\begin{problem*}
Following the method of section V.1 of Simon's \emph{Representations of
Finite and Compact Groups}, compute the character table for $\mcV_4 \rtimes
\mcS_3$.

\begin{answer}
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 the previous homework assignment 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 $\mcV_4 \rtimes \mcS_3$:
$$
\begin{tabular}{|l||r|r|r|r|r|}
	\hline {} & $e1$ & $a1$ & $e12$ & $a12$ & $e123$ \\
	\hline
	\hline $\chi_{U1}$ &  1 &  1 &  1 &  1 &  1 \\
	\hline $\chi_{U2}$ &  1 &  1 & -1 & -1 &  1 \\
	\hline $\chi_{U3}$ &  2 &  2 &  0 &  0 & -1 \\
	\hline $\chi_{U4}$ &  3 & -1 &  1 & -1 &  0 \\
	\hline $\chi_{U5}$ &  3 & -1 & -1 &  1 &  0 \\
	\hline
\end{tabular}
$$



\end{answer}
\end{problem*}
