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Let $X$ be a non-empty set. Prove that $F_X$, the free group on $X$ is solvable if and only if $|X| = 1$.

We can see that if $|X| = 1$, then $F_X$ is abelian, and hence solvable. However, the other direction stumps me. Any suggestions?

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The statement is not quite right: the free group of rank $0$ is also solvable. –  Arturo Magidin Jul 22 '12 at 0:32
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2 Answers

up vote 13 down vote accepted

Yes: a quotient of a solvable group is solvable. Every free group of rank greater than $1$ has the free group of rank $2$ as a quotient (just kill all but two of the free generators), so it suffices to find a nonsolvable group which can be generated by two elements. Try $A_5$ for instance...

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Every non-abelian free group also contains every countably generated free group as a subgroup, so if a non-abelian free group was solvable, every countable group would be solvable! –  user641 Jul 23 '12 at 20:02
@Steve D: Thanks, this is a nice observation: using a little more theory about free groups one does not need to exhibit a nonsolvable $2$-generated group. Let me say though that the $2$-generation of $A_5$ is close to my heart ($A_5$ is the von Dyck triangle group $\Delta(2,3,5)$), and in fact every finite simple group is $2$-generated...although of course this, relying as it does on CFSG, is a much less elementary fact! –  Pete L. Clark Jul 24 '12 at 0:49
@PeteL.Clark: Thanks for the hint about using $A_5$, which indeed can be generated by 2 elements. Does it matter that $A_5$ is finite? Wouldn't most quotients of free groups be infinite? –  phenomenalwoman4 Jul 24 '12 at 2:46
@Emily: You're welcome. What do you mean when you ask "Does it matter that $A_5$ is finite?" It doesn't disturb anything in the argument, if that's what you mean. (Why would it?) Note that every group is a quotient of some free group, so there are a lot of examples both finite and infinite! For the case of $2$-generated groups, there are certainly only countably isomorphism classes of finite such groups (or of all finite groups!), whereas I believe there are uncountably many isomorphism classes of infinite $2$-generated groups. So yes, in that sense, "most of them" are infinite. –  Pete L. Clark Jul 24 '12 at 2:51
I haven't worked too much with free groups, so thanks for reminding me that "every group is a quotient of some free group". I see it now! –  phenomenalwoman4 Jul 26 '12 at 2:17
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Another possibility is to show that the commutator subgroup of a non-abelian free group $F$ is itself a non-abelian free group; thus, the sequence $D^n(F)$ cannot stabilizes at $\{1\}$, and $F$ is not solvable.

In fact, $D(F)$ is a free group by Nielsen-Schreier theorem, and you can notice that $[x,y]^n[x,y^2]^m \neq 1$ for $nm \neq 0$ so it is non-cyclic (and consequently non-abelian).

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