Let \(G\) be a finite group. Suppose \(G\times G\) has exactly four normal subgroups. Prove that \(G\) is simple and nonabelian.
Solution
1. The group \(G\) is nontrivial, and the following four normalsubgroups of \(G\times G\) are distinct:\[ \{1\}\times\{1\}, \qquad G\times\{1\}, \qquad \{1\}\times G, \qquad G\times G. \]
If \(G\) were trivial, then \(G\times G\) would be trivial and would have only one normal subgroup, contrary to the hypothesis.
For nontrivial \(G\), the four displayed subgroups are pairwise distinct. Each is normal because each is a direct product of normal subgroups of the two factors.
2. The group \(G\) is simple.
Let \[ N\trianglelefteq G. \] Then \[ N\times\{1\}\trianglelefteq G\times G. \] By step 1, the four displayed subgroups already account for all normal subgroups of \(G\times G\). Hence \(N\times\{1\}\) must be one of them.
Because its second coordinate is always \(1\), it cannot equal \(\{1\}\times G\) or \(G\times G\) unless \(G\) is trivial, which step 1 excludes. Therefore \[ N\times\{1\} = \{1\}\times\{1\} \quad\text{or}\quad N\times\{1\} = G\times\{1\}. \] Thus \[ N=\{1\} \quad\text{or}\quad N=G. \] So \(G\) has no proper nontrivial normal subgroup and is simple.
3. The group \(G\) is nonabelian.
Suppose instead that \(G\) were abelian. Then \(G\times G\) would be abelian, so every subgroup would be normal. In particular, the diagonal subgroup \[ \Delta \coloneqq \{(g,g):g\in G\} \] would be normal.
Since \(G\) is nontrivial, \(\Delta\) is neither \(\{1\}\times\{1\}\) nor \(G\times G\). It is also distinct from each coordinate subgroup: if \(g\ne1\), then \[ (g,g)\in\Delta \] has both coordinates nontrivial, whereas elements of \(G\times\{1\}\) or \(\{1\}\times G\) have one coordinate equal to \(1\). Thus \(\Delta\) would be a fifth normal subgroup of \(G\times G\), a contradiction.
4. Q.E.D.
Step 2 proves that \(G\) is simple, and step 3 proves that it is nonabelian.