= Solution
{c}
Take \b[$p=5$]. The unit $3$ modulo $121$ has order $5$, because
$$
3^5=243\equiv1\pmod{121},\qquad3\not\equiv1\pmod{121}.
$$
Let a generator $b$ of $C_{25}$ act on $C_{121}=\langle a\rangle$ by $a\mapsto a^3$. The action is well defined since its order $5$ divides $25$. The <semidirect product>
$$
\boxed{G=C_{121}\rtimes C_{25}=\langle a,b\mid a^{121}=b^{25}=1,\ bab^{-1}=a^3\rangle}
$$
has order $121\cdot25=11^2 5^2$. 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.
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