Show that if \(|G| = 12\) and \(G\) is a non-abelian group with a normal subgroup of order \(4\) (or equivalently, \(n_3 = 4\)), then \(G \cong A_4\).
Solution
Let \(V\trianglelefteq G\) have order \(4\), and let \(P\in\operatorname{Syl}_3(G)\). Then \(|P|=3\), \(V\cap P=1\), and \[ |VP|=\frac{|V||P|}{|V\cap P|}=12=|G|. \] Hence \[ G\cong V\rtimes P. \]
There are two possibilities for \(V\): \(C_4\) and \(V_4\).
If \(V\cong C_4\), then \[ \operatorname{Aut}(V)\cong C_2. \] Every homomorphism \(P\cong C_3\to C_2\) is trivial, so the semidirect product is direct: \[ G\cong C_4\times C_3\cong C_{12}, \] contrary to the hypothesis that \(G\) is nonabelian. Thus \(V\cong V_4\).
Now \[ \operatorname{Aut}(V_4)\cong S_3. \] The conjugation action \(P\to S_3\) cannot be trivial, again because that would make \(G\cong V_4\times C_3\) abelian. Hence its image is the unique subgroup of order \(3\), acting cyclically on the three nonidentity elements of \(V_4\). Up to automorphisms of \(V_4\) and \(C_3\), this gives a unique nontrivial semidirect product \[ V_4\rtimes C_3. \]
In \(A_4\), the normal Klein four subgroup \[ \{1,(12)(34),(13)(24),(14)(23)\} \] is acted on cyclically by conjugation with a \(3\)-cycle. Therefore \[ \boxed{G\cong V_4\rtimes C_3\cong A_4.} \]