$\operatorname{Aut}(S_4)$ is isomorphic to $S_4$

Question

I already proved this, but I think I can reduce my solution.

My solution : There are 4 Sylow 3-subgroup of $S_4$, and denote the set of Syl 3-subgroups by $P=\{P_1,P_2,P_3,P_4\}$.

Then, by a group action $\operatorname{Aut}(S_4) \times P \to P$ defined by $(f,P_i) \to f(P_i)$, obtain a homomorphism $\phi:\operatorname{Aut}(S_4)\to S_4(=\operatorname{Perm}(P))$.

To show that $\ker(\phi)=0$, I suppose $f \in \operatorname{Aut}(S_4)$ fix every Sylow 3-subgroup.

Then I can derive that $f$ fixes every 3-cycle rather easily.

But proving that $f$ fixes $2$-cycle is very long and not seems good, and my question arises here. (I consider every case that $f(ab)$ can be)

Does anyone have a smart idea for proving $f$ fixes 2-cycle?

Answer 1

Assuming that you know that $f$ fixes every $3$-cycle, it follows that $f$ also fixes every even permutation (because $A_4$ is generated by $3$-cycles).

Now $A_n$ is the commutator subgroup of $S_n$ when $n \geq 3$ (proof: write a $3$-cycle as the commutator of two transpositions). So the parity of an permutation is invariant under an automorphism of $S_n$.

Thus $f$ permutes the transpositions $(12), (13), (14), (23), (24), (34)$.

Now $(123) = f(123) = f((13)(12)) = f(13)f(12)$ so $f(12)$ is $(12)$ or $(13)$ or $(23)$.

Also $(12)(34) = f((12)(34)) = f(12)f(34)$ so $f(12)$ is $(12)$ or $(34)$.

Thus $f(12)$ is $(12)$ hence $f = 1$.

Answer 2

Let's denote a conjugacy class of representative $\sigma\in S_4$ as: $$\operatorname{cl}(\sigma,o(\sigma),\text{cycle type},\text{size})$$ The conjugacy classes of $S_4$ are then: \begin{alignat}{1} &\operatorname{cl}((),1,(1,1,1,1),1) \\ &\operatorname{cl}((12),2,(2,1,1),6) \\ &\operatorname{cl}((12)(34),2,(2,2),3) \\ &\operatorname{cl}((123),3,(3,1),8) \\ &\operatorname{cl}((1234),4,(4),6) \\ \tag1 \end{alignat} Since automorphisms send conjugacy classes to conjugacy classes, while preserving elements' order and classes' size, from $(1)$ follows that every element of $\operatorname{Aut}(S_4)$ is class-preserving, and hence it is inner. Therefore: $$\operatorname{Aut}(S_4)=\operatorname{Inn}(S_4)\stackrel{Z(S_4)=1}{\cong}S_4$$