In the Combinatorial Group Theor-Lyndon, Schup, the authors say (start with)

Informally, a group is free on a set of generators if no relation holds among these generators except trivial relations that hold among any set of elements in any group.

With this definition of Free group, I tried to see why the commutator subgroup of free group $F_2=\langle a,b\rangle$ on two generators is not finitely generated. The hint given was that the set $\{[a^i,b^j]\colon i,j\in\mathbb{Z}\}$ is a (free) generating set for $[F_2,F_2]$.

As an example, I tried to see why $[a^2,b]$ and $[a,b]$ are free? Informally, if they were not free, then they would a non-trivial relation between them, and hence between $a,b$.

But, I am not satisfied myself with this informal proof of $[a,b]$ and $[a^2,b]$ are free.

How can we prove precisely this last statement? Should we go back to the precise definition of free groups?


1 Answer 1


Let $x=[a^2,b]=a^{-2}b^{-1}ab$ and $y=[a,b]=a^{-1}b^{-1}ab$.

Conisder all words of the form $uvw$, where $u,v,w \in \{x,x^{-1},y,y^{-1}\}$, and $v\ne u^{-1}$, $v \ne w^{-1}$. You will find that there is no such product $uvw$ in which the whole of $v$ cancels.

So the reduction of an reduced word $u_1u_2\cdots u_k$ with $u_i \in \{x,x^{-1},y,y^{-1}\}$ has length at least $k$ and so cannot reduce to the identity. So the subgroup generated by $x$ and $y$ is free.

The Nielsen proof that a subgroup of a free group is free involves an algorithmic method for reducing a generating set of a subgroup to a free generating set, so you could apply that method to examples like this.

  • $\begingroup$ You dropped the second $2$ from $x=a^{-2}b^{-1}a^2b.$ $\endgroup$ Commented Sep 15, 2021 at 20:00
  • $\begingroup$ Yes thanks! I don't think I will edit the post, because that would take it to the top of the active questions list, which seems like overkill. $\endgroup$
    – Derek Holt
    Commented Sep 16, 2021 at 7:51

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