Past exam of the short options course of the University of Oxford / 2021 / 2 / e / Solution

Take $p=5$. The unit $3$ modulo $121$ has order $5$, becauseLet a generator $b$ of $C_{25}$ act on $C_{121}=⟨a⟩$ by $a\mapsto a^3$. The action is well defined since its order $5$ divides $25$. The semidirect producthas order $121\cdot 25=11^2{5}^2$. Its construction as a semidirect product ensures that $a$ retains order $121$. Since $a^3\ne a$, the groups generated by $a$ and $b$ do not commute, so $G$ is nonabelian.