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.
Best Answer
Let $p$ be the smallest prime dividing $n=|G|$. Then there is a subgroup of order $k=\frac{n}{p}$, which is a divisor of $n$. However, it is well-known that every such subgroup of index $p$ is normal:
Normal subgroup of prime index
Hence $G$ is not simple.