Schreier transversal and a basis for commutator subgroup of $F_3$ I've seen the calculation for a Schreier transversal and basis for $[F_2,F_2]\lhd F_2=\langle x,y\rangle$, 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, $F:=F_3=\langle a,b,c\rangle$ and $F'=[F,F]$ give us a free abelian group $F/F'=\langle x,y,z|[x,y]=[x,z]=[y,z]=1\rangle$. The cosets are uniquely given by $a^ib^jc^kF'$, where $i,j,k\in\mathbb{Z}$, 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 $h_{t,x}=\ell(Ht)x\ell(Htx)^{-1}$ that are distinct from 1, where $x\in X$" [$\ell(Ht)$ 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.
 A: You have a confusing notation clash, because you have used $a$ twice with different meanings, so let me change the notation to $h_{t,x} = \ell(Ht)x\ell(Htx)^{-1}$.
Here are a few examples of how to calculate $h_{t,x}$.

*

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


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


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


*$t = b^5$. 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!
A: 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.
