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Consider the homomorphism$ \ $ $f:\ F\{x,y\} \to <x,y|x^2, y^3, xyx^{-1}=y^{-1}>$, find the free generators of $kerf$.

I know that we should first consider the wedge sum of circles whose fundamental group is $F\{x,y\}$, then consider the covering space of the wedge sum of the circles which corresponds to the subgroup $kerf$. But how should I find the corresponding covering space, is there a general algorithm for this sort of problem?

Edit: I tried the following approach. Step 1: Draw the Caley graph of the cosets enter image description here Step 2: finding the free generators according to the caley graph.

If I start at Hx, then the free generators should be $y^3,x^2,yx^2y^{-1},yxy^2x^{-1},y,yxyxy^{-1},xyx^{-1}y^{-1}$

Am I right?

Adam
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  • Do you know what the Cayley complex of a group is ? – Captain Lama Apr 10 '16 at 14:20
  • The Reidermeister-Shrier algorithm is what you're after. (I'm almost certain I've misspelt it!) – user1729 Apr 10 '16 at 14:27
  • I know the Caley graph of a group. @CaptainLama – Adam Apr 10 '16 at 17:10
  • Can you recommend a resource where I can learn the algorithm? @user1729 – Adam Apr 10 '16 at 17:11
  • @Adam The Cayley complex is a little elaboration on that, it's like the Cayley graph but you add $2$-dimensional cells for each relation in the rpesentation. It is made so that its fundamental group is $G$ (which is not the case of the Cayley graph since the fundamental group of a graph is always free). All this and the Reidemeister-Schreirer algorithm can be found in the classic book "Combinatorial Group Theory" by Lyndon and Schupp. – Captain Lama Apr 10 '16 at 17:16
  • @CaptainLama The Cayley complex is the universal cover of the presentation complex, so it has trivial fundamental group. (Also...your italicised "the"...there are actually two classic books entitled "Combinatorial Group Theory"; Lyndon and Schupp named their book in honour of Magnus, Karrass and Solitar. From memory, contain Reidemeister-Schreirer, but if either contains a more geometric or conception version it will be Lyndon and Schupp.) – user1729 Apr 10 '16 at 21:17
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    Also see here. There is a nice "conceptual" version of the algorithm, which uses covering spaces. The issue with it is that is does not give you explicit generators. I am sure you can modify it to find these generators also, but I have never quite gathered up the motivation to do this. The algebraic version from Magnus, Karrass and Solitar gives you the generators, but is a "drier" algorithm. – user1729 Apr 10 '16 at 21:20
  • @user1729 It appears you're right, I got my terminology mixed up, I meant the $2$-dimensional complex associated to the presentation, whose universal cover is the Cayley complex. – Captain Lama Apr 10 '16 at 21:20
  • @CaptainLama That's okay - the only reason I know which is the Cayley Complex and which is the Presentation Complex is that I kept getting them mixed up so made a conscious effort to not make this mistake anymore! – user1729 Apr 10 '16 at 21:22
  • How to find the covering space corresponding to the $ker f$? @CaptainLama – Adam Apr 11 '16 at 01:24

1 Answers1

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In general, let $S$ be a set, and $X_S$ a wedge sum of circles labeled with $S$ (which we see as a graph). Then for any group $G$ generated by $S$ (as a monoid), you have a covering map $\Gamma_G\to X_S$ where $\Gamma_G$ is the Cayley graph of $G$ (of course $\Gamma_G$ depends on $S$) given by sending all vertices of $\Gamma_G$ to the only vertex of $X_S$, and all edges labeled by $s\in S$ to the only edge of $X_S$ labeled with $s$. Then the group of this cover is $G$.

Then in your case, $S=\{x,y\}$, and you have two groups generated by $S$ : your $G$, and the free group $F_S$. Clearly the Cayley graph $Y_S$ of $F_S$ is the universal cover of $X_S$, since it's a tree. So you have a tower of coverings $Y_S\to \Gamma_G\to X_S$.

Since the group of $Y_S\to X_S$ is $F_S$ and the group of $\Gamma_G\to X_S$ is $G$, the group of $Y_S\to \Gamma_S$ is $\ker(f)$, so $\ker(f) = \pi_1(\Gamma_G)$.

Now you have to find generators of $\pi_1(\Gamma_G)$. As explained for instance in the Lyndon-Schupp, this is done by choosing a maximal subtree in $\Gamma_G$ and contracting it to a point : the edges that remain are your generators. Looking at the labels along each one should give you explicit elements of $F_S$ that generate $\ker(f)$.

Captain Lama
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