# Group automorphism of $S_4$

I want to prove that $Aut(S_4) \cong S_4$. I saw $\operatorname{Aut}(S_4)$ is isomorphic to $S_4$ but i was troubled.

I know $Z(S_4)=1$, and so $S_4 \cong Inn(S_4)$. Thus it is sufficien to show that every automorphism of $S_4$ is inner.

Now suppose that ${ \rm Syl}_3(S_4)= \lbrace P_1,P_2, P_3, P_4 \rbrace$. Thus $Aut(S_4)$ acts on ${ \rm Syl}_3(S_4)$.

$\textbf{Watson}$ say that this action is as follow: $(P_i,f)=f(P_i)=P_{\sigma_f(i)}$. I still dont understan this action. Also why every automorphism of $S_4$ is inner?

• ${\rm Aut}(S_4)$ acts on ${ \rm Syl}_3(S_4)$ via $f \bullet P_j := f(P_j) = P_{\sigma_f(j)} \in { \rm Syl}_3(S_4)$ for every $3$-Sylow subgroup $P_j ≤ S_4$. – Watson Sep 7 '16 at 9:01
• I know but $f:S_4 \longrightarrow S_4$ how defined? – Hana Sep 7 '16 at 9:04
• Here $f$ is any automorphism of $S_4$. Any automorphism of $S_4$ will induce a permutation of the $4$ $3$-Sylow subgroups of $S_4$, i.e. a permutation of $\{P_1,P_2,P_3,P_4\}$, i.e. an element of $S_4$, which I denoted by $\sigma_f$. – Watson Sep 7 '16 at 9:04
• So $\phi : f \mapsto \sigma_f$ is a map from ${\rm Aut}(S_4) \to S_4$, which can be verified to be an homomorphism. Does this answer your question? – Watson Sep 7 '16 at 9:09
• While $f:S_4 \to S_4$ is an automorphism, $\sigma_f$ is an element of $S_4$ which maps $j$ to $\sigma_f(j)$ such that $f(P_j) = P_{\sigma_f(j)}$. – Watson Sep 7 '16 at 9:12

To move this old question out of the unanswered list, here's an argument which works up to $$n=5$$.
In general, given a group $$G$$, the group $$\operatorname{Aut}(G)$$ acts on the set of the conjugacy classes of $$G$$, $$\mathscr{C}:=\{\operatorname{cl}(a), a\in G\}$$, via $$\varphi\cdot \operatorname{cl}(a):=\varphi(\operatorname{cl}(a))$$. Essentially, the proof that this is indeed an action boils down to proving that $$\varphi(\operatorname{cl}(a))=\operatorname{cl}(\varphi(a))$$. Since inner automorphisms fix each conjugacy class, we have that $$\operatorname{Inn}(G)\le\operatorname{ker}\phi$$, where $$\phi\colon\operatorname{Aut}(G)\to S_\mathscr{C}$$ is the homomorphism equivalent to the said action.
Now, if $$G=S_n$$, then $$\operatorname{Inn}(G)=\operatorname{ker}\phi$$, because the stabilizer of the conjugacy class of the transpositions is precisely equal to $$\operatorname{Inn}(S_n)$$ (see e.g. here). Therefore (first homomorphism theorem), $$\operatorname{Aut}(S_n)/\operatorname{Inn}(S_n)\cong\phi(\operatorname{Aut}(S_n))$$. If distinct conjugacy classes are "tagged" by the distinct orders of their elements, then the action is trivial, because any automorphism (order-preserving) is then class-preserving, and finally $$\operatorname{Aut}(S_n)=\operatorname{Inn}(S_n)$$. This happens for $$n=3$$, being the three distinct conjugacy classes of $$S_3$$ made of elements of order $$1$$ (the identity), $$2$$ (the three $$2$$-cycles) and $$3$$ (the two $$3$$-cycles). Therefore $$\operatorname{Aut}(S_3)=\operatorname{Inn}(S_3)\cong S_3$$. By relaxing such a "strong" sufficient condition ("each conjugacy class, an order of its elements"), we can encompass $$n=4$$ and $$n=5$$ cases, too. In fact, if $$S_n$$ has two conjugacy classes of elements of the same order, but their sizes are different, than again all the automorphisms are class-preserving (being $$|\varphi(\operatorname{cl}(\sigma))|=|\operatorname{cl}(\sigma)|$$), and hence again the action is trivial (whence again $$\operatorname{Aut}(S_n)=\operatorname{Inn}(S_n)$$). For $$n=4$$, the conjugacy classes of the transpositions and of the double transpositions (both whose elements have order $$2$$) have sizes $$6$$ and $$3$$, respectively, and hence no one automorphism can "swap" each other. Likewise, for $$n=5$$, the conjugacy classes of the transpositions and of the double transpositions have sizes $$10$$ and $$15$$, respectively, and the same conclusion as for $$n=4$$ does hold.
This simple approach can't support any further in the $$n=6$$ case, where we have conjugacy classes of elements of the same order, which have also the same size.