Let \(S_4\) be the group of permutations of \(\{1,2,3,4\}\). Determine the order of \(\operatorname{Aut}(S_4)\). Justify your answer.
Solution
The center of \(S_4\) is trivial, so conjugation embeds \(S_4\) into \(\operatorname{Aut}(S_4)\). It remains to show every automorphism is inner.
There are exactly four subgroups \(H_i\) of order \(3\), where \(H_i\) consists of the identity and the two \(3\)-cycles fixing \(i\). Any automorphism permutes these four subgroups. Inner automorphisms realize every permutation of them, so after composing with an inner automorphism we may assume an automorphism \(\sigma\) fixes every \(H_i\).
The transpositions form the unique conjugacy class of six elements of order \(2\), so \(\sigma\) preserves that class. A transposition \((ij)\) is characterized among transpositions by the property that it and \(H_k\) generate \(S_4\) exactly when \(k\in\{i,j\}\). Since \(\sigma\) fixes each \(H_k\), it fixes every transposition. The transpositions generate \(S_4\), hence \(\sigma\) is the identity.
Therefore every automorphism of \(S_4\) is inner, and \[ \abs{\Aut(S_4)}=\abs{S_4}=24. \]