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.

2 Answers

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:

enter image description here

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

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