Can we say something about the index of these subgroups? Let M, K, and L be isomorphic normal subgroups of a group G. Suppose $M \cap K \leq M \cap L$. Can we conclude anything about the index of MK in G and the index of ML in G? I would like it to be the case that MK has smaller index than ML, but I'm having a hard time convincing myself either way.
(EDIT: I initially left out the word "normal".)
 A: It's true for finite groups, since $|MK| = |M| |K|/|M \cap K|$. It will not be true in general for infinite groups. The example that springs to mind is $G={\mathbb Z}^{\infty}$, where it is easy to find subgroups $M,K,L$ all isomorphic to $G$ with $M \cap K = M \cap L = 1$, $ML=G$ and $|G:MK|$ infinite.
A: 
Lemma Let $H,K \leq G$ with $H \unlhd K$ and let $N \unlhd G$ then the following hold.
$(a)$ $HN \unlhd KN$
$(b)$ $KN/HN$ is a quotient of $K/H$ 
$(c)$ If $G$ is finite, $|KN:HN|$ divides $|K:H|$ with equality if and only if $K \cap N \subseteq H$.

Proof (a) let $h \in H, k \in K, m, n \in N$, we need to show that $(hm)^{kn} \in HN$. Now observe that
$$(hm)^{kn}=n^{-1}k^{-1}hmkn=n^{-1}(k^{-1}hk)(k^{-1}mk)n=(k^{-1}hk)(k^{-1}h^{-1}k)n^{-1}(k^{-1}hk)(k^{-1}mk)n=h^k(n^{-1})^{k^{-1}hk}m^kn \in HN.$$
(b) $KN/HN \cong K/(K \cap HN)$. Note that $H \subseteq K \cap HN$. So $K/(K \cap HN) \cong (K/H)/((K \cap HN)/H)$ which is a quotient of $K/H$.
(c) From (b) it follows that $|KN:HN| \mid |K:H|$ and that we have equality if and only if $K \cap HN=H$. But by Dedekind's Modular Law, we have $K \cap HN=H(K \cap N)$. Hence (c) follows. Note that $K \cap N \subseteq H$ is equivalent to $K \cap N=H \cap N$.
