Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 1 / 10G (Groups, Rings and Modules)

The group $G$ is a nontrivial finite p-group, so the nontrivial center of a finite p-group gives $Z(G)\ne \{e\}$. If $Z(G)\ne G$, then $Z(G)$ is already a nontrivial proper normal subgroup, so $G$ is not a simple group. If $Z(G)=G$, then $G$ is abelian. By Cauchy's theorem for finite groups, it has a subgroup of order $p$; every subgroup of an abelian group is normal, and this subgroup is nontrivial and proper because $|G|=p^4$. Thus in every case

The Sylow theorems state that if $|G|=p^{a}m$ with $p\nmid m$, then:

Interchange the two primes if necessary so that $p<q$. By the Sylow theorems, the number $n_q$ of Sylow $q$-subgroups satisfiesIf $n_q\ne 1$, then $n_q$ is $p$ or $p^2$. The first is impossible because $1<p<q$. The second would implySince $q>p$ is prime, it cannot divide $p-1$, so it must divide $p+1$. But $q>p$ then forces $q=p+1$, impossible because $p$ and $q$ are both odd. Therefore $n_q=1$. The unique Sylow $q$-subgroup is a nontrivial proper normal subgroup, and hence