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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.

Shaun
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Walker
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  • This is the set of functions that stabilize $4$. You don't need to show that $H$ is a normal subgroup, so the condition $S_4/\ker \varphi = H$ is not required, you just need to find a homomorphism from $S_3$ into $S_4$. To give you an idea of what's happening this will show that if you exclude the fourth element of the set, it's just like there are three elements in the set. Can you take it from here? – CyclotomicField Aug 15 '22 at 16:38
  • I have hard time finding what the homomorphism should be. It's should be sending functions $f: {1,2,3} \to {1,2,3}$ to functions $g: {1,2,3,4} \to {1,2,3,4}$, but with the restriction that $f(4)=4$? – Walker Aug 15 '22 at 17:15

1 Answers1

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$f\in H$ 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 $H\cong S_3$.

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.

Sourav Ghosh
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