Is the free product of arbitrarily many copies of `${\mathbb{Z}}$` and `${\mathbb{Z}}/2$` residually nilpotent?

Question

A group is residually nilpotent if the intersection of the terms in its lower central series is the trivial group.

Is the free product of arbitarily (possibly infinite) many abelian groups residually nilpotent? In particular, this would imply that the free product of abelian groups with any free group is residually nilpotent.

---

Later: Thank you for all the comments and answers. It is good to know that the free product of abelian groups is residually solvable.

The group I am interested in is the free product $F*G$ where $F$ is a free group and $G$ is a free product of arbitrarily many ${\mathbb{Z}}/2$. In the special case ${\mathbb{Z}}*{\mathbb{Z}}/2$, I know that this group is residually nilpotent. This follows from a result of Mal'cev on adjoint groups of residually nilpotent algebras.

Answer 1

The group $PSL(2,\mathbb{Z})=\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/3\mathbb{Z}$ is not residually nilpotent. Indeed, if it was residually nilpotent, then it would be residually (finite $p$-)group since every finitely generated nilpotent group is residually finite (Malcev) and every finite nilpotent group is a direct product of finite $p$-groups. But in every finite $p$-group either the image of the generator $a$ of order 2 or the image of generator $b$ of order $3$ of $PSL(2,\mathbb{Z})$ is equal to 1 (since either 2 or 3 is co-prime with $p$). Hence the image of the commutator $[a,b]$ is 1. Thus one cannot separate $[a,b]$ from 1 in any nilpotent homomorphic image of $PSL(2,\mathbb{Z})$.

On the other hand, for every $p$, any free product of two residually (finite $p$-) groups is residually (finite $p$-) by a result of Gruenberg (see, for example, http://www.ams.org/journals/bull/1969-75-02/S0002-9904-1969-12149-X/S0002-9904-1969-12149-X.pdf), and hence residually nilpotent.

Update 1. This is an answer to Andreas Thom's comment. Every free product $G=* G_i$ of Abelian groups is residually solvable. Indeed, $G$ has a homomorphism onto the direct product $A$ of $G_i$. The kernel $K$ certainly contains the derived subgroup (actually these two subgroups coincide but it does not matter here). The subgroup $K$ is free (hence the derived subgroup of $G$ is free). Indeed, if $K$ was not free, it would have a non-trivial intersection with one of the conjugates of $G_i$ by Kurosh's theorem, and hence a non-trivial image in $A$. Since $K$ is free, and $G/K$ is Abelian, $G$ is residually solvable. Moreover $G$ is residually (torsion-free nilpotent-by-Abelian).

Update 2. The question changed. Here is the answer to the new question. The free product of $F∗G$ where $F$ is free and $G$ is a free product of $\mathbb{Z}/2\mathbb{Z}$ is residually (finite 2-)group, hence residually nilpotent. For finitely many factors it follows from Gruenberg's result I mentioned above. For infinitely many factors take any word $w\ne 1$ in generators of the free product. Let $X$ be the finite set of generators that appear in $w$. Let $U$ be the finitely generated free factor of $F*G$ generated by $X$. Then $U$ is residually nilpotent by Gruenberg. Hence there exists a homomorphism $\phi$ from $U$ onto a nilpotent group separating $w$ from 1. Since $U$ is a retract of $F*G$, the composition of the retraction and $\phi$ maps $F*G$ into a nilpotent group and separates $w$ from 1.

Answer 2

Mark Sapir has already given you a definitive answer, but I think it is interesting to note that his argument generalises to show something stronger. A finitely-generated non-trivial free product of groups is residually nilpotent only if it is $p^{\prime}$-torsion-free, for some prime $p$. (Here, $p^{\prime}$ denotes the set of primes not equal to $p$ so, in this case, we are allowed only $p$-torsion.) For, in a finitely-generated residually nilpotent group, elements of finite coprime order commute. And, if $G = A\ast B$ is such a free product (with $A\neq 1\neq B$), and $G$ is not $p^{\prime}$-torsion-free, for any prime $p$, there are distinct primes $p$ and $q$, and elements $u$, of order $p$, and $v$, of order $q$ in $G$. These may be taken (up to conjugation) to be members of $A\cup B$. Now, if $u$ and $v$ belong to different free factors ($A$ or $B$) then they cannot commute, so it must be that $u$ and $v$ belong to the same factor, say $A$. But then, taking a non-trivial element $b$ in $B$ and forming the conjugate $w = b^{-1}vb$, we get an element of order $q$ which again does not commute with $u$. (The commutator $[u,w] = u^{-1}b^{-1}v^{-1}bub^{-1}vb\neq 1$, by the normal form theorem for free products.) This again contradicts the fact that elements of coprime order commute.