Let $F$ be a free group, and let $H,K \leq F$ be finitely generated subgroups of infinite index in $F$. Is it possible that for the set of products we have $HK = F$ ?
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6$\begingroup$ I'm not sure about the votes to close - this seems a reasonable question to me. Anyway, the answer is 'no'. I'll try to write down an answer when I have time. $\endgroup$– HJRWCommented Dec 18, 2014 at 10:52
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1$\begingroup$ There are lots of ways to prove that $HK$ is a proper subset of $F$. For example, Theorem 1.1 in arxiv.org/abs/1308.3192 claims that there is a finite index subgroup $K'$ of $K$, s.t. $M:=\langle H,K'\rangle$ still has infinite index in $F$. Then, clearly $HK=\cup_{i=1}^k HK'g_i \subseteq \cup_{i=1}^k Mg_i \neq F$ (where $K=\sqcup_{i=1}^k K'g_i$). $\endgroup$– Ashot MinasyanCommented Dec 18, 2014 at 16:53
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$\begingroup$ Can you not use Marshall Hall's theorem to reduce to the case where one of the subgroups is a free factor? If one of the subgroups is a proper free factor then you can use the finiteness of the Stallings graph of the other to show you can't reach every vertex in the Schreier graph using generators from the first factor. $\endgroup$– Benjamin SteinbergCommented Dec 18, 2014 at 18:00
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1 Answer
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It's actually easier to prove the stronger statement that there are infinitely many double cosets $H\backslash F/K$.
First, note that if $F'$ is a subgroup of finite index in $F$, with $H'$ and $K'$ its intersections with $H$ and $K$ respectively, then the natural map $H'\backslash F'/K'\to H\backslash F/K$ is finite-to-one. So we may pass to finite-index subgroups.
EDIT: (I was a little glib in translating from topology to group theory before. Here's a corrected version of the final paragraph.)
Therefore, by Marshall Hall's theorem, we may take $F$ to be the fundamental group of a graph $X$ and $H$ and $K$ to be carried by embedded subgraphs $Y$ and $Z$, say. But now it's easy. Indeed, let $a$ be a based loop not contained in $Y$ and $b$ a based loop not contained in $Z$. Then the double cosets $Ha^mb^nK$ are all distinct.
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1$\begingroup$ How do you use M. Hall's theorem to find a single finite index subgroup $F' \leqslant F$ such that both $F' \cap H$ and $F' \cap K$ are free factors of $F'$? $\endgroup$ Commented Dec 18, 2014 at 21:31
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$\begingroup$ Ashot, you need the observation that if $H$ is a free factor of $F$ then $H'$ is a free factor of $F'$. Topologically, this is obvious. $\endgroup$– HJRWCommented Dec 18, 2014 at 21:44
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$\begingroup$ Actually, sorry, you're right, I should be a little more careful. The question is whether, given two embedded subgraphs in a graph $X$, one can find a maximal tree that restricts to a maximal tree in each. But this issue isn't really important. One should just work in the graph $X$ (without contracting a maximal tree), and the same argument goes through. $\endgroup$– HJRWCommented Dec 18, 2014 at 21:58
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$\begingroup$ This argument is not correct. $\endgroup$ Commented Dec 19, 2014 at 7:39
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$\begingroup$ @AndreasThom, since there's already some discussion of how to modify the argument in comments, and you give no details of your critique, your comment is not very helpful. Anyway, I've now edited the answer to fix my small mistake. $\endgroup$– HJRWCommented Dec 19, 2014 at 8:33