Exercise 4.5.15 (A group of order $351$ has a normal Sylow $p$-subgroup)

Proof. Suppose that $|G|=351=3^3\cdot 13$.

By Sylow’s Theorem,

Therefore

Similarly $n_3\mid 13$, thus

Assume that $n_{13}\ne 1$. Then $n_{13}=27$. Therefore there are $27$ subgroups of order $13$, and distinct $13$-subgroups intersect in the identity. Thus there are $27\cdot 12=324$ elements of order $13$ in $G$. It remains a set $S$ of $27$ elements, whose orders are not $13$. Then every Sylow $3$-subgroup, of order $27$, is included in this set $S$, so is equal to $S$, hence there is only one Sylow $3$-subgroup, and so $n_3=1$.

This shows that $n_{13}=1$ or $n_3=1$ : a group of order $56$ has a normal Sylow $p$-subgroup for $p=3$ or $p=13$. □