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Let G be a finite abelian group. Prove that G is a cyclic p-group if and only if for every two subgroups $H, F\le G$, we have $H\cap F\neq \{e\}$.

I came up with one implication ($\Rightarrow$), and in the other direction I have the following:

$H\cap F \le G$, so if $d = |H\cap F|$, then $d$ divides all of $|G|, |H|, |F|$. If $|G|$ had two distinct prime factors, this wouldn't be possible for every two subgroups. Hence $G$ is a p-group. Moreover, there exists a subgroup $H_0\le G$, isomorphic to $C_p$, such that for every $F\le G$, we have $H_0\le F$ (since $H_0\cap F$ is non-trivial).

I feel it's close to proving that $|G|$ is cyclic, but I haven't managed to do it.

Shaun
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jrg
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1 Answers1

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If $|G|$ has two different prime factors $p,q$, then by Cauchy's theorem we may take subgroups $H$, $F$ of $G$ of orders $p$ and $q$ respectively. But this is not possible as $H\cap F$ has to be nontrivial.

We claim that $G$ has one subgroup of order $d$ for every divisor $d$ of $|G|$ (this condition implies that $G$ is cyclic. See here for example). To prove the claim, we argue by induction over $|G|$. As you said, there exists only one subgroup of $G$ of order $p$, say $H_0$. Then, reasoning by induction over $G/H_0$ the claim follows.

Deif
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