Retract of $F_2$ onto $[F_2,F_2]$

Question

I'm trying to show that the commutator subgroup  $[F_2,F_2]$ is not a retract of $F_2$. I was trying to do the proof by contradiction so I assume there is a retract $f: F_2 to [F_2,F_2]$, then I know this is surjective and then I tried to apply abelianization to get the map $f^{ab}: mathbb{Z}^2 to {1}$ but this doesn't give me any contradictions since this map is surjective. Then I was thinking to apply the universal property of quotient groups somehow because I know $F_2/ [F_2,F_2] = mathbb{Z}^2$ and that is a surjective map but I can't find anything that gives me a contradiction. I'm not sure where to go from here.

Thomas · Answer

We can make the previous argument more precise. The group $[F_2,F_2]$ is a free group. It is freely generated by elements of the form $a^n b^m [a,b]b^{-m}a^{-n}$, for $n, min bf Z$, as it is the fundmantal group of the Cayley graph of $bf Z^2$ viewed as a quotient $F_2/[F_2,F_2]$, so it is not finitely generated.

Answered by Thomas on January 8, 2021

Qiaochu Yuan · Answer

Abelianization does give you a contradiction; $[F_2, F_2]$ is a free group on countably many generators, so its abelianization is the free abelian group on countably many generators, and $f^{ab}$ can't be surjective.

Answered by Qiaochu Yuan on January 8, 2021

Retract of $F_2$ onto $[F_2,F_2]$

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