Determine the number of non-isomorphic abelian groups of order \(144\).
Note \(144=2^4 \times 3^2\). The number of abelian groups of order \(2^4\) is the number of partitions of \(4\), which is \(5\). The number of abelian groups of order \(3^2\) is the number of partitions of \(2\), which is \(2\). Thus there are \(10\) non-isomorphic abelian groups of order \(144\).
Prove that a group of order \(2907\) is not simple.
We find \(2907=3^2 \times 17 \times 19\). Let \(n_p\) denote the number of Sylow \(p\)-subgroups. We have \(n_3 \equiv 1 \mod 3\) and \(n_3 | 17 \times 19\); thus \(n_3 = 1,19\). We have \(n_{17} \equiv 1 \mod {17}\) and \(n_{17} | 3^2 \times 19\); thus \(n_{17} = 1,171\). We have \(n_{19} \equiv 1 \mod {19}\) and \(n_{19} | 3^2 \times 17\); thus \(n_{19} = 1,153\). Suppose the group is simple; then \(n_p \ne 1\) for any \(p\). There are \(171\times (17-1)=2736\) elements of order \(17\) and \(132\times(19-1)=2754\) elements of order \(19\). This is more elements in total than there are in the group so it is not simple.
Let \(G\) be a finite group and let \(H\) be a subgroup of \(G\) of index \(n\). Prove that there is a normal subgroup \(K\) of \(G\) such that \(K \subseteq H\) and \(|G:K| \le n!\).
The action of \(G\) on \(G/H\) is equivalent to a permutation representation \(\alpha: G \to S_n\) where the stabilizer of \(1\) is \(H\). Let \(K\) be the kernel of \(\alpha\). Note that \(K\) fixes \(1\), so \(K\) is in \(H\). Moreover, \(\alpha\) induces an injective map \(G/K \subseteq S_n\), so \(|G:K| \le |S_n| = n!\).
A subgroup \(H\) of a group \(G\) is characteristic if \(\phi(H)=H\) for every automorphism \(\phi: G \to G\).
Let \(H\) be a characteristic subgroup of \(H\). Recall that \(\phi_g(h)=ghg^{-1}\) is a group automorphism \(G \to G\) for every \(g \in G\). Thus \(\phi_g(H)=H\) for all \(g \in G\). Thus \(gHg^{-1}=H\) for all \(g \in G\) so \(H\) is normal in \(G\).
Suppose \(\alpha: G \to G\) is an automorphism of \(G\). Then \(\alpha(H)=H\) since \(H\) is characteristic in \(G\). Thus the restriction \(\beta := \alpha|_H : H \to H\) is well-defined. Since \(\beta\) is a bijection and a group homomorphism it is an automorphism of \(H\). Thus \(\beta(K)=K\) since \(K\) is characteristic in \(H\). Thus \(\alpha(K)=\beta(K)=K\) and we conclude \(K\) is characteristic in \(G\).
Consider \(G=A_4\) with \(H=\langle (12)(34), (13)(24) \rangle\) and \(K=\langle (12)(34) \rangle\). Note that the subgroup \(H\) is isomorphic to the Klein four-group and contains exactly the elements of order \(1\) or \(2\) in \(G\). The subgroup \(H\) is characteristic in \(G\) since orders of elements must be preserved by automorphisms. We see that \(K\) is normal in \(H\) since \(H\) is abelian. However \((123)(12)(34)(132)=(14)(23) \notin K\) so \(K\) is not normal in \(G\).