% infinite-dim stuff

% concrete matrix rep for clifford group?
% concrete matrix rep for clifford algebra?
% concrete matrix rep for quat unit group and division algebra
%   ... real and complex ... invertibility proofs.


%% ================================================================
\section{Basic formulas and orthogonality relations}

$\chi(g)=\Tr(U(g))$

\begin{eqnarray*}
	\langle \chi_i, \chi_j \rangle &=& \frac{1}{o(G)} \sum_{g\in G}
		\overline{\chi_i(g)} \chi_j(g) = \delta_{ij} \\
	&=& \frac{1}{o(G)} \sum_{C_k}
		\#(C_k) \overline{\chi_i(C_k)} \chi_j(C_k) = \delta_{ij} \\
\end{eqnarray*}

$$
	\frac{1}{o(G)} \sum_{g \in G}
		\overline{D_{ij}^{(\alpha)}(g)} D_{k\ell}^{(\beta)}(g)
	= \frac{1}{d_\alpha} \delta_{\alpha\beta} \delta_{ik} \delta_{j\ell}
$$

$$\{\sqrt{d_\alpha} D_{ij}^{(\alpha)}(g)\}$$ is an o.n. basis for $\C G$,
with inner product
$$
	\langle f, g \rangle = \frac{1}{o(G)} \sum_{x\in G} \overline{f(x)}g(x).
$$

$\dim(\C G) = o(G)$

$\#(\hat(G)) = $ number of cjg classses of $G$.

$\{\chi^{(\alpha)}\}$ form an o.n. basis for $Z(\C G)$.

characters are class functions

Inner product on characters:
$$
	\langle \chi_i, \chi_j \rangle = \frac{1}{o(G)}
		\sum_{g\in G} \overline{\chi_i(g)} \chi_j(g) = \delta_{ij}.
$$

Orthogonality:

$$
	\sum_{\chi} \overline{\chi(C_i)} \chi(C_j) = \frac{\delta_{ij} o(G)}{\#C_j};
$$

$$
	\sum_C \chi^{(\alpha)}(C) \chi^{(\beta)}(C) \frac{\#C}{o(G)} = \delta_{\alpha\beta}.
$$

%% ================================================================
\newpage
\section{Character table for $\mcS_3$}

%% ================================================================
\newpage
\section{Induced representations}

To write or not to write?

%% ================================================================
\newpage
\section{Induced characters}

Type up the nice homework here.

\bfidx{Frobenius character formula} \ldots .

%% ================================================================
\newpage
\section{The little-group method}

Following the method of section V.1 of \cite{bib:simon}, we compute the
character table for $\mcV_4 \rtimes \mcS_3$.

In general, the method of the constructive proof of Simon's proposition V.1.1.
applies to $G = N \rtimes_\alpha H$ where $N \lhd G$, $N$ is abelian, and $H
\to \Aut(N)$.  Here we have $N=\mcV_4 = \{e,a,b,c\}$, and $\alpha$ has
$H=\mcS_3$ permuting $\{a,b,c\}$.

Recall that the dual of $\mcV_4$, i.e. $\hat{\mcV}_4$, is
$$
	\begin{tabular}{|l||r|r|r|r|}
		\hline {} & $e$ & $a$ & $b$ & $c$ \\
		\hline
		\hline $\chi_0$ &  1 &  1 &  1 &  1 \\
		\hline $\chi_1$ &  1 &  1 & -1 & -1 \\
		\hline $\chi_2$ &  1 & -1 &  1 & -1 \\
		\hline $\chi_3$ &  1 & -1 & -1 &  1 \\
		\hline
	\end{tabular}
$$
Then we form $\beta:H \to \Aut(\hat{N})$ by
$$
	(\beta_h\chi)(n) = \chi(\alpha_h^{-1} n).
$$
Here, we have
$$
	\begin{tabular}{|l||r|r|r|r|}
		\hline {} & $e$ & $a$ & $b$ & $c$ \\
		\hline
		\hline $\beta_{1}\chi_0$ &  1 &  1 &  1 &  1 \\
		\hline $\beta_{1}\chi_1$ &  1 &  1 & -1 & -1 \\
		\hline $\beta_{1}\chi_2$ &  1 & -1 &  1 & -1 \\
		\hline $\beta_{1}\chi_3$ &  1 & -1 & -1 &  1 \\
		\hline
	\end{tabular}
	\quad
	\begin{tabular}{|l||r|r|r|r|}
		\hline {} & $e$ & $a$ & $b$ & $c$ \\
		\hline
		\hline $\beta_{(12)}\chi_0$ &  1 &  1 &  1 &  1 \\
		\hline $\beta_{(12)}\chi_1$ &  1 & -1 &  1 & -1 \\
		\hline $\beta_{(12)}\chi_2$ &  1 &  1 & -1 & -1 \\
		\hline $\beta_{(12)}\chi_3$ &  1 & -1 & -1 &  1 \\
		\hline
	\end{tabular}
	\quad
	\begin{tabular}{|l||r|r|r|r|}
		\hline {} & $e$ & $a$ & $b$ & $c$ \\
		\hline
		\hline $\beta_{(123)}\chi_0$ &  1 &  1 &  1 &  1 \\
		\hline $\beta_{(123)}\chi_1$ &  1 & -1 &  1 & -1 \\
		\hline $\beta_{(123)}\chi_2$ &  1 & -1 & -1 &  1 \\
		\hline $\beta_{(123)}\chi_3$ &  1 &  1 & -1 & -1 \\
		\hline
	\end{tabular}
$$
and likewise for the remaining three elements of $\mcS_3$, namely, $\beta_h$
acts by permuting the second through fourth columns of the character table.
Thus we read off the action of $\beta$ on $\hat{\mcV}_4$:
$$
	\begin{tabular}{|l||r|r|r|r|}
		\hline {}               & $\chi_0$ & $\chi_1$ & $\chi_2$ & $\chi_3$ \\
		\hline
		\hline $\beta_{1}$     & $\chi_0$ & $\chi_1$ & $\chi_2$ & $\chi_3$ \\
		\hline $\beta_{(12)}$  & $\chi_0$ & $\chi_2$ & $\chi_1$ & $\chi_3$ \\
		\hline $\beta_{(13)}$  & $\chi_0$ & $\chi_3$ & $\chi_2$ & $\chi_1$ \\
		\hline $\beta_{(23)}$  & $\chi_0$ & $\chi_1$ & $\chi_3$ & $\chi_2$ \\
		\hline $\beta_{(123)}$ & $\chi_0$ & $\chi_2$ & $\chi_3$ & $\chi_1$ \\
		\hline $\beta_{(132)}$ & $\chi_0$ & $\chi_3$ & $\chi_1$ & $\chi_2$ \\
		\hline
	\end{tabular}
$$
Thus the orbits of $\hat{\mcV}4$ under $\beta$ are
$$
	\mcO_0 = \{\chi_0\}, \quad
	\mcO_1 = \{\chi_1, \chi_2, \chi_3\}.
$$
For these two orbits I choose representatives $\chi_0$ and $\chi_1$,
respectively.

Next we compute the isotropy subgroups $H_i$ for $\chi_i$.  From the table
above, we see that all elements of $\mcS_3$ fix the trivial character of
$\mcV_4$; only the identity permutation and $(23)$ fix $\chi_1$.  That is,
$$
	H_0 = \mcS_3, \qquad H_1 = \langle (23) \rangle.
$$

The construction of irreducible \emph{representations} $U$ for $N \rtimes H$
requires knowledge of irreducible \emph{representations} $L$ for $H_i$.
However, we wish only to compute irreducible \emph{characters} $\chi_U$ for $N
\rtimes H$ and it happens that we only need irreducible \emph{characters}
$\chi_L$ for $H_i$.  Thus we skip most of step 1 of Simon's proof, which
involves the construction of $U$'s.

Recall the character tables for $H_0=\mcS_3$ and $H_1 =
\langle(23)\rangle\cong\Z_2$:
$$
	\begin{tabular}{|l||r|r|r|}
	\hline {} & $1$ & $(12)$ & $(123)$ \\
	\hline
	\hline $\chi_{L{\mcS_3,1}}$ & $1$ &  $1$ &  $1$ \\
	\hline $\chi_{L{\mcS_3,2}}$ & $1$ & $-1$ &  $1$ \\
	\hline $\chi_{L{\mcS_3,3}}$ & $2$ &  $0$ & $-1$ \\
	\hline
	\end{tabular}
	\qquad
	\begin{tabular}{|l||r|r|}
	\hline {} & $1$ & $(23)$ \\
	\hline
	\hline $\chi_{L{\langle(23)\rangle,1}}$ & $1$ &  $1$ \\
	\hline $\chi_{L{\langle(23)\rangle,2}}$ & $1$ & $-1$ \\
	\hline
	\end{tabular}
$$

Following Simon's notation, we compute cosets $x$ of $H/H_i$:
\begin{eqnarray*}
	H_0:&& x = \mcS_3; \\
	H_1:&& x = \{1, (23)\}, \{(12),(123)\}, \{(13),(132)\}.
\end{eqnarray*}
I select the following coset representatives:
\begin{eqnarray*}
	H_0:&& h_x = \mcS_3; \\
	H_1:&& h_x = (23), (12), (13).
\end{eqnarray*}
We should have $\chi_x = (\beta_{h_x})\chi_i \in \mcO_i$ being distinct and
exhausting all of $\mcO_i$.  In fact, we compute
\begin{eqnarray*}
	H_0:&& \chi_x = \chi_0; \\
	H_1:&& \chi_x = \chi_1, \chi_2, \chi_3
\end{eqnarray*}
which furnishes a sanity check.

For each choice of $\mcO_i$ and $\chi_L$ we will have an irreducible character
$\chi_U$ of $\mcV_4 \rtimes \mcS_3$.  For $\mcO_0$ there are three $\chi_L$'s;
for $\mcO_1$ there are two $\chi_L$'s.  The dimensions should turn out to be
$\dim(L) \cdot \#(H/H_i)$.  These dimensions are then $1\cdot1=1$, $1\cdot
1=1$, and $2\cdot 1=2$ for $\mcO_0$, and $1\cdot 3=3$ and $1\cdot 3=3$ for
$\mcO_1$.  This will furnish another sanity check below.

Next we compute $H_x = h_x H_i h_x^{-1}$.  Recalling the values of $h_x$ above,
we have
\begin{eqnarray*}
	H_0: 1 \mcS_3 1 &=& \mcS_3, \\
	H_1: (23)\langle(23)\rangle (23) &=& \langle(23)\rangle,\\
	H_1: (12)\langle(23)\rangle (12) &=& \langle(13)\rangle,\\
	H_1: (13)\langle(23)\rangle (13) &=& \langle(12)\rangle.
\end{eqnarray*}

Simon's formula for $\chi_U$ requires the indicator function $Ch_x: \mcS_3 \to
\{0,1\}$:
$$
	Ch_x(h) = \begin{cases}
		1, & h \in H_x; \\
		0, & h \not\in H_x.
	\end{cases}
$$
Here, we have
$$
	\begin{tabular}{|l||r|r|r|r|r|r|}
	\hline            & 1 & (12) & (13) & (23) & (123) & (132) \\
	\hline
	\hline $H_1$      & 1 & 1 & 1 & 1 & 1 & 1 \\
	\hline
	\hline $H_{(23)}$ & 1 & 0 & 0 & 1 & 0 & 0 \\
	\hline $H_{(12)}$ & 1 & 0 & 1 & 0 & 0 & 0 \\
	\hline $H_{(13)}$ & 1 & 1 & 0 & 0 & 0 & 0 \\
	\hline
	\end{tabular}
$$

Having tabulated these data, we are almost ready to apply Simon's formula V.1.7
for $\chi_U$'s:
$$
	\chi_U(n,h) \;=\; \sum_{x\in\mcO_i} \;\chi_x(n) \;Ch_x(h) \;\chi_L(h_x^{-1} h h_x).
$$
Since characters are class functions, we do not need to compute these sums for
all 24 elements of $\mcV_4 \rtimes \mcS_3$ (i.e. 4 values of $n$ and 6 values
of $h$).  Recall from appendix \ref{appx:v4s3_arithmetic} that we have the
following conjugacy classes for $\mcV_4 \rtimes \mcS_3$:
\begin{eqnarray*}
	C_1 &=& \{e1\}, \\
	C_2 &=& \{a1, b1, c1\}, \\
	C_3 &=& \{e12, e13, e23, a23, b13, c12\}, \\
	C_4 &=& \{a12, b12, a13, c13, b23, c23\}, \\
	C_5 &=& \{e123, a123, b123, c123, e132, a132, b132, c132\}.
\end{eqnarray*}
I select the first element of each conjugacy class as representatives.  Now we
may apply the formula.  Having already carefully tabulated the needed data, we
may simply (and with low risk of computational error) read off the desired
values.

\textbf{(1)}  $\chi_{U1}$ comes from $\mcO_0$ and $L$ the trivial representation
on $\mcS_3$; $h_x=1$.
$$
\begin{tabular}{rrrrrrrrrr}
          $n$  & $h$&      & $\chi_x(n)$ &       & $Ch_x(h)$ &       & $\chi_L(h_x^{-1} h h_x)$ \\
  \hline
$\chi_{U1}(e,$ & $1)$  &$=$& $\chi_0(e)$ &$\cdot$& 1         &$\cdot$& 1                      & $=1$\\
$\chi_{U1}(a,$ & $1)$  &$=$& $\chi_0(a)$ &$\cdot$& 1         &$\cdot$& 1                      & $=1$\\
$\chi_{U1}(e,$ & $12)$ &$=$& $\chi_0(e)$ &$\cdot$& 1         &$\cdot$& 1                      & $=1$\\
$\chi_{U1}(a,$ & $12)$ &$=$& $\chi_0(a)$ &$\cdot$& 1         &$\cdot$& 1                      & $=1$\\
$\chi_{U1}(e,$ & $123)$&$=$& $\chi_0(e)$ &$\cdot$& 1         &$\cdot$& 1                      & $=1$\\
\end{tabular}
$$


\textbf{(2)}  $\chi_{U2}$ comes from $\mcO_0$ and $L$ the parity representation
on $\mcS_3$; $h_x=1$.
$$
\begin{tabular}{rrrrrrrrrr}
          $n$  & $h$&      & $\chi_x(n)$ &       & $Ch_x(h)$ &       & $\chi_L(h_x^{-1} h h_x)$ \\
  \hline
$\chi_{U2}(e,$ & $1)$  &$=$& $\chi_0(e)$ &$\cdot$& 1         &$\cdot$&  1                     & $= 1$\\
$\chi_{U2}(a,$ & $1)$  &$=$& $\chi_0(a)$ &$\cdot$& 1         &$\cdot$&  1                     & $= 1$\\
$\chi_{U2}(e,$ & $12)$ &$=$& $\chi_0(e)$ &$\cdot$& 1         &$\cdot$& -1                     & $=-1$\\
$\chi_{U2}(a,$ & $12)$ &$=$& $\chi_0(a)$ &$\cdot$& 1         &$\cdot$& -1                     & $=-1$\\
$\chi_{U2}(e,$ & $123)$&$=$& $\chi_0(e)$ &$\cdot$& 1         &$\cdot$&  1                     & $= 1$\\
\end{tabular}
$$

\textbf{(3)}  $\chi_{U3}$ comes from $\mcO_0$ and $L$ the two-dimensional representation
on $\mcS_3$; $h_x=1$.
$$
\begin{tabular}{rrrrrrrrrr}
          $n$  & $h$&      & $\chi_x(n)$ &       & $Ch_x(h)$ &       & $\chi_L(h_x^{-1} h h_x)$ \\
  \hline
$\chi_{U3}(e,$ & $1)$  &$=$& $\chi_0(e)$ &$\cdot$& 1         &$\cdot$&  2                     & $= 2$\\
$\chi_{U3}(a,$ & $1)$  &$=$& $\chi_0(a)$ &$\cdot$& 1         &$\cdot$&  2                     & $= 2$\\
$\chi_{U3}(e,$ & $12)$ &$=$& $\chi_0(e)$ &$\cdot$& 1         &$\cdot$&  0                     & $= 0$\\
$\chi_{U3}(a,$ & $12)$ &$=$& $\chi_0(a)$ &$\cdot$& 1         &$\cdot$&  0                     & $= 0$\\
$\chi_{U3}(e,$ & $123)$&$=$& $\chi_0(e)$ &$\cdot$& 1         &$\cdot$&  1                     & $=-1$\\
\end{tabular}
$$

\textbf{(4)}  $\chi_{U4}$ comes from $\mcO_1$ and $L$ the trivial representation
on $\langle(23)\rangle$; $h_x=(23),(12),(13)$.
$$
\begin{tabular}{rrrrrrrrl}
          $n$  & $h$&      & $\chi_x(n)$ &       & $Ch_x(h)$ &       & $\chi_L(h_x^{-1} h h_x)$ \\
\hline

$\chi_{U4}(e,$ & $1)$  &$=$& $\chi_1(e)$ &$\cdot$& 1         &$\cdot$&  1                     & \\
               &       &$+$& $\chi_2(e)$ &$\cdot$& 1         &$\cdot$&  1                     & \\
               &       &$+$& $\chi_3(e)$ &$\cdot$& 1         &$\cdot$&  1                     & $=1+1+1 = 3$\\
\\

$\chi_{U4}(a,$ & $1)$  &$=$& $\chi_1(a)$ &$\cdot$& 1         &$\cdot$&  1                     & \\
               &       &$+$& $\chi_2(a)$ &$\cdot$& 1         &$\cdot$&  1                     & \\
               &       &$+$& $\chi_3(a)$ &$\cdot$& 1         &$\cdot$&  1                     & $=1-1-1 = -1$\\
\\

$\chi_{U4}(e,$ & $12)$ &$=$& $\chi_1(e)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_2(e)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_3(e)$ &$\cdot$& 1         &$\cdot$&  $\chi_L((13)(12)(13)$ & $=\chi_L(23) = 1$\\
\\

$\chi_{U4}(a,$ & $12)$ &$=$& $\chi_1(a)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_2(a)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_3(a)$ &$\cdot$& 1         &$\cdot$&  $\chi_L((13)(12)(13)$ & $=-\chi_L(23) = -1$\\
\\

$\chi_{U4}(e,$ & $123)$&$=$& $\chi_1(e)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_2(e)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_3(e)$ &$\cdot$& 0         &       &                        & $=0$\\
\\

\end{tabular}
$$

\textbf{(5)}  $\chi_{U5}$ comes from $\mcO_1$ and $L$ the parity representation
on $\langle(23)\rangle$; $h_x=(23),(12),(13)$.
$$
\begin{tabular}{rrrrrrrrl}
          $n$  & $h$&      & $\chi_x(n)$ &       & $Ch_x(h)$ &       & $\chi_L(h_x^{-1} h h_x)$ \\
\hline

$\chi_{U5}(e,$ & $1)$  &$=$& $\chi_1(e)$ &$\cdot$& 1         &$\cdot$&  1                     & \\
               &       &$+$& $\chi_2(e)$ &$\cdot$& 1         &$\cdot$&  1                     & \\
               &       &$+$& $\chi_3(e)$ &$\cdot$& 1         &$\cdot$&  1                     & $=1+1+1 = 3$\\
\\

$\chi_{U5}(a,$ & $1)$  &$=$& $\chi_1(a)$ &$\cdot$& 1         &$\cdot$&  1                     & \\
               &       &$+$& $\chi_2(a)$ &$\cdot$& 1         &$\cdot$&  1                     & \\
               &       &$+$& $\chi_3(a)$ &$\cdot$& 1         &$\cdot$&  1                     & $=1-1-1 = -1$\\
\\

$\chi_{U5}(e,$ & $12)$ &$=$& $\chi_1(e)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_2(e)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_3(e)$ &$\cdot$& 1         &$\cdot$&  $\chi_L((13)(12)(13)$ & $=\chi_L(23) = -1$\\
\\

$\chi_{U5}(a,$ & $12)$ &$=$& $\chi_1(a)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_2(a)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_3(a)$ &$\cdot$& 1         &$\cdot$&  $\chi_L((13)(12)(13)$ & $=-\chi_L(23) = 1$\\
\\

$\chi_{U5}(e,$ & $123)$&$=$& $\chi_1(e)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_2(e)$ &$\cdot$& 0         &       &                        & \\
               &       &$+$& $\chi_3(e)$ &$\cdot$& 0         &       &                        & $=0$\\
\\

\end{tabular}
$$

Collecting these results, we have the character table for $\mcV_4 \rtimes \mcS_3$:
$$
\begin{tabular}{|l||r|r|r|r|r|}
	\hline {} & $e1$ & $a1$ & $e12$ & $a12$ & $e123$ \\
	\hline
	\hline $\chi_{U1}$ &  1 &  1 &  1 &  1 &  1 \\
	\hline $\chi_{U2}$ &  1 &  1 & -1 & -1 &  1 \\
	\hline $\chi_{U3}$ &  2 &  2 &  0 &  0 & -1 \\
	\hline $\chi_{U4}$ &  3 & -1 &  1 & -1 &  0 \\
	\hline $\chi_{U5}$ &  3 & -1 & -1 &  1 &  0 \\
	\hline
\end{tabular}
$$

%% ================================================================
\appendix
\newpage
\section{Semidirect products}
\label{sec:semi}

\emph{(This section is merely review material from abstract algebra.)}

\begin{defn}
\label{def:semi_operation}
Let $G$ be a group with $N \lhd G$ and $K \subseteq G$.  Furthermore suppose
that $G=NK$ and $N \cap K = \{1\}$.  If $K$ acts on $N$ (formally, if there is
a homomorphism $K \to \Aut(N)$), we write
$$
	N \rtimes K
$$
and call this the \bfidx{semidirect product} of $N$ and $K$.  We sometimes
write the action of $k\in K$ on $n\in N$ as $n^k$.  Since $G=NK$, any $g \in G$
may be written in the form $(n,k)$.  Multiplication in $G$ is
$$
	(n_1, k_1)(n_2, k_2) = (n_1 {n_2}^{k_1}, k_1 k_2).
$$
\end{defn}

\begin{rem}
\label{rem:semi_inversion}
How does inversion work?  Given $(n, k)$ we need to find $(m, j)$
such that $(n,k)(m,j)=(1,1)$.  This gives
$$
	(1,1) = (n,k)(m,j) = (n {m}^{k}, k j)
$$
which forces $j = k^{-1}$.  Then, in the first slot,
$n {m}^{k} = 1$ forces
\begin{eqnarray*}
	{m}^{k} &=& {n}^{-1} \\
	m &=& ({n}^{-1})^{k^{-1}}.
\end{eqnarray*}
Thus,
$$
	(n,k)^{-1} = ((n^{-1})^{k^{-1}}, k^{-1}).
$$
\end{rem}

\beginex
Dihedral groups may be written as
$$
	\mcD_n = \mcC_n \rtimes \mcC_2 = \pig{\rho} \rtimes \pig{\phi}
$$
where $\rho$ is the rotation of order $n$ and $\phi$ is the flip.
The flip acts on the rotations by inverting them:
$$
	\rho^i \phi^j \rho^k \phi^\ell =
	\begin{cases}
		\rho^{i+k} \phi^{j+\ell}, & j \textrm{ even} \\
		\rho^{i-k} \phi^{j+\ell}, & j \textrm{ odd} \\
	\end{cases}
$$
which is to say (since there are only two distinct powers of $\phi$)
\begin{eqnarray*}
	\rho^i      \rho^k \phi^\ell &=& \rho^{i+k} \phi^{\ell} \\
	\rho^i \phi \rho^k \phi^\ell &=& \rho^{i-k} \phi^{1+\ell}.
\end{eqnarray*}
This is the familiar transposition rule
$$
	\phi\rho^k = \rho^{-k}\phi.
$$
\mendex

\beginex
The $T$ group (the other non-abelian group of order 12, besides
$\mcA_4$ and $\mcD_6$) is
$$
	T = \Z_3 \rtimes \Z_4
$$
where $\Z_4$ acts on $\Z_3$ by inversion.  That is, $1$ and $3$ in $\Z_4$
negate elements of $\Z_3$; $0$ and $2$ in $\Z_4$ leave elements of $\Z_3$
alone.
\mendex

%% ================================================================
\newpage
\section{Arithmetic in $\mcV_4 \rtimes \mcS_3$}
\label{appx:v4s3_arithmetic}

Some of the examples in this note make use of $\mcV_4 \rtimes \mcS_3$.  Here I
work out arithmetic for this group.

Let $$\mcV_4=\{e,a,b,c\}$$ be the \bfidx{Klein-four group} as usual.
Recall that \bfidx{symmetric group} $\mcS_3$ is
$$
	\mcS_3 = \{1, (12),(13),(23), (123), (132)\}.
$$
Let $\mcS_3$ act on $\mcV_4$ by permuting the symbols $a$, $b$, and $c$.  For
example, $a^{(123)}=b$.  Then we can write the semidirect product
$$
	\mcV_4 \rtimes \mcS_3.
$$
It can be shown (I won't here) that this is isomorphic to $\mcS_4$.

Let $u,v \in \mcV_4$ and $\sigma,\tau \in \mcS_3$.
Using definition \ref{def:semi_operation}, we can write the product
$$
	(u,\sigma)(v,\tau) = (u v^\sigma, \sigma\tau).
$$
For example,
\begin{eqnarray*}
	(a,(123))(b,(12)) &=& (ab^{123}),(123)(12)) \\
	&=& (ac,(13)) \\
	&=& (b, 13).
\end{eqnarray*}

In particular, we have a \bfidx{square formula}
$$
	(u,\sigma)^2 = (u u^\sigma,\sigma^2)
$$
and a \bfidx{cube formula}
\begin{eqnarray*}
	(u,\sigma)^3 &=& (u u^\sigma,\sigma^2)(u,\sigma) \\
	&=& (u u^\sigma u^{\sigma^2}, \sigma^3).
\end{eqnarray*}

Also from remark \ref{rem:semi_inversion}, and recalling that all elements of
$\mcV_4$ are their own inverses, we have an \bfidx{inverse formula}
$$
	(u,\sigma)^{-1} = (u^{\sigmainv}, \sigmainv).
$$

From this we obtain a \bfidx{conjugation formula}
\begin{eqnarray*}
	(v,\tau)^{-1} (u,\sigma)(v,\tau)
	&=& (v^{\tauinv}, \tauinv) (u v^\sigma, \sigma\tau) \\
	&=& (v^{\tauinv} (u v^\sigma)^{\tauinv}, \tauinv\sigma\tau).
\end{eqnarray*}

xxx conjugacy classes:  first compute element orders \ldots .
