TITLE: Is the free product of arbitrarily many copies of `${\mathbb{Z}}$` and `${\mathbb{Z}}/2$` residually nilpotent?
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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.