Let $F$ be a (finitely generated) free group, $H \leq F$ of infinite index. Is it possible that $$ \bigcup_{g \in F} gHg^{-1} = F?$$
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3$\begingroup$ mathoverflow.net/questions/34044/… shows a group can be a union of conjugates of proper subgroups. Write that group as a quotient of a free group and take preimages. $\endgroup$– Benjamin SteinbergCommented Sep 13, 2014 at 11:53
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$\begingroup$ You are right (I have edited the question now), but I am mainly interested in the finitely generated case. Can something like that happen there? $\endgroup$– PabloCommented Sep 13, 2014 at 12:47
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1$\begingroup$ The answer is "No" is $H$ is finitely generated. Namely, a f.g. free group cannot be equal to the union of conjugates of a proper f.g. subgroup. This is because any f.g. subgroup $H$ of a free group $F$ is closed in the profinite topology, hence there is a map $\phi$ from $F$ onto a finite group $K$ s.t. $\phi(H) \neq K$. From the result for finite groups, mentioned in mathoverflow.net/questions/34044 , it follows that $K$ cannot be equal to the union of conjugates of $\phi(H)$, which implies the same about $F$ and $H$. $\endgroup$– Ashot MinasyanCommented Sep 15, 2014 at 15:43
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$\begingroup$ Yes this was clear to me. In fact, this is true if $H$ is merely contained in a proper open subgroup. $\endgroup$– PabloCommented Sep 15, 2014 at 15:56
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1 Answer
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The problem of characterizing groups that are union of conjugates of a proper subgroup was considered in some papers by Wiegold and others. In particular, you can look at
Transitive groups with fixed point free permutations, Archiv der Mathematik 27 (1976), 473-475.
Groups covered by conjugated of proper subgroups, Journal of Algebra 293 (2005), 261–268.
It turns out that the answer to your question is yes.
In fact, already the free group on two generators $F_2 = \langle a, b \rangle$ can be covered by the conjugates of one of its proper subgroups. This is shown in the first paper linked above, see Example 3.1 page 474.
answered Sep 13, 2014 at 12:52