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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".)

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"Index" is only defined for subgroups of groups, so do you want to make some extra assumptions (such as M being a normal subgroup of G) to ensure that MK and ML are subgroups of G? –  Derek Holt Nov 4 '10 at 14:12
    
Right, I meant to stipulate that these are all normal. I'll update the question. –  Gabe Cunningham Nov 4 '10 at 14:30
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up vote 6 down vote accepted

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.

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I thought something like that might happen. The case I'm really interested in is somewhat more specific -- I'll think about it more and maybe post a follow-up. Thanks! –  Gabe Cunningham Nov 4 '10 at 14:53
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