 1}$.
Then:
:$\beta \beta^k = \beta^k \beta$
Thus:
:$\forall k \in \set {0, 1}: \beta^k \in \map {N_{D_n} } {\set \beta}$
Now let $g = \alpha^j \beta^k$ for $0 < j < n, k \in \set {0, 1}$.
Suppose $g \alpha = \alpha g$.
Then:
{{begin-eqn}}
{{eqn | l = \alpha^j \beta^k \beta
| r = \beta \alpha^j \beta^k
| c =
}}
{{eqn | ll= \leadsto
| l = \alpha^j \beta \beta^k
| r = \beta \alpha^j \beta^k
| c =
}}
{{eqn | ll= \leadsto
| l = \alpha^j \beta
| r = \beta \alpha^j
| c = Cancellation Laws
}}
{{eqn | ll= \leadsto
| l = \alpha^j \beta
| r = \alpha^{n - j} \beta
| c = Product of Generating Elements of Dihedral Group
}}
{{eqn | ll= \leadsto
| l = \alpha^j
| r = \alpha^{n - j}
| c =
}}
{{eqn | ll= \leadsto
| l = \alpha^2 j
| r = \alpha^n
| c =
}}
{{eqn | ll= \leadsto
| l = 2 j
| r = n
| c =
}}
{{end-eqn}}
Thus when $n$ is odd, there is no such $j$ such that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$.
But when $n$ is even such that $n = 2 j$, we have that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$.
Hence the result.
{{qed}}
Category:Dihedral Groups
Category:Normalizers
\end{proof}
|
23726
|
\section{Normalizer of Rotation in Dihedral Group}
Tags: Normalizers, Dihedral Groups
\begin{theorem}
Let $n \in \N$ be a natural number such that $n \ge 3$.
Let $D_n$ be the dihedral group of order $2 n$, given by:
:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
Let $\map {N_{D_n} } {\set \alpha}$ denote the normalizer of the singleton containing the rotation element $\alpha$.
Then:
:$\map {N_{D_n} } {\set \alpha} = \gen \alpha$
where $\gen \alpha$ is the subgroup generated by $\alpha$.
\end{theorem}
\begin{proof}
By definition, the normalizer of $\set \alpha$ is:
:$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \set \alpha g^{-1} = \set \alpha}$
That is:
:$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \alpha = \alpha g}$
First let $g = \alpha^k$ for some $k \in \Z$.
Then:
:$\alpha \alpha^k = \alpha^k \alpha$
which includes $k = 0$, that is $e$.
Thus:
:$\forall k \in \Z: \alpha^k \in \map {N_{D_n} } {\set \alpha}$
Now let $g = \alpha^k \beta$.
Suppose $g \alpha = \alpha g$.
Then:
{{begin-eqn}}
{{eqn | l = \alpha^k \beta \alpha
| r = \alpha \alpha^k \beta
| c =
}}
{{eqn | ll= \leadsto
| l = \beta \alpha
| r = \alpha \beta
| c = Cancellation Laws
}}
{{end-eqn}}
But $\beta \alpha \ne \alpha \beta$ in $D_n$ except for $n < 3$.
Hence the result:
:$\map {N_{D_n} } {\set \alpha} = \set {e, \alpha, \alpha^2, \ldots, \alpha^{n - 1} }$
Hence the result, by definition of generator of subgroup.
{{qed}}
Category:Dihedral Groups
Category:Normalizers
\end{proof}
|
23727
|
\section{Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup}
Tags: Normal Subgroups, Normalizers
\begin{theorem}
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then $\map {N_G} H$, the normalizer of $H$ in $G$, is the largest subgroup of $G$ containing $H$ as a normal subgroup.
\end{theorem}
\begin{proof}
From Subgroup is Subgroup of Normalizer, we have that $H \le \map {N_G} H$.
Now we need to show that $H \lhd \map {N_G} H$.
For $a \in \map {N_G} H$, the conjugate of $H$ by $a$ in $\map {N_G} H$ is:
{{begin-eqn}}
{{eqn | l = H^a
| r = \set {x \in \map {N_G} H: a x a^{-1} \in H}
| c = {{Defof|Conjugate of Group Subset}}
}}
{{eqn | r = H^a \cap \map {N_G} H
| c = {{Defof|Set Intersection}}
}}
{{eqn | r = H \cap \map {N_G} H
| c = {{Defof|Normalizer}}
}}
{{eqn | r = H
| c = Intersection with Subset is Subset
}}
{{end-eqn}}
so:
:$\forall a \in \map {N_G} H: H^a = H$
and so by definition of normal subgroup:
:$H \lhd \map {N_G} H$
Now we need to show that $\map {N_G} H$ is the largest subgroup of $G$ containing $H$ such that $H \lhd \map {N_G} H$.
That is, to show that any subgroup of $G$ in which $H$ is normal is also a subset of $\map {N_G} H$.
Take any $N$ such that $H \lhd N \le G$.
In $N$, the conjugate of $H$ by $a \in N$ is $N \cap H^a = H$.
Therefore:
:$H \subseteq H^a$
Similarly, $H \subseteq H^{a^{-1} }$, so:
:$H^a \subseteq \paren {H^a}^{a^{-1} } = H$
Thus:
:$\forall a \in N: H^a = H, a \in \map {N_G} H$
That is:
:$N \subseteq \map {N_G} H$
So what we have shown is that any subgroup of $G$ in which $H$ is normal is a subset of $\map {N_G} H$, which is another way of saying that $\map {N_G} H$ is the largest such subgroup.
{{Qed}}
\end{proof}
|
23728
|
\section{Normalizer of Subgroup of Symmetric Group that Fixes n}
Tags: Normalizers, 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:
:$\map \pi n = n$
The normalizer of $H$ is given by:
:$\map {N_{S_n} } H = \map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$
\end{theorem}
\begin{proof}
We have from Subgroup of Symmetric Group that Fixes n that $N = S_{n - 1}$.
By definition of normalizer:
:$\map {N_{S_n} } {S_{n - 1} } := \set {\rho \in S_n: \rho S_{n - 1} \rho^{-1} = S_{n - 1} }$
We have from Group is Normal in Itself that:
:$\forall \rho \in S_{n - 1}: \rho S_{n - 1} \rho^{-1} \in S_{n - 1}$
and so:
:$S_{n - 1} \subseteq \map {N_{S_n} } {S_{n - 1} }$
It remains to be shown that $\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$.
This will be done by demonstrating that:
: $S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$
where $\setminus$ denotes set difference.
Let $\rho \in S_n$ such that $\rho \notin S_{n - 1}$.
Thus:
:$\map \rho n \ne n$
and so:
:$\exists a \in S_n, a \ne n: \map \rho a = n$
for some $a \in S_{n - 1}$.
Then:
:$\map {\rho^{-1} } n = a$
Let $\pi \in S_{n - 1}$ such that:
: $\map \pi a = b$
for some $b \ne a$.
As $\rho$ is a permutation, $\rho$ is by definition a bijection.
Hence:
:$\map \rho b \ne n$
We have:
{{begin-eqn}}
{{eqn | l = \map {\rho \pi \rho^{-1} } n
| r = \map {\rho \pi} a
| c = as $\map {\rho^{-1} } n = a$
}}
{{eqn | r = \map \rho b
| c = as $\map \pi a = b$
}}
{{eqn | o = \ne
| r = n
| c = as $\map \rho b \ne n$
}}
{{end-eqn}}
Thus $\rho \pi \rho^{-1}$ does not fix $n$.
That is:
:$\rho \pi \rho^{-1} \notin S_{n - 1}$
Since this is the case for all arbitrary $\rho \in S_n \setminus S_{n - 1}$, it follows that:
:$S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$
So from Intersection with Complement is Empty iff Subset:
:$\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$
and so by definition of set equality:
:$\map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$
{{qed}}
\end{proof}
|
23729
|
\section{Normalizer of Sylow p-Subgroup}
Tags: P-Groups, Subgroups, Normalizers, Group Theory, Sylow p-Subgroups
\begin{theorem}
Let $P$ be a Sylow $p$-subgroup of a finite group $G$.
Let $\map {N_G} P$ be the normalizer of $P$.
Then any $p$-subgroup of $\map {N_G} P$ is contained in $P$.
In particular, $P$ is the unique Sylow $p$-subgroup of $\map {N_G} P$.
\end{theorem}
\begin{proof}
Let $Q$ be a $p$-subgroup of $N = \map {N_G} P$.
Let $\order Q = p^m, \order P = p^n$.
By Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:
:$P \lhd \map {N_G} P$
thus by Subset Product with Normal Subgroup as Generator:
:$\gen {P, Q} = P Q$
Thus by Order of Subgroup Product:
:$P Q \le G: \order {P Q} = p^{n + m - s}$
where $\order {P \cap Q} = p^s$.
Since $n$ is the highest power of $p$ dividing $\order G$, this is possible only when $m \le s$.
Since $P \cap Q \le Q, s \le m$ thus we conclude that $m = s$ and therefore $P \cap Q = Q$.
Thus from Intersection with Subset is Subset:
:$Q \subseteq P$
In particular, if $Q$ is a Sylow $p$-subgroup of $\map {N_G} P$, then $Q = P$.
{{qed}}
\end{proof}
|
23730
|
\section{Normed Division Algebra is Division Algebra}
Tags: Algebras
\begin{theorem}
Let $A = \struct {A_F, \oplus}$ be a normed divison algebra over a field $F$.
Let the unit of $A$ be $1_A$, and the zero of $A$ be $0_A$.
Then $A$ is a unitary division algebra.
Also:
:$\norm {1_A} = 1$
where $\norm {1_A}$ denotes the norm of $1_A$.
\end{theorem}
\begin{proof}
Let $A = \struct {A_F, \oplus}$ be a normed divison algebra as defined in the hypothesis.
The fact that $A$ is a unitary algebra is a consequence of the definition of normed divison algebra.
From the definition of a norm, we have that:
:$\forall a \in A: \norm a = 0 \iff a = 0_A$
So, let $a, b \in A \setminus \set {0_A}$.
We have:
{{begin-eqn}}
{{eqn | l = \norm a \norm b
| o = \ne
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = \norm {a \oplus b}
| o = \ne
| r = 0
| c = {{Defof|Normed Division Algebra}}: $\norm {a \oplus b} = \norm a \norm b$
}}
{{eqn | ll= \leadsto
| l = a \oplus b
| o = \ne
| r = 0_A
| c = {{Defof|Norm on Vector Space}}
}}
{{end-eqn}}
Thus for any arbitrary $a, b \ne 0_A$ we have shown that $a \oplus b \ne 0_A$.
Thus $A$ is a division algebra.
{{qed|lemma}}
Next:
{{begin-eqn}}
{{eqn | o =
| r = \norm a \norm {1_A}
| c =
}}
{{eqn | r = \norm {a \oplus 1_A}
| c = {{Defof|Normed Division Algebra}}: $\norm {a \oplus b} = \norm a \norm b$
}}
{{eqn | r = \norm a
| c = {{Defof|Norm on Vector Space}}
}}
{{end-eqn}}
demonstrating that $\norm {1_A} = 1$.
{{qed}}
\end{proof}
|
23731
|
\section{Normed Division Ring Completions are Isometric and Isomorphic}
Tags: Complete Metric Spaces, Definitions: Norm Theory, Normed Division Rings, Completion of Normed Division Ring, Definitions: Division Rings, Definitions: Complete Metric Spaces
\begin{theorem}
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be normed division ring completions of $\struct {R, \norm {\, \cdot \,} }$
Then there exists an isometric isomorphism $\psi: \struct {S_1, \norm {\, \cdot \,}_1 } \to \struct {S_2, \norm {\, \cdot \,}_2 }$
\end{theorem}
\begin{proof}
By the definition of a normed division ring completion then:
:there exists a distance-preserving ring monomorphisms $\phi_1: R \to S_1$
:$R_1 = \map {\phi_1} R$ is a dense subring of $S_1$
:$S_1$ is a complete metric space
:there exists a distance-preserving ring monomorphisms $\phi_2: R \to S_2$
:$R_2 = \map {\phi_2} R$ is a dense subring of $S_2$
:$S_2$ is a complete metric space
\end{proof}
|
23732
|
\section{Normed Division Ring Completions are Isometric and Isomorphic/Lemma 1}
Tags: Complete Metric Spaces, Definitions: Norm Theory, Normed Division Rings, Completion of Normed Division Ring, Definitions: Division Rings, Definitions: Complete Metric Spaces
\begin{theorem}
Let $\struct {R, \norm {\, \cdot \,} }, \struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be normed division rings.
Let $\phi_1: R \to S_1, \phi_2: R \to S_2$ be distance-preserving ring monomorphisms.
Let $\psi' = \phi_2 \circ \phi_1^{-1}:\phi_1 \paren R \to \phi_2 \paren R$ be the composition of $\phi_1^{-1}$ with $\phi_2$.
Then $\psi': \struct {\map {\phi_1} R, \norm {\, \cdot \,}_1 } \to \struct {\map {\phi_2} R, \norm {\, \cdot \,}_2 }$ is an isometric ring isomorphism.
\end{theorem}
\begin{proof}
By Monomorphism Image is Isomorphic to Domain, $\phi_1:R \to \map {\phi_1} R$ and $\phi_2:R \to \map {\phi_2} R$ are ring isomorphisms.
By Distance-Preserving Image Isometric to Domain for Metric Spaces, $\phi_1:R \to \map {\phi_1} R$ and $\phi_2:R \to \map {\phi_2} R$ are isometries.
By Inverse of Algebraic Structure Isomorphism is Isomorphism, $\phi_1^{-1}:\map {\phi_1} R \to R$ is a ring isomorphism.
By Inverse of Isometry of Metric Spaces is Isometry, $\phi_1^{-1}:\map {\phi_1} R \to R$ is a ring isomorphism is an isometry.
By Composition of Ring Isomorphisms is Ring Isomorphism, $\psi'$ is a ring isomorphism.
By Composition of Isometries is Isometry, $\psi'$ is an isometry.
{{qed}}
Category:Completion of Normed Division Ring
\end{proof}
|
23733
|
\section{Normed Division Ring Completions are Isometric and Isomorphic/Lemma 2}
Tags: Complete Metric Spaces, Definitions: Norm Theory, Normed Division Rings, Completion of Normed Division Ring, Definitions: Division Rings, Definitions: Complete Metric Spaces
\begin{theorem}
Let $\struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be complete normed division rings.
Let $R_1$ be a dense subring of $S_1$.
Let $R_2$ be a dense subring of $S_2$.
Let $\psi': R_1 \to R_2$ be an isometric ring isomorphism.
Let $\psi: S_1 \to S_2$ be defined by:
:$\forall x \in S_1: \map \psi x = \ds \lim_{n \mathop \to \infty} \map {\psi'} {x_n}$
where $\ds x = \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$
Then $\psi$ is a well-defined mapping.
\end{theorem}
\begin{proof}
Let $x \in S_1$.
By the definition of dense subset:
:$\map \cl {R_1} = S_1$
By Closure of Subset of Metric Space by Convergent Sequence, there exists a sequence $\sequence {x_n} \subseteq R_1 $ that converges to $x$, that is:
:$\ds \lim_{n \mathop \to \infty} x_n = x$
By Isometric Image of Cauchy Sequence is Cauchy Sequence, $\sequence {\map {\psi'} {x_n} }$ is a Cauchy sequence in $R_2 \subseteq S_2$.
Since $S_2$ is complete then the sequence $\sequence {\map {\psi'} {x_n} }$ has a limit, say $y$.
Let $\sequence {x_n}$ and $\sequence {x'_n}$ be sequences in $\map {\phi_1} R$ such that:
:$\ds \lim_{n \mathop \to \infty} x_n = \lim_{n \mathop \to \infty} x'_n = x$
Then:
{{begin-eqn}}
{{eqn | ll=
| l = \lim_{n \mathop \to \infty} x_n - x'_n
| r = \paren {\lim_{n \mathop \to \infty} x_n} - \paren {\lim_{n \mathop \to \infty} x'_n}
| c = Difference Rule for Sequences in Normed Division Ring
}}
{{eqn | ll=
| r = x - x
}}
{{eqn | ll=
| r = 0_{S_1}
}}
{{eqn | o =
}}
{{eqn | ll= \leadsto
| r = \lim_{n \mathop \to \infty} \norm {x_n - x'_n}_1
| l = 0
| c = {{Defof|Convergent Sequence in Normed Division Ring}}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \norm {\map {\psi'} {x_n} - \map {\psi'} {x'_n} }_2
| c = {{Defof|Isometric Metric Spaces}}
}}
{{eqn | o =
}}
{{eqn | ll= \leadsto
| r = \lim_{n \mathop \to \infty} \map {\psi'} {x_n} - \map {\psi'} {x'_n}
| l = 0_{S_2}
| c = {{Defof|Convergent Sequence in Normed Division Ring}}
}}
{{eqn | r = \paren {\lim_{n \mathop \to \infty} \map {\psi'} {x_n} } - \paren {\lim_{n \mathop \to \infty} \map {\psi'} {x'_n} }
| c = Difference Rule for Sequences in Normed Division Ring
}}
{{eqn | o =
}}
{{eqn | ll= \leadsto
| l = \lim_{n \mathop \to \infty} \map {\psi'} {x_n}
| r = \lim_{n \mathop \to \infty} \map {\psi'} {x'_n}
}}
{{end-eqn}}
The result follows.
{{qed}}
Category:Completion of Normed Division Ring
\end{proof}
|
23734
|
\section{Normed Division Ring Completions are Isometric and Isomorphic/Lemma 3}
Tags: Complete Metric Spaces, Definitions: Norm Theory, Normed Division Rings, Completion of Normed Division Ring, Definitions: Division Rings, Definitions: Complete Metric Spaces
\begin{theorem}
Let $\struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be complete normed division rings.
Let $R_1$ be a dense subring of $S_1$.
Let $R_2$ be a dense subring of $S_2$.
Let $\psi': R_1 \to R_2$ be an isometric ring isomorphism.
Let $\psi: S_1 \to S_2$ be defined by:
:$\forall x \in S_1: \map \psi x = \ds \lim_{n \mathop \to \infty} \map {\psi'} {x_n}$
where $x = \ds \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$.
Then:
:$\psi$ is a surjective mapping.
\end{theorem}
\begin{proof}
Let $y \in S_2$.
By the definition of dense subset:
:$\map \cl {R_2} = S_2$
By Closure of Subset of Metric Space by Convergent Sequence:
:there exists a sequence $\sequence {y_n} \subseteq R_2 $ that converges to $y$, that is, $\ds \lim_{n \mathop \to \infty} y_n = y$
By Isometric Image of Cauchy Sequence is Cauchy Sequence, $\sequence {\map {\psi'^{-1} } {y_n} }$ is a Cauchy sequence in $R_1 \subseteq S_1$.
Because $S_1$ is complete, the sequence $\sequence {\map {\psi’^{-1} } {y_n} }$ has a limit, say $x$.
By the definition of $\psi$:
{{begin-eqn}}
{{eqn | l = \map \psi x
| r = \lim_{n \mathop \to \infty} \map {\psi'} {\map {\psi'^{-1} } {y_n} }
}}
{{eqn | r = \lim_{n \mathop \to \infty} y_n
}}
{{eqn | r = y
}}
{{end-eqn}}
{{qed}}
Category:Completion of Normed Division Ring
\end{proof}
|
23735
|
\section{Normed Division Ring Completions are Isometric and Isomorphic/Lemma 4}
Tags: Complete Metric Spaces, Definitions: Norm Theory, Normed Division Rings, Completion of Normed Division Ring, Definitions: Division Rings, Definitions: Complete Metric Spaces
\begin{theorem}
Let $\struct {S_1, \norm {\, \cdot \,}_1 }$ and $\struct {S_2, \norm {\, \cdot \,}_2 }$ be complete normed division rings.
Let $R_1$ be a dense subring of $S_1$.
Let $R_2$ be a dense subring of $S_2$.
Let $\psi': R_1 \to R_2$ be an isometric ring isomorphism.
Let $\psi:S_1 \to S_2$ be defined by:
:$\forall x \in S_1: \map \psi x = \ds \lim_{n \mathop \to \infty} \map {\psi'}{x_n}$
:where $x = \ds \lim_{n \mathop \to \infty} x_n$ for some sequence $\sequence {x_n} \subseteq R_1$
Then:
:$\psi$ is an isometry.
\end{theorem}
\begin{proof}
Let $x, y \in S_1$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $R_1$ such that $\ds \lim_{n \mathop \to \infty} x_n = x, \lim_{n \mathop \to \infty} y_n = y$.
Then:
{{begin-eqn}}
{{eqn | l = x - y
| r = \lim_{n \mathop \to \infty} x_n - y_n
| c = Difference Rule for Sequences in Normed Division Ring
}}
{{eqn | o =
}}
{{eqn | ll= \leadsto
| l = \norm {x - y}
| r = \lim_{n \mathop \to \infty} \norm {x_n - y_n}
| c = Modulus of Limit
}}
{{eqn | r = \lim_{n \mathop \to \infty} \norm {\map {\psi'} {x_n} - \map {\psi'} {y_n} }
| c = {{Defof|Isometry (Metric Spaces)}}
}}
{{end-eqn}}
On the other hand:
{{begin-eqn}}
{{eqn | l = \map \psi x - \map \psi y
| r = \paren {\lim_{n \mathop \to \infty} \map {\psi'} {x_n} } - \paren {\lim_{n \mathop \to \infty} \map {\psi'} {y_n} }
| c = Definition of $\psi$
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map {\psi'} {x_n} - \map {\psi'} {y_n}
| c = Difference Rule for Sequences in Normed Division Ring
}}
{{eqn | o =
}}
{{eqn | ll= \leadsto
| l = \norm {\map \psi x - \map \psi y}
| r = \lim_{n \mathop \to \infty} \norm {\map {\psi'} {x_n} - \map {\psi'} {y_n} }
| c = Modulus of Limit
}}
{{end-eqn}}
By Convergent Sequence in Metric Space has Unique Limit:
:$\norm {x - y} = \norm {\map \psi x - \map \psi y}$
It follows that $\psi$ is distance-preserving.
By Distance-Preserving Surjection is Isometry of Metric Spaces then $\psi$ is an isometry.
{{qed}}
Category:Completion of Normed Division Ring
\end{proof}
|
23736
|
\section{Non-Finite Cardinal is equal to Cardinal Product/Corollary}
Tags: Cardinals
\begin{theorem}
Let $S$ be a set that is equinumerous to its cardinal number.
Let $\left|{ S }\right|$ denote the cardinal number of $S$.
Suppose that $\left|{ S }\right| \ge \omega$, where $\omega$ denotes the minimal infinite successor set.
Then:
:$\left|{ S \times S }\right| = \left|{ S }\right|$
\end{theorem}
\begin{proof}
By hypothesis:
:$S \sim \card S$
By Cartesian Product Preserves Cardinality:
:$S \times S \sim \card S \times \card S$
Therefore:
{{begin-eqn}}
{{eqn | l = \card {S \times S}
| r = \card {\card S \times \card S}
| c = Equivalent Sets have Equal Cardinal Numbers
}}
{{eqn | r = \card S
| c = Non-Finite Cardinal is equal to Cardinal Product
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
23737
|
\section{Non-Forking Types have Non-Forking Completions}
Tags: Model Theory
\begin{theorem}
Let $T$ be a complete $\LL$-theory.
Let $\mathfrak C$ be a monster model for $T$.
Let $A\subseteq B$ be subsets of the universe of $\mathfrak C$.
Let $\map \pi {\bar x}$ be an $n$-type over $B$.
If $\pi$ does not fork over $A$, then there is a complete $n$-type $p$ over $B$ such that $\pi \subseteq p$ and $p$ does not fork over $A$.
\end{theorem}
\begin{proof}
Suppose $\pi$ does not fork over $A$.
We will use Zorn's Lemma to find a candidate for the needed complete type.
Consider the collection $\Pi$ of all non-forking sets $\pi'$ of $\LL$-formulas with parameters from $B$ such that $\pi'$ contains $\pi$.
Order $\Pi$ by subset inclusion.
Since a set forks iff a finite subset forks, the union over any chain is still a non-forking set, and hence is an upper bound for the chain.
Thus, by Zorn's Lemma, there is a maximal (with respect to subset inclusion) $p$ in $\Pi$.
{{AimForCont}} $p$ is not complete.
:By definition, for some $\map \phi {\bar x, \bar b}$, $p$ contains neither $\map \phi {\bar x, \bar b}$ nor $\neg \map \phi {\bar x, \bar b}$.
:Since $p$ is non-forking, by Formula and its Negation Cannot Both Cause Forking, at least one of $p \cup \map \phi {\bar x, \bar b}$ or $p \cup \neg \map \phi {\bar x, \bar b}$ is non-forking as well.
:Hence $p$ is not maximal in $\Pi$, contradicting the choice of $p$.
Thus $p$ is complete.
{{qed}}
{{AoC|Zorn's Lemma}}
Category:Model Theory
\end{proof}
|
23738
|
\section{Non-Homeomorphic Sets may be Homeomorphic to Subsets of Each Other}
Tags: Homeomorphisms
\begin{theorem}
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H_1 \subseteq S_1$ and $H_2 \subseteq S_2$.
Then it is possible for:
:$(1): \quad T_1$ to be homeomorphic to $H_2$
:$(2): \quad T_2$ to be homeomorphic to $H_1$
but:
:$(3): \quad T_1$ and $T_2$ to not be homeomorphic.
\end{theorem}
\begin{proof}
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
Let $S_1 := \closedint 0 1$ be the closed unit interval.
Let $S_2 := \openint 0 1$ be the open unit interval.
Let $H_1 := \openint 0 1$ and $H_2 := \closedint {\dfrac 1 4} {\dfrac 3 4}$.
Then:
:$(1): \quad \struct {S_1, \tau_d}$ is homeomorphic to $\struct {H_2, \tau_d}$ by the mapping $f: S_1 \to H_2$ defined as $\map f x = \dfrac x 2 + \dfrac 1 4$
:$(2): \quad \struct {S_2, \tau_d}$ is trivially homeomorphic to $\struct {H_1, \tau_d}$
From Continuous Image of Compact Space is Compact, if $S_1$ and $S_2$ are homeomorphic then both must be compact.
From Closed Real Interval is Compact in Metric Space, $S_1$ is compact.
But from Open Real Interval is not Compact, $S_2$ is not compact.
Hence:
:$(3): \quad \struct {S_1, \tau_d}$ and $\struct {S_2, \tau_d}$ are not homeomorphic.
{{qed}}
\end{proof}
|
23739
|
\section{Non-Injective Mapping may be Strictly Order-Preserving and Order-Reversing}
Tags: Increasing Mappings
\begin{theorem}
Let $\struct {S, \prec_1}$ and $\struct {T, \prec_2}$ be strictly ordered sets.
Let $\phi: S \to T$ be a mapping.
Let $\pi: S \to T$ be a mapping with the property that:
:$\forall x, y \in S: x \prec_1 y \iff \map \pi x \prec_2 \map \pi y$
Then it is not necessarily the case that $\pi$ is an injection.
\end{theorem}
\begin{proof}
Proof by Counterexample:
Let $S = \set {\O, \set a, \set b, \set {a, b} }$ and $T = \set {1, 2, 3}$.
Let $\prec_1$ be the proper subset relation:
:$\forall x, y \in S: x \prec_1 y \iff x \subsetneq y$
Let $\prec_2$ be the usual strict ordering on the integers $1, 2, 3$:
:$\forall x, y \in T: x \prec_2 y \iff x < y$
Let $\pi = \set {\tuple {\O, 1}, \tuple {\set a, 2}, \tuple {\set b, 2}, \tuple {\set {a, b}, 3} }$.
By inspection it is seen that $\pi: S \to T$ is a mapping which is specifically not an injection, as $\map \pi {\set a} = \map \pi {\set b}$.
Equally by inspection it can be seen that:
:$\forall x, y \in S: x \prec_1 y \iff \map \pi x \prec_2 \map \pi y$
{{qed}}
\end{proof}
|
23740
|
\section{Non-Maximal Element of Well-Ordered Class has Immediate Successor}
Tags: Well-Orderings
\begin{theorem}
Let $\left({C, \le}\right)$ be a well-ordered class.
Let $x \in C$.
Suppose that $x$ is maximal in $C$.
{{mistake|maximal?}}
Then $x$ has an immediate successor in $C$.
\end{theorem}
\begin{proof}
Let $x$ be a non-maximal element of $C$.
Let $S$ be the class of successors of $x$ in $C$.
$S$ is non-empty because $x$ is not maximal.
By Proper Well-Ordering Determines Smallest Elements, $S$ has a minimal element, $y$.
Then $x < y$ by the definition of $S$.
Suppose that for some $z \in C$, $x < z < y$.
Then by the definition of $S$, $z \in S$.
This contradicts the minimality of $y$.
Thus $y$ is the immediate successor element of $x$.
{{qed}}
Category:Well-Orderings
\end{proof}
|
23741
|
\section{Non-Negative Additive Function is Monotone}
Tags: Set Systems
\begin{theorem}
Let $\SS$ be an algebra of sets.
Let $f: \SS \to \overline \R$ be an additive function, that is:
:$\forall A, B \in \SS: A \cap B = \O \implies \map f {A \cup B} = \map f A + \map f B$
If $\forall A \in \SS: \map f A \ge 0$, then $f$ is monotone, that is:
:$A \subseteq B \implies \map f A \le \map f B$
\end{theorem}
\begin{proof}
Let $A \subseteq B$.
Then:
{{begin-eqn}}
{{eqn | l = B
| r = \paren {B \setminus A} \cup \paren {A \cap B}
| c = Set Difference Union Intersection
}}
{{eqn | r = \paren {B \setminus A} \cup A
| c = Intersection with Subset is Subset
}}
{{eqn | l = \O
| r = \paren {B \setminus A} \cap A
| c = Set Difference Intersection with Second Set is Empty Set
}}
{{eqn | l = 0
| o = \le
| r = \map f {B \setminus A}
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = \map f A
| o = \le
| r = \map f {B \setminus A} + \map f A
}}
{{eqn | ll= \leadsto
| l = \map f A
| o = \le
| r = \map f B
| c = {{Defof|Additive Function (Measure Theory)}}
}}
{{end-eqn}}
{{qed}}
Category:Set Systems
\end{proof}
|
23742
|
\section{Non-Negative Signed Measure is Measure}
Tags: Signed Measures
\begin{theorem}
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$ such that:
:$\map \mu A \ge 0$
for each $A \in \Sigma$.
Then $\mu$ is a measure on $\struct {X, \Sigma}$.
\end{theorem}
\begin{proof}
We verify each of the conditions given in the definition of a measure.
From the definition of a signed measure, $\mu$ is a function $\Sigma \to \overline \R$.
\end{proof}
|
23743
|
\section{Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary}
Tags: Normed Division Rings
\begin{theorem}
Let $\struct {R, \norm{\,\cdot\,} }$ be a non-Archimedean normed division ring with zero $0_R$
Let $\sequence {x_n}$ be a Cauchy sequence such that $\sequence {x_n}$ does not converge to $0_R$.
Then:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n} = \norm {x_m}$
\end{theorem}
\begin{proof}
By Cauchy Sequence Is Eventually Bounded Away From Non-Limit then:
:$\exists N_1 \in \N$ and $C \in \R_{\gt 0}: \forall n \ge N_1: \norm {x_n} \gt C$
Since $\sequence {x_n}$ is a Cauchy sequence then:
:$\exists N_2 \in \N: \forall n, m \ge N_2: \norm {x_n - x_m} < C$
Let $N = \max \set {N_1, N_2}$.
Let $n, m \ge N$.
Then:
:$\norm {x_n - x_m} < C < \norm {x_n}$
By Corollary to Three Points in Ultrametric Space have Two Equal Distances then:
:$\norm {x_n} = \norm {x_m}$
The result follows.
{{qed}}
\end{proof}
|
23744
|
\section{Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary/P-adic Norm}
Tags: Normed Division Rings, P-adic Number Theory
\begin{theorem}
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\sequence {x_n}$ be a Cauchy sequence such that $\sequence {x_n}$ does not converge to $0$.
Then:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n}_p = \norm {x_m}_p$
\end{theorem}
\begin{proof}
From:
:P-adic Numbers form Non-Archimedean Valued Field
:Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary
it follows that:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n}_p = \norm {x_m}_p$
{{qed}}
\end{proof}
|
23745
|
\section{Non-Palindromes in Base 2 by Reverse-and-Add Process}
Tags: 43, Palindromes, Palindromic Numbers, Reverse-and-Add, Recreational Mathematics
\begin{theorem}
Let the number $22$ be expressed in binary: $10110_2$.
When the reverse-and-add process is performed on it repeatedly, it never becomes a palindromic number.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 10110_2 + 01101_2
| r = 100011_2
| c =
}}
{{eqn | ll= \leadsto
| l = 100011_2 + 110001_2
| r = 1010100_2
| c =
}}
{{end-eqn}}
It remains to be shown that a binary number of this form does not become a palindromic number.
Let $d_n$ denote $n$ repetitions of a binary digit $d$ in a number.
Thus:
:$10111010000$
can be expressed as:
:$101_3010_4$
'''Beware''' that the subscript, from here on in, does not denote the number base.
It is to be shown that the reverse-and-add process applied to:
:$101_n010_{n + 1}$
leads after $4$ iterations to:
:$101_{n + 1}010_{n + 2}$
Thus:
{{begin-eqn}}
{{eqn | l = 101_n010_{n + 1} + 0_{n + 1}101_k01
| r = 110_n101_{n - 1}01
| c =
}}
{{eqn | ll= \leadsto
| l = 110_n101_{n - 1}01 + 101_{n - 1}010_n11
| r = 101_{n + 1}010_{n + 1}
| c =
}}
{{eqn | ll= \leadsto
| l = 101_{n + 1}010_{n + 1} + 0_{n + 1}101_{n + 1}01
| r = 110_{n - 1}10001_{n - 1}01
| c =
}}
{{eqn | ll= \leadsto
| l = 110_{n - 1}10001_{n - 1}01 + 101_{n - 1}00010_{n - 1}11
| r = 101_{n + 1}010_{n + 2}
| c =
}}
{{end-eqn}}
As neither $101_n010_{n + 1}$ nor $101_{n + 1}010_{n + 2}$ are palindromic numbers, nor are any of the intermediate results, the result follows.
{{qed}}
\end{proof}
|
23746
|
\section{Non-Successor Element of Peano Structure is Unique}
Tags: Natural Numbers, Peano's Axioms
\begin{theorem}
Let $\struct {P, s, 0}$ be a Peano structure.
Then:
:$P \setminus s \sqbrk P$ is a singleton
where:
:$\setminus$ denotes set difference
:$s \sqbrk P$ denotes the image of the mapping $s$.
It follows that the non-successor element $0$ is the only element of $P$ with this property.
\end{theorem}
\begin{proof}
Let $T = P \setminus s \sqbrk P$.
From Axiom $(\text P 4)$ we know that $T \ne \O$.
Now suppose that $t_1 \in T$ and $t_2 \in T$.
{{AimForCont}} $t_1 \ne t_2$.
Define $A = P \setminus \set {t_2}$.
Thus $t_1 \in A \ne P$.
Moreover, by the nature of $t_2$:
:$x \in A \implies \map s x \in A$
Thus, by the induction axiom $(\text P 5)$, $A = P$.
From this contradiction it follows that $P \setminus s \sqbrk P$ cannot contain two different elements.
{{qed}}
\end{proof}
|
23747
|
\section{Non-Trivial Annihilator Contains Positive Integer}
Tags: Rings, Rings with Unity
\begin{theorem}
Let $\left({R, +, \times}\right)$ be a ring with unity.
Let $A = \operatorname{Ann} \left({R}\right)$ be the annihilator of $R$.
Let $a \in A$ such that $a \ne 0$.
Then $A$ contains at least one strictly positive integer.
\end{theorem}
\begin{proof}
Let the zero of $R$ be $0_R$ and the unity of $R$ be $1_R$.
First we note that:
{{begin-eqn}}
{{eqn | l = a
| o = \in
| r = \operatorname{Ann} \left({R}\right)
| c =
}}
{{eqn | ll= \leadsto
| l = a \cdot 1_R
| r = 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = \left({-a}\right) \cdot 1_R
| r = 0_R
| c =
}}
{{eqn | ll= \leadsto
| l = -\left({a \cdot 1_R}\right)
| r = 0_R
| c =
}}
{{end-eqn}}
So:
: $a \in A \implies -a \in A$
But if $a \ne 0$ then either $a > 0$ or $-a > 0$.
That is, either $a$ or $-a$ is positive.
So, if $\operatorname{Ann} \left({F}\right)$ contains at least one non-zero element, it contains at least one strictly positive integer.
{{qed}}
Category:Rings with Unity
\end{proof}
|
23748
|
\section{Non-Trivial Commutative Division Ring is Field}
Tags: Division Rings, Division Ring, Field Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a non-trivial division ring such that $\circ$ is commutative.
Then $\struct {R, +, \circ}$ is a field.
Similarly, let $\struct {F, +, \circ}$ be a field.
Then $\struct {F, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.
\end{theorem}
\begin{proof}
Suppose $\struct {R, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.
By definition $\struct {R, +}$ is an abelian group.
Thus {{Field-axiom|A}} are satisfied.
Also by definition, $\struct {R, \circ}$ is a semigroup such that $\circ$ is commutative.
Thus {{Field-axiom|M0}} to {{Field-axiom|M2}} are satisfied.
As $\struct {R, +, \circ}$ is a ring with unity, {{Field-axiom|M3}} is satisfied.
{{Field-axiom|D}} is satisfied by dint of $\struct {R, +, \circ}$ being a ring.
Finally note that by definition of division ring, {{Field-axiom|M4}} is satisfied.
Thus all the field axioms are satisfied, and so $\struct {R, +, \circ}$ is a field.
{{qed|lemma}}
Suppose $\struct {F, +, \circ}$ is a field.
Then by definition $\struct {F, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.
{{qed}}
\end{proof}
|
23749
|
\section{Non-Trivial Connected Set in T1 Space is Dense-in-itself}
Tags: Connectedness, Connected Spaces, T1 Spaces, Denseness, Connected Sets
\begin{theorem}
Let $T = \struct {S, \tau}$ be a $T_1$ (Fréchet) topological space.
Let $H \subseteq S$ be connected in $T$.
If $H$ has more than one element, then $H$ is dense-in-itself.
\end{theorem}
\begin{proof}
{{AimForCont}} $H$ is not dense-in-itself.
Then $\exists x \in H$ such that $x$ is isolated in $H$.
That is, $\exists U \in \tau: U \cap H = \set x$.
Since $H$ has more than one element, we can find $y \in H$ with $y \ne x$.
Since $T$ is $T_1$:
:$\forall y \in H: y \ne x \implies \paren {\exists V_y \in \tau: y \in V_y, x \notin V_y}$
Define $V = \ds \bigcup_{y \mathop \in H \\ y \mathop \ne x} V_y$.
Then $\forall z \in H$:
:$z = x \implies z \in U$
:$z \ne x \implies z \in V_z \subseteq V$
Showing that $H \subseteq U \cup V$.
This also shows that both $U \cap H$ and $V \cap H$ are non-empty.
Since $x \notin V_y$ for any $y \in H$,
we have $x \notin V$.
This implies $\set x \cap V = \O$
So $H \cap U \cap V = \O$.
Thus we have:
:$H \subseteq U \cup V$
:$H \cap U \cap V = \O$
:$U \cap H \ne \O$
:$V \cap H \ne \O$
Showing that $H$ is disconnected.
This is a contradiction.
Hence we must have $H$ is dense-in-itself.
{{qed}}
\end{proof}
|
23750
|
\section{Non-Trivial Discrete Space is not Arc-Connected}
Tags: Arc-Connected Spaces, Non-Trivial Discrete Space is not Connected, Discrete Topology
\begin{theorem}
Let $T = \left({S, \tau}\right)$ be a non-trivial discrete topological space.
$T$ is not arc-connected.
\end{theorem}
\begin{proof}
{{AimForCont}} $T$ is arc-connected.
From:
:Arc-Connected Space is Path-Connected
:Path-Connected Space is Connected
we have that $T$ is connected.
But this directly contradicts Non-Trivial Discrete Space is not Connected.
The result follows from Proof by Contradiction.
{{qed}}
\end{proof}
|
23751
|
\section{Non-Trivial Discrete Space is not Connected}
Tags: Connectedness, Connected Spaces, Non-Trivial Discrete Space is not Connected, Discrete Topology
\begin{theorem}
Let $T = \struct {S, \tau}$ be a non-trivial discrete topological space.
Then $T$ is not connected.
Thus also:
\end{theorem}
\begin{proof}
Let $T = \struct {S, \tau}$ be a non-trivial discrete space.
Let $a \in S$.
Let $A = \set a$ and $B = \relcomp S {\set a}$, where $\relcomp S {\set a}$ is the complement of $A$ in $S$.
As $T$ is not trivial, $B \ne \O$.
Then from Set in Discrete Topology is Clopen, $A$ and $B$ are both open.
So $A \mid B$ is a separation of $S$.
It follows, by definition, that $T$ is not connected.
{{qed}}
\end{proof}
|
23752
|
\section{Non-Trivial Discrete Space is not Irreducible}
Tags: Irreducible Spaces, Non-Trivial Discrete Space is not Connected, Discrete Topology
\begin{theorem}
Let $T = \left({S, \tau}\right)$ be a non-trivial discrete topological space.
$T$ is not irreducible.
\end{theorem}
\begin{proof}
{{AimForCont}} $T$ is irreducible.
From Irreducible Space is Connected, we have that $T$ is connected.
But this directly contradicts Non-Trivial Discrete Space is not Connected.
The result follows from Proof by Contradiction.
{{qed}}
Category:Discrete Topology
Category:Irreducible Spaces
Category:Non-Trivial Discrete Space is not Connected
\end{proof}
|
23753
|
\section{Non-Trivial Discrete Space is not Path-Connected}
Tags: Path-Connected Spaces, Non-Trivial Discrete Space is not Connected, Discrete Topology
\begin{theorem}
Let $T = \left({S, \tau}\right)$ be a non-trivial discrete topological space.
$T$ is not path-connected.
\end{theorem}
\begin{proof}
{{AimForCont}} $T$ is path-connected.
From Path-Connected Space is Connected, we have that $T$ is connected.
But this directly contradicts Non-Trivial Discrete Space is not Connected.
The result follows from Proof by Contradiction.
{{qed}}
\end{proof}
|
23754
|
\section{Non-Trivial Discrete Space is not Ultraconnected}
Tags: Non-Trivial Discrete Space is not Connected, Ultraconnected Spaces, Discrete Topology
\begin{theorem}
Let $T = \left({S, \tau}\right)$ be a non-trivial discrete topological space.
$T$ is not ultraconnected.
\end{theorem}
\begin{proof}
{{AimForCont}} $T$ is ultraconnected.
From Ultraconnected Space is Connected, we have that $T$ is connected.
But this directly contradicts Non-Trivial Discrete Space is not Connected.
The result follows from Proof by Contradiction.
{{qed}}
Category:Discrete Topology
Category:Ultraconnected Spaces
Category:Non-Trivial Discrete Space is not Connected
\end{proof}
|
23755
|
\section{Non-Trivial Event is Union of Simple Events}
Tags: Events
\begin{theorem}
Let $\EE$ be an experiment.
Let $e$ be an event in $\EE$ such that $e \ne \O$.
That is, such that $e$ is non-trivial.
Then $e$ can be expressed as the union of a set of simple events in $\EE$.
\end{theorem}
\begin{proof}
By definition of event, $e$ is a subset of the sample space $\Omega$ of $\EE$.
By hypothesis:
:$e \ne \O$
and so:
:$\exists s \in \Omega: s \in e$
Let $S$ be the set defined as:
:$S = \set {\set s: s \in e}$
By Union is Smallest Superset: Set of Sets it follows that:
:$\ds \bigcup S \subseteq e$
Let $x \in e$.
Then by Singleton of Element is Subset:
:$\set x \subseteq e$
and by definition of $S$ it follows that:
:$\set x \in S$
and so by definition of set union:
:$x \in \ds \bigcup S$
Thus we have:
:$e \subseteq \ds \bigcup S$
The result follows by definition of set equality.
{{qed}}
\end{proof}
|
23756
|
\section{Non-Trivial Excluded Point Topology is not T1}
Tags: Excluded Point Topology, T1 Spaces
\begin{theorem}
Let $T = \struct {S, \tau_{\bar p} }$ be a excluded point space such that $S$ is not a singleton.
Then $T$ is not a $T_1$ (Fréchet) space.
\end{theorem}
\begin{proof}
Follows directly from:
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Open Extension Topology is not $T_1$
{{qed}}
\end{proof}
|
23757
|
\section{Non-Trivial Particular Point Topology is not T1}
Tags: T1 Spaces, Particular Point Topology
\begin{theorem}
Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton.
Then $T$ is not a $T_1$ (Fréchet) space.
\end{theorem}
\begin{proof}
Follows directly from:
:Particular Point Topology is Closed Extension Topology of Discrete Topology
:Closed Extension Topology is not $T_1$.
{{qed}}
\end{proof}
|
23758
|
\section{Non-Trivial Particular Point Topology is not T3}
Tags: T3 Spaces, Particular Point Topology
\begin{theorem}
Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton.
Then $T$ is not a $T_3$ space.
\end{theorem}
\begin{proof}
We have that there are at least two distinct elements of $S$.
So, consider $x, p \in S: x \ne p$.
Then $X = \set x$ is closed in $T$ and $p \notin X$.
Suppose $U \in \tau_p$ is an open set in $T$ such that $X \subseteq U$.
We have that $\set p \in \tau_p$ such that $p \in \set p$.
But as $p \in U, p \in \set p$ we have that $U \cap \set p \ne \O$.
So $T$ is not a $T_3$ space.
{{qed}}
\end{proof}
|
23759
|
\section{Non-Trivial Ultraconnected Space is not T1}
Tags: T1 Spaces, Ultraconnectedness, Separation Axioms, Ultraconnected Spaces, Connectedness
\begin{theorem}
Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
If $S$ has more than one element, then $T$ is not a $T_1$ (Fréchet) space.
That is, if $T$ is a $T_1$ (Fréchet) space with more than one element, it is not ultraconnected.
\end{theorem}
\begin{proof}
$T = \struct {S, \tau}$ be ultraconnected.
Thus by definition:
:$(1): \quad \forall x, y \in S: \set x^- \cap \set y^- \ne \O$
Let $a, b \in S$ such that $a \ne b$.
{{AimForCont}} $T$ is a $T_1$ (Fréchet) space.
By definition of $T_1$ Space, $\set a$ and $\set b$ are closed.
From Closed Set Equals its Closure we have that $\set a^- = \set a$ and $\set b^- = \set b$.
It immediately follows that:
:$\set a^- \cap \set b^- = \O$
But that contradicts $(1)$ above.
The result follows by Proof by Contradiction.
{{qed}}
\end{proof}
|
23760
|
\section{Non-Zero-Sum Game as Zero-Sum Game}
Tags: Zero-Sum Games
\begin{theorem}
Let $G$ be a non-zero-sum game for $n$ players.
Then $G$ can be modelled as a zero-sum game for $n + 1$ players.
\end{theorem}
\begin{proof}
At each outcome, the total payoff of $G$ will be an amount which will (for at least one outcome) not be zero
Let an $n + 1$th player be introduced to $G$ who has one move:
:$(1): \quad$ Select any player $m$.
:$(2): \quad$ If the total payoff of $G$ is $+k$, receive in payment $k$ from player $m$.
:$(3): \quad$ If the total payoff of $G$ is $-k$, pay $k$ to player $m$.
This new game is the same as $G$ but with an extra (dummy) player, and is now zero-sum.
{{qed}}
\end{proof}
|
23761
|
\section{Non-Zero Complex Numbers are Closed under Multiplication}
Tags: Non-Zero Complex Numbers Closed under Multiplication, Algebraic Closure, Non-Zero Complex Numbers are Closed under Multiplication, Real Numbers, Complex Multiplication, Complex Numbers
\begin{theorem}
The set of non-zero complex numbers is closed under multiplication.
\end{theorem}
\begin{proof}
We have that the complex numbers form a field under the operations of addition and multiplication.
By definition of a field, the algebraic structure $\left({\C_{\ne 0}, \times}\right)$ is a group.
Thus, by definition, $\times$ is closed in $\left({\C_{\ne 0}, \times}\right)$.
{{qed}}
\end{proof}
|
23762
|
\section{Non-Zero Complex Numbers under Multiplication form Group}
Tags: Complex Numbers, Complex Multiplication, Multiplication, Examples of Groups
\begin{theorem}
Let $\C_{\ne 0}$ be the set of complex numbers without zero, that is:
:$\C_{\ne 0} = \C \setminus \set 0$
The structure $\struct {\C_{\ne 0}, \times}$ is a group.
\end{theorem}
\begin{proof}
Taking the group axioms in turn:
\end{proof}
|
23763
|
\section{Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group}
Tags: Multiplication, Examples of Abelian Groups, Group Examples, Abelian Groups, Abelian Groups: Examples, Infinite Groups: Examples, Complex Multiplication, Examples of Infinite Groups, Abelian Group Examples, Complex Numbers
\begin{theorem}
Let $\C_{\ne 0}$ be the set of complex numbers without zero, that is:
:$\C_{\ne 0} = \C \setminus \set 0$
The structure $\struct {\C_{\ne 0}, \times}$ is an infinite abelian group.
\end{theorem}
\begin{proof}
From Non-Zero Complex Numbers under Multiplication form Group, $\struct {\C_{\ne 0}, \times}$ is a group.
Then we have:
:Complex Multiplication is Commutative
and:
:Complex Numbers are Uncountable.
{{qed}}
\end{proof}
|
23764
|
\section{Non-Zero Elements of Division Ring form Group}
Tags: Division Rings, Rings, Groups, Group Theory
\begin{theorem}
Let $\struct {R, +, \circ}$ be a division ring.
Then $\struct {R^*, \circ}$ is a group.
\end{theorem}
\begin{proof}
A division ring by definition is a ring with unity, and therefore not null.
A division ring by definition has no zero divisors, so $\struct {R^*, \circ}$ is a semigroup.
$1_R \in \struct {R^*, \circ}$ and so the identity of $\circ$ is in $\struct {R^*, \circ}$.
By the definition of a division ring, each element of $\struct {R^*, \circ}$ is a unit, and therefore has a unique inverse in $\struct {R^*, \circ}$.
Thus $\struct {R^*, \circ}$ is a semigroup with an identity and inverses and so is a group.
{{qed}}
Category:Division Rings
Category:Group Theory
\end{proof}
|
23765
|
\section{Non-Zero Integer has Finite Number of Divisors}
Tags: Number Theory, Divisors
\begin{theorem}
Let $n \in \Z_{\ne 0}$ be a non-zero integer.
Then $n$ has a finite number of divisors.
\end{theorem}
\begin{proof}
Let $S$ be the set of all divisors of $n$.
Then from Absolute Value of Integer is not less than Divisors:
:$\forall m \in S: -n \le m \le n$
Thus $S$ is finite.
{{qed}}
Category:Number Theory
Category:Divisors
\end{proof}
|
23766
|
\section{Non-Zero Integer has Unique Positive Integer Associate}
Tags: Integers
\begin{theorem}
Let $a \in \Z$ be an integer such that $a \ne 0$.
Then $a$ has a unique associate $b \in \Z_{>0}$.
\end{theorem}
\begin{proof}
Let $a \in \Z_{\ne 0}$.
By Integer Divides its Absolute Value:
:$a \divides \size a$ and $\size a \divides a$
Hence $\size a$ is an associate of $a$.
Now we prove its uniqueness.
Let $b, c \in \Z_{\ne 0}$ such that $b > 0$ and $c > 0$.
Let $a \sim b$ and $a \sim c$ where $\sim$ denotes the relation of associatehood.
By definition of associatehood:
:$a \divides b$ and $b \divides a$
and:
:$a \divides c$ and $c \divides a$
From Divisor Relation is Antisymmetric/Corollary/Proof 2:
:$a = \pm b$
and
:$a = \pm c$
That is:
:$\pm b = \pm c$
which means:
:$b = c$ or $b = -c$
But as both $b > 0$ and $c > 0$:
:$b = c$
Hence the result.
{{qed}}
\end{proof}
|
23767
|
\section{Non-Zero Integers Closed under Multiplication}
Tags: Integer Multiplication, Algebraic Closure, Integers
\begin{theorem}
The set of non-zero integers is closed under multiplication.
\end{theorem}
\begin{proof}
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ as:
:$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested.
Integer multiplication is defined as:
:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {a c + b d, a d + b c} {}$
From Integer Multiplication is Closed, we have that $x, y \in \Z \implies x y \in \Z$.
From Ring of Integers has no Zero Divisors, we have that $x, y \in \Z: x, y \ne 0 \implies x y \ne 0$.
Therefore multiplication on the non-zero integers is closed.
{{qed}}
\end{proof}
|
23768
|
\section{Non-Zero Integers are Cancellable for Multiplication}
Tags: Integers, Non-Zero Integers are Cancellable for Multiplication
\begin{theorem}
Every non-zero integer is cancellable for multiplication.
That is:
:$\forall x, y, z \in \Z, x \ne 0: x y = x z \iff y = z$
\end{theorem}
\begin{proof}
Let $x y = x z$.
There are two cases to investigate: $x > 0$ and $x < 0$.
Let $x > 0$.
From Natural Numbers are Non-Negative Integers, $x \in \N_{> 0}$.
By the Extension Theorem for Distributive Operations and Ordering on Naturally Ordered Semigroup Product, $x$ is cancellable for multiplication.
{{qed|lemma}
Let $x < 0$ and
We know that the Integers form Integral Domain and are thus a ring.
Then $-x > 0$ and so:
{{begin-eqn}}
{{eqn | l = x y
| r = x z
| c =
}}
{{eqn | ll= \implies
| l = \left({- \left({-x}\right)}\right) y
| r = \left({- \left({-x}\right)}\right) z
| c = Negative of Ring Negative
}}
{{eqn | ll= \implies
| l = - \left({\left({-x}\right) y}\right)
| r = - \left({\left({-x}\right) z}\right)
| c = Product with Ring Negative
}}
{{eqn | ll= \implies
| l = - y
| r = - z
| c = as $-x$ is (strictly) positive, the above result holds
}}
{{eqn | ll= \implies
| l = y
| r = z
| c = $\left({\Z, +}\right)$ is a group: axiom $G3$
}}
{{end-eqn}}
{{qed|lemma}}
So whatever non-zero value $x$ takes, it is cancellable for multiplication.
{{Qed}}
\end{proof}
|
23769
|
\section{Non-Zero Integers under Multiplication are not Subgroup of Reals}
Tags: Real Numbers, Integer Multiplication, Integers, Real Multiplication
\begin{theorem}
Let $\struct {\Z_{\ne 0}, \times}$ denote the algebraic structure formed by the set of non-zero integers under multiplication.
Let $\struct {\R_{\ne 0}, \times}$ denote the algebraic structure formed by the set of non-zero real numbers under multiplication.
Then, while $\struct {\Z_{\ne 0}, \times}$ is closed, it is not a subgroup of $\struct {\R_{\ne 0}, \times}$.
\end{theorem}
\begin{proof}
We have that Non-Zero Real Numbers under Multiplication form Group.
We also have that the set of non-zero integers $\Z_{\ne 0}$ form a subset of $\R_{\ne 0}$.
From Non-Zero Integers Closed under Multiplication:
:$\forall a, b \in \Z_{\ne 0}: a \times b \in \Z_{\ne 0}$
We have that:
:$\forall x \in \Z_{\ne 0}: 1 \times x= x = x \times 1$
and so $1$ is the identity of $\struct {\Z_{\ne 0}, \times}$.
But for $x \in \Z_{\ne 0}: x \ne 1$, there exists no $y \in \Z_{\ne 0}: x \times y = 1$.
Thus $\struct {\Z_{\ne 0}, \times}$ does not have inverses for all $x \in \Z_{\ne 0}$.
Thus, by definition, $\struct {\Z_{\ne 0}, \times}$ is not a group.
It follows that $\struct {\Z_{\ne 0}, \times}$ is not a subgroup of $\struct {\R_{\ne 0}, \times}$.
{{qed}}
\end{proof}
|
23770
|
\section{Non-Zero Modulo Numbers Closed under Multiplication then Modulo is Prime}
Tags: Real Numbers, Modulo Arithmetic
\begin{theorem}
Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$ for $m > 1$.
Let $\Z'_m$ be the set of non-zero integers modulo $m$.
Let $\struct {\Z_m, \times_m}$ be closed under modulo multiplication.
Then $m$ is prime.
\end{theorem}
\begin{proof}
Suppose $m$ is not prime.
Then $m = r s$ for some $r, s \in \Z: 1 < r < m, 1 < s < m$.
So $r, s \in \Z'_m$.
But:
:$r \times_m s \equiv 0 \pmod m$
and so $r \times_m s \notin \Z'_m$.
So if $m$ is not prime, $\struct {\Z_m, \times_m}$ is not closed.
The result follows from the Rule of Transposition.
{{qed}}
\end{proof}
|
23771
|
\section{Non-Zero Natural Numbers under Addition do not form Monoid}
Tags: Natural Numbers, Monoids
\begin{theorem}
Let $\N_{>0}$ be the set of natural numbers without zero, i.e. $\N_{>0} = \N \setminus \set 0$.
The structure $\struct {\N_{>0}, +}$ does ''not'' form a monoid.
\end{theorem}
\begin{proof}
From Natural Numbers under Addition form Commutative Monoid, $\struct {\N, +}$ forms a commutative monoid.
From Natural Numbers Bounded Below under Addition form Commutative Semigroup, $\struct {\N_{>0}, +}$ forms a commutative semigroup.
From Identity Element of Natural Number Addition is Zero, $0$ is the identity of $\struct {\N, +}$
From Natural Number Addition is Cancellable, all elements of $\struct {\N, +}$ are cancellable.
{{AimForCont}} $\struct {\N_{>0}, +}$ is a monoid.
Then by definition $\struct {\N_{>0}, +}$ has an identity $e$.
From Identity of Cancellable Monoid is Identity of Submonoid, $e$ the same as the identity of $\struct {\N, +}$.
That is:
:$e = 0$
But $0 \notin \N_{>0}$.
Thus it follows that $\struct {\N_{>0}, +}$ has no identity.
Hence by Proof by Contradiction it follows that $\struct {\N_{>0}, +}$ is not a monoid.
{{qed}}
\end{proof}
|
23772
|
\section{Non-Zero Natural Numbers under Addition form Semigroup}
Tags: Natural Numbers, Examples of Semigroups, Semigroup Examples
\begin{theorem}
Let $\N_{>0}$ be the set of natural numbers without zero, that is:
:$\N_{>0} = \N \setminus \set 0$
Let $+$ denote natural number addition.
The structure $\struct {\N_{>0}, +}$ forms a semigroup.
\end{theorem}
\begin{proof}
This is a specific instance of Natural Numbers Bounded Below under Addition form Commutative Semigroup.
{{qed}}
\end{proof}
|
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