On the Magnus Representation of Free Metabelian Group

Question

Let $F=\langle x_1,x_2\rangle$ be a free group of rank $2$ and $\Phi=F/F''=\langle \overline{x}_1, \overline{x}_2\rangle$ where $F''$ is second derived subgroup of $F$ (i.e. $F'=[F,F]$ and $F''=[F',F']$).

Let $M=\mathbb{Z}[s_i,s_i^{-1},t_i,t_i^{-1}]_{i=1,2}$, the polynomial ring in eight (commuting) variables over $\mathbb{Z}$ with $s_i.s_i^{-1}=t_i.t_i^{-1}=1$. By universal property (definition) of $F$, the map

$$\rho\colon x_i\mapsto \begin{bmatrix} s_i & t_i\\ 0 & 1\end{bmatrix}$$ extends to a homomorphism from $F$ to ${\rm GL}(M)$, and it can be shown that the subgroup $F''$ of $F$ is in the kernel of homomorphism. Thus we get a homomorphism $$\overline{\rho}\colon F/F''\rightarrow {\rm GL}(M)\,\,\,\, \mbox{ i.e. }\,\,\,\, \overline{\rho}\colon \Phi\rightarrow {\rm GL}(M).$$ Question: Whay $\overline{\rho}$ is faithful? (In other words, why $\ker\rho$ is exactly $F''$?)

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$\overline{\rho}$ is called the Magnus representation of the free metabelian group $\Phi$. In the paper

"Automorphisms of Free Matabelian Groups"-Bachmuth,

is was pointed that $\overline{\rho}$ is faithful representation of $\Phi$. But how to prove it? I didn't understand the faithfulness. If this is elementary, then please, give at least a hint.

Answer 1

It is a general result, due to Jorge Almeida, that if $\mathfrak V$ and $\mathfrak W$ are varieties of groups (actually he does semigroups but it is irrelevant) and $\mathfrak V\mathfrak W$ is the variety generated by all semidirect products of groups in $\mathfrak V$ with groups in $\mathfrak W$ (which by the Krasner-Kaloujnine embedding is the same as the variety of all groups $G$ with a normal subgroup $N\in \mathfrak V$ with $G/N\in \mathfrak W$), then the relatively free group in $\mathfrak V\mathfrak W$ on a set $A$ is the subgroup of $$F_{\mathfrak V}(F_{\mathfrak W}(A)\times A)\rtimes F_{\mathfrak W}(A)$$ generated by the elements of the form $((1,a),a)$. Here $F_{\mathfrak W}(A)$ acts on $F_{\mathfrak W}(A)\times A$ in the natural way and then extends to the relatively free group via the universal property.

This is not too hard to prove once somebody tells you it should be true. Then you should verify that when $\mathfrak V=\mathfrak W$ is the variety of abelian groups, then this yields your Magnus embedding above in disguise.

Answer 2

You may find a proof (that was done by Romanovsky) of a more general result of Magnus in: J.S. Wilson, Free subgroups in groups with few relations. L’Enseignement Math. (2) 56 (2010), 173–185.

See also Sect. 3--4 in arXiv:1407.3447 [math.GR] for variants of the Magnus embedding.

By the way, you don't need $t_i^{-1}$ (and actually may get rid of $t_i$ as well if you slightly modify the homomorphism/embedding).