Show that the free group on three generators is a subgroup of the free group on two generators

In the free group $F(a,b)$ on 2 generators $a,b$, the family $(a^nba^{-n})_{n\in\mathbf{Z}}$ is free.

One way to prove this is to consider the free group $F(\mathbf{Z})$ on the generators $(b_n)_{n\in\mathbf{Z}}$, its automorphism $f$ induced by the assignment $b_n\mapsto b_{n+1}$, and consider the semidirect product $G=\mathbf{Z}\ltimes_fF(\mathbf{Z})$, where the positive generator $t$ of $\mathbf{Z}$ acts by $f$. Then there is a unique homomorphism $u:F(a,b)\to G$ mapping $a\mapsto t$, $b\mapsto b_0$. Then $u$ maps $a^nba^{-n}$ to $b_n$. Since $(b_n)_{n\in\mathbf{Z}}$ is free, it follows that $(a^nba^{-n})_{n\in\mathbf{Z}}$ is free too.

Remark (not used above): one can show that $u$ is an isomorphism $F(a,b)\to G$.


It is no coincidence this was in an algebraic topology textbook. I will show the stronger statement "The free group on countably infinite generators is a subgroup of the free group on two generators."

Take the usual universal cover of $\mathbb{R}$ which is given by the product of the universal covers for $S^1$, notably it's domain is $\mathbb{R}^2$. Now if we restrict this map to $\mathbb{Z}^2$ (Edit: this should be the grid lines through $\mathbb{Z}^2$) we get a universal cover for $S^1 \vee S^1$. Pick your favorite spanning tree for the grid and contract it to see that it has the homotopy type of a wedge of countably many circles. Such a thing has fundamental group a free group on the inclusion of each circle, so is free on countably infinite generators. $S^1 \vee S^1$ has fundamental group that is free on two generators. The last thing you need is that any covering map induces an injection on fundamental groups.

If you actually pick a spanning tree, you can use it to write exactly what the basis of this group is. I think a basis element looks like $a^n b^m aba^{-1}b^{-1} a^{-n}b^{-m}$.

Tags:

Group Theory

Free Groups

Algebraic Topology

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