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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.2012-07-22

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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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    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!2012-07-26
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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).