Index of intersection of two sugroups If $H,K$ are subgroups of index $m$ and $n$ respectively in $G$, then we have 
$$\operatorname{lcm}(m,n)\leq [G\colon H\cap K]\leq mn.$$
Question. What are necessary and sufficient conditions to hold the equality (separately, or simultaneously)?
[If $m$ and $n$ are relatively prime, then equality at right holds; is converse true? Also, if $m$ and $n$ are relatively prime, then equality at left holds; is converse true?]
 A: There is nothing more we can say about $m$ and $n$ than what you've said.
For the left side: Let $G$ be a cyclic group of order $mn$, and let $H,K$ be the unique subgroups with index $m$ and $n$.  Then $[G:H\cap K] = \operatorname{lcm}(m,n)$, which holds without any assumptions on $m$ and $n$.
For the right side: Let $G$ be the direct product of cyclic groups $\mathbb{Z}_m \times \mathbb{Z}_n$, and let $H$ and $K$ be its factors.  Then $H\cap K = 1$, so $[G:H\cap K] = mn$, again with no assumptions on $m$  and $n$.
More generally, you can find an example where $[G:H\cap K] = d$, for any $d$ with $\operatorname{lcm}(m,n)\mid d \mid mn$.  In other words, the only way we can guarantee equality, on either side, is if $\operatorname{lcm}(m,n)=mn$, i.e. $(m,n)=1$.
On the other hand, I don't think it's at all realistic to ask for necessary and sufficient conditions on $G,H,K$ that will cause these equalities to hold.  There are a lot of groups, and a lot of different ways that two subgroups can have a very large or very small intersection.  If $G$ is finitely generated abelian, you can probably use the above thinking to develop necessary and sufficient conditions, but I would not be very optimistic about a generalization.
