## In a finite group, normal subgroups whose order and index are coprime are unique up to order

Let $G$ be a finite group, $N \leq G$ a normal subgroup, and suppose that $|N|$ and $[G:N]$ are relatively prime. Prove that $N$ is the unique subgroup of order $|N|$ in $G$.

Let $H \leq G$ be a subgroup of order $|N|$. By the previous exercise, we have $H \leq N$. Because $H$ and $N$ are finite sets of the same cardinality, we in fact have $H = N$.

a typo: $H$ be a subgroup of order $|N|$