Let $G$ be a group and $H,K$ nontrivial subgroups of $G$. Does $H \cap K \cong K$ imply $H \cap K = K$ (and then $K \le H$)? If not in general, are there conditions on $H,K$ making this implication to hold?

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    $\begingroup$ @Hongyi Huang, it seems to me that this is an example in which the statement trivially holds, being $K$ chosen to be a subgroup of $H$, but doesn't prove the statement in general. Did I misunderstand? $\endgroup$ – Luca Jul 3 at 6:26
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    $\begingroup$ @Luca He probably meant it the other way around. $H = 2\mathbb{Z}$ and $K = \mathbb{Z}$. Then $H \cap K = 2\mathbb{Z} \cong \mathbb{Z} = K$ but obviously $H \cap K = 2\mathbb{Z} \neq \mathbb{Z} = K$. This is a counterexample to your question. $\endgroup$ – 0XLR Jul 3 at 6:29
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    $\begingroup$ @Arthur Yes I mean $H< K$ and I wrote as an answer. $\endgroup$ – Hongyi Huang Jul 3 at 6:30

Counterexample: $K = \mathbb{Z}$ and $H$ be the group of all even integers under addition. Then $H\cap K = H\cong \mathbb{Z}\cong K$ but $H\cap K\ne K$.

If $K$ is a finite group, then $H\cap K\cong K$ implies $H\cap K = K$ because $H\cap K$ is a subgroup of $K$.


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