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 "finite-index" part, so ignore my deleted comment). $\endgroup$– Andy PutmanCommented Nov 3, 2013 at 19:32
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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 graduate-student 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? $\endgroup$– HJRWCommented Nov 3, 2013 at 19:39
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$\begingroup$ While this is certainly not a great question, why does it get THAT many downvotes? $\endgroup$– Stefan Kohl ♦Commented Nov 3, 2013 at 19:50
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1$\begingroup$ @StefanKohl I am guessing because it looks like homework. $\endgroup$– Igor RivinCommented Nov 3, 2013 at 20:00
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1 Answer
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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.
answered Nov 3, 2013 at 20:02
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2$\begingroup$ @StephanKohl The first response was knee-jerk, the second was a result of some thought. There is no way to retract a close vote, as far as I know. $\endgroup$ Commented Nov 3, 2013 at 20:29
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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. $\endgroup$ Commented Nov 3, 2013 at 21:29
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4$\begingroup$ @WillJagy What do you mean "wanders off"? The question was asked just three hours ago. $\endgroup$ Commented Nov 3, 2013 at 21:49
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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. $\endgroup$ Commented Nov 4, 2013 at 0:41
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3$\begingroup$ By the way, Friedman also gave a correct proof of the Hanna Neumann Conjecture (after a couple of incorrect attempts). $\endgroup$– HJRWCommented Nov 4, 2013 at 15:00