# $G$ group and $H,K \le G$, with $H,K \ne \lbrace e \rbrace$. Does $H \cap K \cong K$ imply $H \cap K = K$?

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?

• @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? – Luca Jul 3 at 6:26
• @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. – 0XLR Jul 3 at 6:29
• @Arthur Yes I mean $H< K$ and I wrote as an answer. – 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 $$K$$.