# When is $G \ast H$ solvable?

In a proof that the lamplighter group $\mathbb{Z}_2 \wr \mathbb{Z}$ is not finitely presented, I showed that $\mathbb{Z}_2 \ast \mathbb{Z}$ is not solvable. More precisely, one can prove that the commutator subgroup of $\mathbb{Z}_2 \ast \mathbb{Z}= \langle a,b | a^2=1 \rangle$ is free on $\{[a,b^n], n \in \mathbb{Z} \backslash \{0\} \}$; my proof is mainly based on Kurosh theorem and it can be extend to any group of the form $A \ast B$ or $A \ast F$ where $A$ and $B$ are abelian and have finite exponents and $F$ is a free group.

Is it right that the commutator subgroup of $A \ast B$ is always free on $\{[a,b] : a \in A \backslash \{1\},b \in B \backslash \{1\}\}$? It seems to be right because there are few cancellations in a product of commutators: I think a combinatorial proof would be possible, but I don't have a rigorous proof.

As a corollary, we would have that $A \ast B$ is solvable iff $A \simeq B \simeq \mathbb{Z}_2$ ($\mathbb{Z}_2 \ast \mathbb{Z}_2=D_{\infty}$ is indeed solvable since it admits an exact sequence $1 \to \mathbb{Z} \to D_{\infty} \to \mathbb{Z}_2 \to 1$).

-
The answers to this question explain that the kernel of the canonical map $A \ast B \to A \times B$ is free on the commutators $[a,b]$ with $a \neq e \neq b$. More generally, one can show that a free product with amalgamation $A \mathbin{\ast_C} B$ contains a free non-abelian subgroup as soon as $([A : C] - 1) ([B : C]-1) \geq 2$. – Martin Jan 15 '13 at 17:25
@Martin: The proof of Serre is exactly what I seeked, thank you. – Seirios Jan 15 '13 at 18:33
@Martin You should post that as an answer! – Alexander Gruber Jan 15 '13 at 19:42
@Alexander: I was hoping some of the strong group theorists here could provide some more insight than I can. If no answer is posted by tomorrow I will do so. – Martin Jan 15 '13 at 19:48
The general case $G\ast H$ reduces to the abelian case by choosing an abelian subgroup $A\le G$ and a nontrivial element $b\in H$, so that $A\ast \langle b\rangle$ is a subgroup of $G\ast H$. When $|A|>2$ - which can always be made the case when $|G|>2$ - the kernel of the map $A\ast\langle b\rangle\rightarrow A\times\langle b\rangle$ is nonabelian free. – user641 Jan 16 '13 at 14:14

If $A$ and $B$ are any two groups then one can deduce from the normal form theorem for elements of $A \ast B$ that the commutators $[a,b]$ with $a \in A \setminus \{e_A\}$ and $b \in B \setminus \{e_B\}$ form a free basis of the kernel of the map $A \ast B \to A \times B$ determined by $a \mapsto (a, e_B)$ and $b \mapsto (e_A,b)$ for $a \in A$ and $b \in B$. Since the only solvable free groups are the ones of rank at most one, this kernel is solvable only if either $\lvert A \rvert, \lvert B \rvert \leq 2$. Thus, your proposed solution is confirmed (in your formulation you implicitly assume that $A$ and $B$ are both non-trivial).
More generally, according to Complement VII.9 on page 191f of de la Harpe's Topics in geometric group theory an amalgamated product $A \ast_{C} B$ contains a non-abelian free group whenever $([A:C]-1)([B:C]-1) \geq 2$. Again, this fact can be proved by appealing to normal forms.