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## A finite group of composite order n having a subgroup of every order dividing n is not simple

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.14

Solution: Let $p$ be the smallest prime dividing $n$, and write $n = pm$. Now $G$ has a subgroup $H$ of order $m$, and $H$ has index $p$. By Corollary 5 in the text, $H$ is normal in $G$.