Prove that a group of order \(351=3^3\cdot 13\) cannot be simple.
Prove that a group of order \(33\) must be cyclic.
Solution
1. A group \(G\) of order \(351\) is not simple.
Sylow’s theorems give \(n_{13}\mid27\) and \(n_{13}\equiv1\pmod{13}\) [@DF04]. The divisors of \(27\) are \(1,3,9,27\), so \(n_{13}=1\) or \(27\). If \(n_{13}=1\), the unique Sylow \(13\)-subgroup is a nontrivial proper normal subgroup, proving the claim.
Suppose \(n_{13}=27\). Two distinct subgroups of order \(13\) have trivial intersection: a shared nonidentity element would generate both. They therefore contain exactly \(27\cdot12=324\) distinct nonidentity elements. Let \(X\) be the complement of this set in \(G\). Then \(|X|=351-324=27\), and the identity lies in \(X\).
Every Sylow \(3\)-subgroup \(P\) is contained in \(X\), since the order of a nonidentity element of \(P\) is a power of three and cannot be thirteen. But \(|P|=27=|X|\), so \(P=X\). Consequently every Sylow \(3\)-subgroup is the same subgroup. It is normal by Sylow conjugacy, and is nontrivial and proper. This proves nonsimplicity also in the second case.
2. A group \(G\) of order \(33\) is cyclic.
Here \(n_{11}\mid3\) and \(n_{11}\equiv1\pmod{11}\), so the Sylow \(11\)-subgroup \(Q\) is unique. Also \(n_3\mid11\) and \(n_3\equiv1\pmod3\); since \(11\equiv2\pmod3\), the Sylow \(3\)-subgroup \(P\) is unique [@DF04]. Thus both are normal, and their intersection is trivial because its order divides both \(3\) and \(11\).
For \(x\in P\) and \(y\in Q\), the commutator \(xyx^{-1}y^{-1}\) belongs to both normal subgroups and hence is one. Multiplication gives a homomorphism \(P\times Q\to G\). Its kernel is trivial, because \(xy=1\) implies \(x=y^{-1}\in P\cap Q\). Both groups have order \(33\), so the homomorphism is an isomorphism. The prime-order groups \(P,Q\) are cyclic. A pair of generators has order \(\operatorname{lcm}(3,11)=33\), and its image generates \(G\).