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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.

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

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

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