Exercise 4.2.2 (Left regular representation of $S_3$)

Question

List the elements of $S_3$ as $1, (1\ 2), (2\ 3) (1\ 3), (1\ 2\ 3), (1\ 3\ 2)$ and label these with the integers $1,2,3,4,5,6$ respectively. Exhibit the image of each element of $S_3$ under the left regular representation of $S_3$ into $S_6$.

Richard Ganaye · Accepted Answer

\begin{proof} To label the elements of $S_3$, we choose the bijection $f$ given by
\begin{center}
\begin{tabular}{c|cccccc}
$k$ & $1$ & $2$ & $3$ & $4$ &$5$ & $6$ \
\hline
$f(k)$ & $()$ &$(1\ 2)$& $(2\ 3)$ &$ (1\ 3)$& $(1\ 2\ 3)$ & $(1\ 3\ 2)$ 
\end{tabular}
\end{center}
For each $\gamma \in S_3$, $\sigma_\gamma$ is defined by
$$\sigma_\gamma(i) = j \iff  \gamma \cdot f(i) = f(j)\qquad (0 \leq i \leq 6,\ 0 \leq j \leq 6),$$
or equivalently, $\varphi _ \gamma \circ f = f \circ \sigma_\gamma$, where $\varphi : \gamma \mapsto \varphi_\gamma$ is the homomorphism afforded by the action of $S_3$ on itself.

Then
\begin{align*}
(1\ 2) \circ () = (1\ 2) &	ext{ and so }\sigma_{(1\ 2)}(1) = 2,\
(1\ 2) \circ (1\ 2) = () &	ext{ and so }\sigma_{(1\ 2)}(2) = 1,\
(1 \ 2) \circ (2\ 3) = (1\ 2\ 3)&	ext{ and so }\sigma_{(1\ 2)}(3) = 5,\
(1\ 2) \circ (1\ 3) = (1\ 3\ 2) &	ext{ and so }\sigma_{(1\ 2)}(4) = 6,\
(1\ 2) \circ (1\ 2\ 3) = (2\ 3) &	ext{ and so }\sigma_{(1\ 2)}(5) = 3,\
(1\ 2) \circ (1\ 3\ 2) = (1\ 3) &	ext{ and so }\sigma_{(1\ 2)}(6) = 4,\
\end{align*}
therefore 
$$\sigma_{(1\ 2)}= \begin{pmatrix} 1& 2 & 3 & 4 & 5 & 6\ 2 & 1 & 5 & 6 & 3 & 4\end{pmatrix} = (1\ 2) (3\ 5) (4\ 6).$$
Similarly
\begin{align*}
(1\ 2\ 3) \circ () = (1\ 2\ 3) &	ext{ and so }\sigma_{(1\ 2\ 3)}(2) = 5,\
(1\ 2\ 3) \circ (1\ 2) = (1\ 3) &	ext{ and so }\sigma_{(1\ 2\ 3)}(2) = 4,\
(1 \ 2 \ 3) \circ (2\ 3) = (1\ 2)&	ext{ and so }\sigma_{(1\ 2\ 3)}(3) = 2,\
(1\ 2\ 3) \circ (1\ 3) = (2\ 3) &	ext{ and so }\sigma_{(1\ 2\ 3)}(4) = 3,\
(1\ 2\ 3) \circ (1\ 2\ 3) = (1\ 3\ 2) &	ext{ and so }\sigma_{(1\ 2\ 3)}(5) = 6,\
(1\ 2\ 3) \circ (1\ 3\ 2) = () &	ext{ and so }\sigma_{(1\ 2\ 3)}(6) = 1,\
\end{align*}
therefore 
$$\sigma_{(1\ 2\ 3)}= \begin{pmatrix} 1& 2 & 3 & 4 & 5 & 6\ 5 & 4 & 2 & 3 & 6 & 1\end{pmatrix} = (1\ 5\ 6)(2\ 4 \ 3).$$

Since $S_3 = \langle (1\ 2), (1\ 2\ 3) angle$, this gives all the maps $\sigma_\gamma$ for $\gamma \in S_3$. For instance
$$\sigma_{(2\ 3)}  = \sigma_{(1\ 2)(1\ 2\ 3)} = \sigma_{(1\ 2)} \sigma_{(1\ 2\ 3)} = (1\ 2) (3\ 5) (4\ 6)\cdot (1\ 5\ 6)(2\ 4 \ 3) = (1\ 3)(2\ 6)(4\ 5)$$
More generally
\begin{align*}
\sigma_{()} &= (),\
\sigma_{(1\ 2)}&= (1\ 2) (3\ 5) (4\ 6),\
\sigma_{(2\ 3)} &= (1\ 3)(2\ 6)(4\ 5),\
\sigma_{(1\ 3)} &=(1\ 4)(2\ 5)(3\ 6),\
\sigma_{(1\ 2\ 3)}&= (1\ 5\ 6)(2\ 4 \ 3),\
\sigma_{(1\ 3\ 2)} &= (1 \ 6\ 5)(2 \ 3\ 4).
\end{align*}
\end{proof}
Verification with Sagemath:
\begin{verbatim}
sage: a , b = [(1,5,6),(2,4,3)], [(1,2),(3,5),(4,6)]
sage: G = PermutationGroup([a,b])
sage: G.list()

[(),
 (1,2)(3,5)(4,6),
 (1,5,6)(2,4,3),
 (1,6,5)(2,3,4),
 (1,3)(2,6)(4,5),
 (1,4)(2,5)(3,6)]
 sage: G.stabilizer(1).list()
[()]
\end{verbatim}

Exercise 4.2.2 (Left regular representation of $S_3$)

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$k$	$1$	$2$	$3$	$4$	$5$	$6$

$f (k)$	$()$	$(1 2)$	$(2 3)$	$(1 3)$	$(1 2 3)$	$(1 3 2)$

Exercise 4.2.2 (Left regular representation of $S_3$)

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