Let $G$ be non-abelian group of order $n$. Also, for every $k$ which is a divisor of $n$ , there is a subgroup of $G$ of order $k$. I want to prove that $G$ is not simple.
Well, from what is given, I see that there is a subgroup of order $p$ for every prime $p$ that divides $n$. However, I don't see how it helps. I am not sure how to use the fact that $G$ is not abelian too.
Help would be appreciated.