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## Normal subgroups whose order and index are coprime are unique up to order

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.2 Exercise 3.2.19

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 $G$.

Solution: Let $H \leq G$ be a subgroup of order $|N|$. By Exercise 3.2.18, we have $H \leq N$. Because $H$ and $N$ are finite sets of the same cardinality, we in fact have $H = N$.