|$
is the order of the corresponding cohomology class. We note that in
\cite{Nat05},   Natale defined the {\em modified exponent }$\exp_\omega
G$ of a group $G$ endowed with a 3-cocycle $\omega$ by same formula
(\ref{FSexp}),
    and proved that $\exp_\omega G$ is a gauge
invariant of $D^\omega(G)$ for groups of odd order. In \cite{NS3} it is
shown that generally that $f$ is a gauge invariant of $D^\omega(G)$ and
that $f=\exp(D^\omega(G))$ or  $2\exp(D^\omega(G))$.

We now have


\begin{thm}\label{FSexpcomp} The following hold for \emph{any}
$3$-cocycle $\omega$ on one of the indicated groups  ($Q$ is an
extra-special group of width $l$):
    \begin{eqnarray*}
    && a) \ f(\mathbb{Z}_2^l, \omega) = \left \{ \begin{array}{ll}
4 & \mbox{if} \ \omega \not = 1, \\
2 & \mbox{if $ \omega = 1$,} \ l \geq 1;
\end{array} \right.\\
&& b) \ f(Q, \omega) = \left \{ \begin{array}{ll}
4 & \mbox{if} \ \omega^2 = 1, \  l \geq 2, \\
8 & \mbox{if $\omega^2 \not = 1, \  l  \geq 2$};
\end{array} \right.\\
    && c) \ f(Q_8, \omega) = \left \{ \begin{array}{lll}
4 & \mbox{if} \ \omega^2 = 1, \\
8 & \mbox{if} \ \omega \ \mbox{has order $4$}, \\
16 & \mbox{if} \ \omega \ \mbox{has order $8$}; \\
\end{array}
\right.  \\
    && d) \ f(D_8, \omega) = \left \{ \begin{array}{lll}
4 & \mbox{if} \ \omega \in  \operatorname{Infl}_V^{D_8}H^3(V, \mathbb{C}^*), \\
8 & \mbox{if} \ \omega \in \Omega H^3(D_8, \mathbb{C}^*) \setminus
\operatorname{Infl}_V^{D_8}H^3(V, \mathbb{C}^*), \\
16 & \mbox{if} \ \omega \ \mbox{has order $4$}. \\
\end{array}
\right.
    \end{eqnarray*}
\end{thm}
\begin{proof}

If $\omega$ is trivial then by (\ref{FSexp}) $f$ coincides with the
exponent of the group in question. Suppose that $\omega$ has order
$2$ in the elementary abelian $2$-group case. Then from \cite{M2}
that there is a subgroup $C$ of order $2$ such that $\omega_C$ is
nontrivial.
    Part a) now follows.
    Parts b) and c) follow immediately from Lemma \ref{cohomrestrict}.


    \medskip
    It remains to handle the case $Q = D_8$. Let $C$ be the cyclic
subgroup of order $4$ and let $I =\operatorname{Infl}_V^{D_8}H^3(V,
\mathbb{C}^*)$.   By Lemma \ref{lemmaD8} we have
    $H^3(D_8, \mathbb{C}^*) = I\langle \zeta \rangle$ for any $\zeta$ of
order $4$.
    If $\omega \in I$ then
    $\omega_C$ is necessarily trivial, and therefore
    $f(D_8, \omega) = 4$ for such $\omega$. If $\omega$ has order $4$ then
    $\omega_C$ also has order $4$ (Lemma \ref{cohomrestrict}), so $f =
16$ in this case.
    Finally, $\zeta^2_C$ has order $2$ for $\zeta$ of order $4$. Since
$\zeta^2 \notin I$ we conclude that $\omega_C$ has order $2$
    whenever $\omega \in \Omega H^3(D_8, \mathbb{C}^*) \setminus I$, so
that $f = 8$ in this case. This completes the proof of all parts
    of the Theorem.
\end{proof}





\section{Twisted Quantum Doubles of Dimension 64}\label{section64}

\hspace{0.475cm} We have seen in the previous Section that the
properties of the groups $H^3(Q, \mathbb{C}^*)$ for $Q = Q_8, D_8$
are exceptional in several ways. Thus the same is true for the
corresponding twisted quantum doubles.
    In this Section we consider the problem of
understanding the gauge equivalence classes of the quantum doubles
$D^{\omega}(G)$ where $G$ has order $8$. In \cite{MN1},  the same
question was treated for \emph{abelian} groups $G$ and
\emph{abelian} cocycles, that is cocycles $\omega$ for which
$D^{\omega}(G)$ is commutative. Here we consider the case when
$D^{\omega}(G)$ is \emph{noncommutative}. This precludes the two
groups $\mathbb{Z}_8$ and $\mathbb{Z}_2 \times \mathbb{Z}_4$, leaving the groups $E_8 =
\mathbb{Z}_2^3, Q_8$ and $D_8$ to be considered. In this Section we denote
these three groups by $E, Q$ and $D$ respectively. Now $E$ has $64$
nonabelian $3$-cohomology classes, while $D$ and $Q$ have $16$ and
$8$ classes respectively (cf. Remark \ref{cohomgroups}).
    So our task is to sort the
$88$ resulting twisted quantum doubles into gauge equivalence
classes.  We will see that there are at least 8, and no more than
20, gauge equivalence classes.

\vspace{0.4cm} A significant reduction in the problem is achieved by
considering automorphisms of the three groups in question. That is
because it is easy to see (\cite{MN1}, Remark 2.1(iii)) that
automorphisms preserve gauge equivalence classes. Consider first the
group $E$. One knows that $H = H^3(E, \mathbb{C}^*)$ has order $2^7$
and exponent $2$. Of the $128$ cohomology classes in $H$, the
abelian classes form a subgroup of order $64$ generated by the Chern
classes of characters of $E$ (\cite{MN1}, Proposition 7.5).


\vspace{0.4cm} The automorphism group of $E$ is the simple group
$SL(3, 2)$ of order $168$, and we regard $H$ as a $7$-dimensional
$\mathbb{F}_2SL(3, 2)$-module.
     We can understand the structure of $H$
    as follows (for more details, see \cite{M2}).
    A cohomology class $\omega \in H$
is characterized by the subset of elements $1 \neq g \in E$ for
which the restriction Res$_{\langle g \rangle}^{E}\omega$ is
\emph{nontrivial}. Let us call this set of elements the
\emph{support} of $\omega$, denoted
    supp $\omega$.
     The \emph{weight} of $\omega$ is the cardinality $|\mbox{supp} \
\omega|$ of its support.
     There is an isomorphism of $SL(3, 2)$-modules
     \begin{eqnarray}\label{cohomisom}
H \stackrel{\cong}{\longrightarrow} \mathbb{F}_2^7, \ \ \omega \mapsto
\mbox{supp} \ \omega.
\end{eqnarray}
So $H$ is a permutation module for $SL(3, 2)$ corresponding to  the
permutation action of $SL(3, 2)$ on the nonzero elements of $E$.
      Thus there is a decomposition
\begin{eqnarray}\label{cohomdecomp}
H = 3 \oplus \overline{3} \oplus 1
\end{eqnarray}
into simple modules. The abelian cocycles are those in the unique
submodule of codimension 1, and they may therefore be alternately
characterized as those classes
    of  \emph{even} weight.
As for the nonabelian classes, the possible weights are $1, 3, 5,
7$, and the number of cohomology classes of each type is $7,35, 21
,1$ respectively. Those of