# A finite non-abelian group of order $n$ that for every divisor of $n$ has a subgroup is not simple

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.

• The abelian case is trivial because in an abelian group , every subgroup is normal. Therefore the assumption that the group is not abelian. – Peter Apr 22 at 12:21

## 1 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.

• Yeah that is sufficient to prove it. Thanks – Gabi G Apr 22 at 13:02