# Prove that $S_4 \cong V_4 \rtimes_\phi S_3$ for any isomorphism $\phi: S_3 \to \text{Aut}(V_4)$

Note that $$\text{Aut}(V_4) \cong S_3$$. I know how to prove that $$S_4$$ isomorphic to some semidirect product of $$V_4$$ and $$S_3$$. I know if it works for an isomphorism it works for any isomorphism. However, I'm having trouble seeing that the $$\phi$$ must be an isomorphism of $$\phi: S_3 \to \text{Aut}(V_4)$$ (i.e. the kernel is trivial). Is there a better way to check this without doing each case (kernel cannot be 3-cycles, or all of $$S_3$$)?

• Can you elaborate what you want to ask better? I have trouble understanding this part of the question: "However, I'm having trouble seeing that the $\phi$ must be an isomorphism of $\phi: S_3 \to \text{Aut}(V_4)$ (i.e. the kernel is trivial)." Are you asking us how to find a good $\phi$?
– user593746
Nov 4 '18 at 19:23
• If some element of $S_3$ were in the kernel of $\phi$, then that element would commute with all of $V_4$. As $V_4$ is itself abelian you would get an abelian subgroup of order 8 or 12. $S_4$ has no such subgroups. Nov 4 '18 at 19:27
• There is of course the direct product $V_4\times S_3$ when $\ker\phi=S_3$ but that isn't isomorphic to $S_4$. Nov 4 '18 at 19:27
• Ah, I see what @Raekye wanted to ask: Why should every $\phi\colon S_3\xrightarrow{\cong}\operatorname{Aut}(V_4)$ gives $V_4\rtimes_\phi S_3\to S_4$ an isomorphism, without having to do some explicit calculation with elements? For that, answer to this question would be very helpful. Nov 4 '18 at 19:32
• @Zvi In general for a semidirect product we just need a homomorphism from $S_3$ to $\text{Aut}(V_4)$. As others have pointed out, the trivial homomorphism gives a direct product of $V_4$ and $S_3$, which is not isomorphic to $S_4$. However, there could be other candidates for homomorphisms that give the semidirect product $V_4 \rtimes S_3 \cong S_4$. The best way I could think of to classify them is by kernels (which have to be normal subgroups). If we show that the kernel of $\phi$ must be trivial, then we know the semidirect product comes from an isomorphism of $S_3 \to \text{Aut}(V_4)$ Nov 4 '18 at 20:02

To show that $$S_4 \cong V_4 \rtimes S_3$$, first note that $$V_4$$ is isomorphic to the double transpositions in $$S_4$$, and this $$V_4$$ is normal in $$S_4$$. Consider an isomorphic copy of $$S_3$$ in $$S_4$$ in the usual way. Note that their intersections are trivial. Denote the two subgroups as $$H$$ and $$K$$, then $$HK$$ is a subgroup of $$S_4$$ of size $$\frac{|H||K|}{|H \cap K|} = 4 \cdot 6 = |S_4|$$, so $$HK$$ is equal to $$S_4$$, meaning $$S_4$$ is a semidirect product of $$V_4$$ and $$S_3$$.
To show that $$S_4 \cong V_4 \rtimes_\phi S_3$$ for some isomorphism of $$S_3 \to \text{Aut}(V_4)$$ (rather than, more generally, some homomorphism), note that the kernel of $$\phi$$ must be a normal subgroup of $$S_3$$. We make use of Jyrki's comment. The only nontrivial normal subgroups of $$S_3$$ are $$C_3$$ and all of $$S_3$$, which includes $$C_3$$. Note that $$V_4$$ is abelian. If $$C_3$$ is in the kernel of $$\phi$$, then
$$\{ (h, k) \in V_4 \rtimes_\phi S_3 | h \in V_4, k \in C_3 \}$$
is an abelian subgroup of order 12. However, $$V_4$$ has no abelian subgroup of order 12. Therefore the kernel of $$\phi$$ must be trivial, i.e. it is an isomorphism of $$S_3 \to \text{Aut}(V_4)$$.
To show that $$S_4$$ is a semidirect product of $$V_4$$ and $$S_3$$ for any isomorphism of $$S_3 \to \text{Aut}(V_4)$$, see the answer to this question.