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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?

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  • $\begingroup$ "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. $\endgroup$
    – Rasmus
    Apr 26 '11 at 22:31
  • $\begingroup$ thanks Rasmus. i corrected it $\endgroup$
    – El Moro
    Apr 26 '11 at 22:36
  • $\begingroup$ Is there any clement soul to help out with this :)? $\endgroup$
    – El Moro
    Apr 26 '11 at 22:58
  • $\begingroup$ I have to add that kernel here is the group of loops in ZxZ. $\endgroup$
    – El Moro
    Apr 26 '11 at 23:02
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$F_2$ 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.

(To compute the fundamental group of a graph, contract it along a spanning tree: the result is a wedge of some number of circles, and then the fundamental group is the free group on the circles by Seifert-van Kampen.)

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  • $\begingroup$ Thanks Yuan for your answer. Could you please do me a favor and explain to me how I could induce the inclusion by the choosing the right covering map and space? $\endgroup$
    – El Moro
    Apr 26 '11 at 23:14
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    $\begingroup$ @El Moro: that depends. Is this homework or not? (Really, in either case, you should try to construct the covering yourself: it's a good exercise, and pretty closely related to what you're saying in the OP.) $\endgroup$ Apr 26 '11 at 23:17
  • $\begingroup$ There are many of them and this why I needed your help. otherwise i wouldn't have asked.. $\endgroup$
    – El Moro
    Apr 26 '11 at 23:58
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    $\begingroup$ @El Moro: so this is homework? My personal policy is not to give complete answers to homework questions. This is already a decent hint. $\endgroup$ Apr 27 '11 at 0:01
  • $\begingroup$ No it is not a homework but I see your point and will try to work on it. this is actually about a fact that was told by arturo at a previous in which there was a question about free groups and he mentioned a proof of that fact (algebraic one) in Rotman's book. I tried to find an easier and intuitive proof by using algebraic topology :) $\endgroup$
    – El Moro
    Apr 27 '11 at 0:19

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