# Schreier transversal and a basis for commutator subgroup of $F_3$

Mathematics Asked by Makenzie on December 25, 2020

I’ve seen the calculation for a Schreier transversal and basis for $$[F_2,F_2]lhd F_2=langle x,yrangle$$, but these groups aren’t so complex that the calculations were particularly illuminating. I was wondering if anyone has a reference for the calculations in $$F_3$$ so I can get a firmer grasp on how these free groups work?

Specifically, $$F:=F_3=langle a,b,crangle$$ and $$F’=[F,F]$$ give us a free abelian group $$F/F’=langle x,y,z|[x,y]=[x,z]=[y,z]=1rangle$$. The cosets are uniquely given by $$a^ib^jc^kF’$$, where $$i,j,kinmathbb{Z}$$, and these yield a Schreier transversal, but what is the basis for $$F’$$?

Rotman’s book on group theory says I should use "all those $$h_{t,x}=ell(Ht)xell(Htx)^{-1}$$ that are distinct from 1, where $$xin X$$" [$$ell(Ht)$$ is his notation for the coset representative and $$X$$ is the generating set for $$F$$], but this statement confuses me because it’s unclear how many calculations I need to do or how to do so systematically.

You have a confusing notation clash, because you have used $$a$$ twice with different meanings, so let me change the notation to $$h_{t,x} = ell(Ht)xell(Htx)^{-1}$$.

Here are a few examples of how to calculate $$h_{t,x}$$.

1. $$t = a^{-3}b^2c^{-4}$$, $$x=a$$. Then $$ell(Htx) = a^{-2}b^2c^{-4}$$, so $$h_{t,x} = a^{-3}b^2c^{-4}ac^4b^{-2}a^2$$.

2. $$t = a^{-3}b^2c^{-4}$$, $$x=b$$. Then $$ell(Htx) = a^{-3}b^3c^{-4}$$, so $$h_{t,x} = a^{-3}b^2c^{-4}bc^4b^{-3}a^3$$.

3. $$t = a^{-3}b^2c^{-4}$$, $$x=c$$. Then $$ell(Htx) = a^{-3}b^2c^{-3}$$, so $$h_{t,x} = 1$$, and we discard this as a generator of $$[F,F]$$.

4. $$t = b^5$$. Then $$h_{t,b}=h_{t,c} = 1$$, but $$h_{t,a} = b^5ab^{-5}a^{-1}$$.

Of course there are infinitely many generators, so you will have to write down the answer using more general notation. You could treat the three cases $$x=a,b,c$$ separately.

For $$t = a^ib^jc^k$$, we have $$h_{t,x}=1$$ if and only if (i) $$x=c$$; (ii) $$k=0$$ and $$x=b$$; or (iii) $$j=k=0$$ and $$x=a$$.

I hope that helps!

Correct answer by Derek Holt on December 25, 2020

Let $$f : F_n to G$$ be any surjection. $$f$$ determines a Galois cover of the wedge $$X = bigvee_{i=1}^n S^1$$ of $$n$$ circles with Galois group $$G$$ which abstractly is the covering of $$X$$ corresponding to the kernel of $$f$$, and which concretely is given by the Cayley graph of $$G$$ with respect to the choice of generators given by $$f$$ applied to the standard generators of $$F_n$$. Every Galois cover of a wedge of circles arises this way.

Applied to $$G = F_n/[F_n, F_n] cong mathbb{Z}^n$$ the abelianization of $$F_n$$ we get that the covering of $$X$$ corresponding to its commutator subgroup is the Cayley graph of $$mathbb{Z}^n$$ corresponding to the generators given by the standard basis $$e_1, dots e_n$$. When $$n = 2$$ this is the "graph paper grid" (I don't know if it has a more standard name) and when $$n = 3$$ it's the obvious 3D generalization of that. Like this but infinite in all directions:

This means we can get a basis of $$[F_n, F_n]$$ by picking a spanning tree of this grid and contracting it to a point. Basis elements correspond to edges not in the spanning tree, and can be constructed as words by constructing a path in the graph from the origin through the spanning tree that passes through the edge and then goes back to the origin. (Is this what a Schreier transversal is?)

Maybe you already know all this in more explicitly group-theoretic language though.

Answered by Qiaochu Yuan on December 25, 2020