Prove that $S_3$ and a subgroup $H$ of $S_4$ are isomorphic.

Question

It is well known that $V＝\{1, t, u, v\}$ is a normal subgroup of $S_4$, where $t＝(12)(34), u＝(13)(24), v＝(14)(23).$ Let $H$ be the set of all $\pi$ in $S_4$ such that $\pi(4)＝4.$ Prove that $H$ is a subgroup of $S_4$, $S_4＝VH$, and also that $H\cong S_3.$

Since $\pi(4)＝4,$ $\pi^{-1} (4)＝4.$ Suppose $\pi_1, \pi_2 \in H$, $\pi_1\pi_2^{-1} (4)＝4$. Therefore, $\pi_1\pi_2^{-1}\in H$.

$H$ is a subgroup of $S_4$.

I know the cycle notations for permutation of elements in $H$ and $S_3$ are one-to-one correspondence but I don't have a neat proof of 'isomorphic'.

Since $V$ is a normal subgroup of $S_4$, $H$ is a subgroup of $S_4$, $VH$ is a subgroup of $S_4$. $V\cap H＝\{1\}$, for $a\in V, b\in H$, $ab$ is unique. Hence $|VH|＝|V|\times |H|＝24＝|S_4|$, $VH＝S_4$.

I wonder if it can be proved without counting the orders.

Thank you.

Answer 1

1. It's straightforward to write up an explicit isomorphism $S_3\to H$: simply extend a permutation $\sigma$ on $\{1,2,3\}$ by $\sigma(4):=4$, and conversely, restrict any $\rho\in H$ to $\{1,2,3\}$.
2. Your counting proof is perfect. Alternatively, it's enough to show that every element of a generator set is in $VH$. Taking e.g. the transpositions (cycles of length $2$), by symmetry, it all boils down to explicitly showing, say $(3\,4)\in VH$.