# Free groups and commutators

Good evening

I was trying to prove that the commutator [F2,F2] of the free group F2 is not finitely generated by using covering spaces (i have to admit that this is the idea of a friend) it seems that F2 is the fundamental group of the plane with two points removed.

I al thinking of linking [F2,F2] to the kernel of the homomorphism that we use to prove that the quotient π1(ZxZ)/[π1(ZxZ),π1(ZxZ)] isomorphic to H1(ZxZ).

The homology group here is the free abelian group with two generators: we can visualise it as the vertices of ZxZ. we can visualise the map from F2 (with generators a and b) to ZxZ as follows: to go from the origin of the grid to a point in ZxZ we move horizontally and vertically by units to go from O to (1,2) we go upward 1 right 1 then upward 1 we model that movement by "bab" we can move right and then upward twice which makes it "abb" then we notice that we have a homomorphism from Fé to ZxZ with a kernel which is exactly [F2,F2].

Any ideas?

• "F2 is the fundamental group of an infinite grid on say ZxZ" seems false to me. $F_2$ is the fundamental group of the plane with two points removed. Apr 26 '11 at 22:31
• thanks Rasmus. i corrected it Apr 26 '11 at 22:36
• Is there any clement soul to help out with this :)? Apr 26 '11 at 22:58
• I have to add that kernel here is the group of loops in ZxZ. Apr 26 '11 at 23:02

$F_2$ is the fundamental group of the wedge of two circles, which we denote by $X$. One idea is to construct the covering $p : Y \to X$ which induces the inclusion $[F_2, F_2] \to F_2$ on fundamental groups. $Y$ is also going to be a graph so this is not hard to do directly, and then it is also not hard to verify that the fundamental group of $Y$ is the free group on countably many generators.