Showing that an automorphism of $S_4$ fixing each Sylow 3-subgroup must be the identity.

Question

I am working on a Hungerford exercise and trying to show that $S_4$ is isomorphic to its automorphism group. I know that there are four Sylow 3-groups in $S_4$ and the four Sylow groups exhaust the 3-cycles of $S_4$. Now, denote the 4 Sylow groups by $P_1$, $P_2$, $P_3$, $P_4$. If $f$ is an automorphism of $S_4$ that sends $P_i$ to $P_i$ for each $i$, then I want to show that $f$ must be the identity map of $S_4$. However it seems trickier than I thought and I cannot find a way to prove it. Could anyone please help me how to show that such $f$ must be the identity?

Answer 1

If $f\in Aut(S_4)$, first note that $f$ fixes (setwise) the set of all $3$-cycles as $f$ preserves elements of order $3$. Thus $f$ fixes $A_4$ as $A_4$ is generated by $3$-cycles. Therefore $f$ preserves the parity of the permutations.

If $\sigma$ is a transposition, then $f(\sigma)$ is an odd permutation of order $2$, hence a transposition as well. Let $P_1=\langle(123)\rangle$, $P_2=\langle(124)\rangle$, $P_3=\langle(134)\rangle$ and $P_4=\langle(234)\rangle$. Consider conjugations by transpositions of $P_i$'s given in the table: $$\begin{array}{c|cccc} &P_1&P_2&P_3&P_4\\\hline (12)&P_1&P_2&P_4&P_3\\ (13)&P_1&P_4&P_3&P_2\\ (14)&P_4&P_2&P_3&P_1\\ (23)&P_1&P_3&P_2&P_4\\ (24)&P_3&P_2&P_1&P_4\\ (34)&P_2&P_1&P_3&P_4 \end{array}$$ Since $(12)$ conjugates $P_3$ to $P_4$, $f((12))$ conjugates $f(P_3)=P_3$ to $f(P_4)=P_4$, so the only possibility for $f((12))$ (by looking in the table) is $(12)$. Similarly, $f$ fixes every transposition. Since transpositions generate $S_4$, $f=id$.