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I am trying to answer the following:

Let $G$ be a finite group, $n$ a divisor of the order $|G|$ of $G$. If there are no subgroups of $G$ of order $n$, prove that $n$ has at least two distinct prime divisors.

I know that if $|n|= p$ for some prime $p$ , then by Cauchy's theorem $\exists g \in G$ such that $|g| = p$ and so the cyclic subgroup $\langle g \rangle$ has order $p$, a contradiction. I also know if $|n| = p^{m}$ , where $p^{m+1} \nmid |G|$, then by Sylow's theorem $\exists$ a subgroup of order $n$, a contradiction.

I am struggling in the case where $|n| = p^{l}$ and $p^{l+k} \mid |G|$ for some natural number $k$. In toy examples I am able to construct subgroups of order $p^{l}$, but don't know how to do it more generally (and thus settle that $n$ has more than one prime divisor).

Any hints would be appreciated!

Shaun
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  • If not, $G$ is a $p$-group, and hence a CLT-group, i.e., the converse of Lagrange is true. But then there would be indeed a subgroup of order $n$, a contradiction. Hence $G$ has at least two distinct prime divisors. – Dietrich Burde Oct 25 '21 at 19:35
  • I don't understand why this was downvoted. It has a decent attempt. – Shaun Oct 25 '21 at 19:38
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    Do you know that $p$-groups have a non-trivial centre? Or any results about $p$-groups? That they are not simple (or have order $p$) is enough to do this. – David A. Craven Oct 25 '21 at 19:42
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    There are several posts here showing that a $p$-group has a subgroup of order $d$ for every divisor $d$ of $|G|$. See for example here - or here. – Dietrich Burde Oct 25 '21 at 19:44
  • Prove the contrapositive statement using Sylow $p$-subgroups. If $n$ has only one prime factor $p$, then there exists one Sylow $p$-subgroup $H$. Inside $H$ you can find anything you need. – Crostul Oct 25 '21 at 19:45
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    Alternatively, use a stronger version of Sylow's theorem that says that a finite groups has a subgroup of order $n$ for any prime power $n$ dividing its order. – Derek Holt Oct 25 '21 at 21:48

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