Let $H$ and $K$ be finitely generated subgroups of a free group $F$, and suppose that $H$ has finite index in $F$. Is it true that $rank(H \cap K)1 \leq (rank(H)1)(rank(K)1)$?
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$\begingroup$ (actually, I missed the "finiteindex" part, so ignore my deleted comment). $\endgroup$ – Andy Putman Nov 3 '13 at 19:32

1$\begingroup$ @mary seva, it looks like your question is about to be closed (and/or migrated) as 'not research level'. In fact, I think it is graduatestudent level, and hence acceptable on MO. But I suspect the formulation, which reads like a homework problem, has irritated the voters to close. Could you tell us how this problem arises in your research? $\endgroup$ – HJRW Nov 3 '13 at 19:39

$\begingroup$ While this is certainly not a great question, why does it get THAT many downvotes? $\endgroup$ – Stefan Kohl Nov 3 '13 at 19:50

1$\begingroup$ @StefanKohl I am guessing because it looks like homework. $\endgroup$ – Igor Rivin Nov 3 '13 at 20:00
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Yes, this statement is true and for a long time was known as the Hanna Neumann Conjecture. It was proved in 2011 by Igor Mineyev.

2$\begingroup$ @StephanKohl The first response was kneejerk, the second was a result of some thought. There is no way to retract a close vote, as far as I know. $\endgroup$ – Igor Rivin Nov 3 '13 at 20:29

5$\begingroup$ Igor, it was asked about the case, where one of the subgroups is of finite index. This is indeed an exercise on Schreier formula. $\endgroup$ – Anton Klyachko Nov 3 '13 at 21:29

4$\begingroup$ @WillJagy What do you mean "wanders off"? The question was asked just three hours ago. $\endgroup$ – Todd Trimble♦ Nov 3 '13 at 21:49

4$\begingroup$ @WillJagy Yes, I do. A "normal person" (not a fanatic like you or me (:) might post, go about their usual day, and maybe check back in the evening when there's a spare moment. Personally I think getting back within 24 hours is fine, but I feel some impatience if it takes more than 48 hours. $\endgroup$ – Todd Trimble♦ Nov 4 '13 at 0:41

3$\begingroup$ By the way, Friedman also gave a correct proof of the Hanna Neumann Conjecture (after a couple of incorrect attempts). $\endgroup$ – HJRW Nov 4 '13 at 15:00