Show that the set $H=\{f \in S_4 \mid f(4)=4\} \subset S_4$ is isomorphic with $S_3$.

Question

Show that the set $H=\{f \in S_4 \mid f(4)=4\} \subset S_4$ is isomorphic with $S_3$.

I think I need to construct a homomorphism $\varphi:S_4 \to S_3$ such that $S_4/\ker \varphi = H$? The problem I'm facing is that I don't quite understand what the condition on $H$ is.

I think that $H$ is the set of permutations in $S_4$ with $4$ being fixed? The notation here is somewhat odd. Usually I've denoted the elements of $S_4$ as either in the "matrix" notation or in cycle.

Answer 1

$f\in H$ implies $f(4) =4$ .

Hence number of permutations / functions that fixes $4$ are exactly $3! =6$

$H$ is non abelian group of order $6$ and there exactly two groups of order $6$ upto isomorphism $S_3$ and $\Bbb{Z_6}$.

Hence $H\cong S_3$.

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Here the identification or isomorphism map from $H$ to $S_3$ is obtained by ignoring $4$ completely fix else as it is.

$\begin{pmatrix} 1&2&3&4\\2&3&1&4\end{pmatrix}\to \begin{pmatrix} 1&2&3\\2&3&1\end{pmatrix}$

Try to construct such map.