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I'm constructing an example of a group which has a particular property on its subgroups, and the property looks like something that might have been considered before.

Fix a group $G$ and a pair of subgroups $L$ and $H$. We consider a (non-symmetric) relation $s$ between $L$ and $H$:

$L\ s\ H$ if and only if for all subgroups $K$, $H\cap K \le L \cap K$ implies $K \le L$.

The example I have in mind is this: Fix a set $M$ and a subset $N\subset M$. Consider also a group $\mathcal{G}$ and subgroups $\mathcal{H}, \mathcal{L} \lt \mathcal{G}$. Let $G = \mathcal{G}^M$, the set of functions from $M$ to $\mathcal{G}$ with pointwise multiplication. The subgroups are $$ H = \lbrace f\in\mathcal{G}^M |\forall n \in N, f(n) \in \mathcal{H} \rbrace $$ $$ L = \lbrace g\in\mathcal{G}^M |\forall m \in M-N, g(m) \in \mathcal{L} \rbrace $$ and it isn't too hard to see that $L\ s\ H$ in this case. I can't think of any other sorts of groups which have subgroups satisfying this relation, but I'm not a group theorist.

(PS if more tags are needed, feel free to add them)


Addendum

Having thought about this a bit more, it seems like I could be addressing this from a sheaf of rings point of view (a case of interest to me is $G=\mathbb{Z}$, or one could take the abelian group underlying a ring). In this case, we consider sheaves of ideals instead of subgroups, and look at the supports of the quotient sheaves. If they are disjoint then they satisfy a relation similar to the one defined above. Put this way, it seems like something quite natural arising from algebraic geometry.

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If $K < L$ then $K \subset L$, so that $L \cap K = K$, which makes me think perhaps you mis-stated the relation you have in mind? –  Benjamin Dickman Nov 5 '12 at 3:22
    
No. That's what I meant. A priori we don't know that $K\lt L$. –  David Roberts Nov 5 '12 at 4:51
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Ah, I think the source of confusion was that I read '$<$' as 'is a proper subgroup'. I usually write $\le$ for 'is a subgroup'. –  Colin Reid Nov 5 '12 at 7:15
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Assuming your '$<$'s are both '$\le$', it is sufficient to check the condition for cyclic subgroups, i.e. $L$ $s$ $H$ if and only if for every $x$ in $G \setminus L$, then $\langle x \rangle \cap H$ is not contained in $L$. –  Colin Reid Nov 5 '12 at 7:42
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So couldn't we write it as: $L\ s\ H$ if and only if ($x\in L \leftrightarrow \langle x \rangle \cap H \le L$)? –  David Roberts Nov 5 '12 at 7:49
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