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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 $1 \to \mathbb{Z} \to D_{\infty} \to \mathbb{Z}_2 \to 1$).

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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
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up vote 2 down vote accepted

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).

A detailed combinatorial proof appears in Serre's Trees, Proposition I.1.4 on page 6 which is based on an exercise in Magnus-Karrass-Solitar. Further references are given in the answers to this question.

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.

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