Exercise 4.5.13 (A group of order $56$ has a normal Sylow $p$-subgroup)

Proof. Suppose that $|G|=56=2^3\cdot 7$.

By Sylow’s Theorem, $n_7\equiv 1(\mathit{𝑚𝑜𝑑}7)$ and $n_7\mid 8$. Thus

Similarly, $n_2\mid 7$, so

Assume that $n_7\ne 1$. Then $n_7=8$. Therefore there are $8$ subgroups of order $7$, and distinct $7$-subgroups intersect in the identity. Thus there are $8\cdot 6=48$ elements of order $7$ in $G$. It remains a set $S$ of $8$ elements, whose orders are not $7$. Then every subgroup of order $8$ is included in this set $S$, so is equal to $S$, hence there is only one Sylow $2$-subgroup, and so $n_2=1$.

This shows that $n_7=1$ or $n_2=1$ : a group of order $56$ has a normal Sylow $p$-subgroup for $p=2$ or $p=7$. □