# Nilpotent of class 2 free product

Question. How is the nilpotent of class 2 (nil-2) free product of groups defined?

I came across this construction reading the following paper.

Alan H. Mekler (1981), Stability of nilpotent groups of class 2 and prime exponent. Journal of Symbolic Logic, 46, pp 781-788. doi:10.2307/2273227.

There the author use the nil-2 free product in order to build the nil-2 exponent $p$ group generated by some nice graph $\Gamma$.

• I don't know, but perhaps, for groups $G,H$, it could be defined as $K/[[K,K],K]$, where $K=G*H$. – Derek Holt May 21 '15 at 8:13
• In Hanna Neumann's book `Varieties of Groups', verbal product of a family of groups is defined as the quotient of the free product by the intersection of the verbal subgroup of the free product and the cartesian subgroup of the free product. In your case, you take the verbal product with respect to the variety defined by the law $[[x,y],z]=1$. – Primoz May 21 '15 at 10:01

For $2$-nilpotent groups $G$ and $H$, the $2$-nilpotent product of $G$ and $H$ is defined to be $$G\amalg^{\mathfrak{N}_2} H = \frac{G*H}{(G*H)_3}$$ where $G*H$ is the free product, and $(G*H)_3$ is the third term of the lower central series of $G*H$. Golovin defines it for arbitrary groups, in which case you mod out by the intersection of $(G*H)_3$ with the cartesian $[G,H]$.
The group has a nice normal form, in that the elements of $G\amalg^{\mathfrak{N}_2} H$ can be written as $ghc$, where $g\in G$, $h\in H$, and $c\in [H,G]$, where $[H,G]$ is isomorphic to the abelian tensor product $H^{\rm ab}\otimes G^{\rm ab}$; the product is given by $(ghc)(g'h'c') = gg'hh'cc'[h,g']$.
The group has the "obvious" universal property relative to nil-2 groups: there are natural inclusions $i_G\colon G\to G\amalg^{\mathfrak{N}_2} H$ and $i_H\colon H\to G\amalg^{\mathfrak{N}_2}H$, and if $K$ is any nil-2 group and $f\colon G\to K$ and $g\colon H\to K$ are group homomorphisms, then there is a unique homomorphism $\varphi\colon G\amalg^{\mathfrak{N}_2}H\to K$ such that $f=\varphi\circ i_G$ and $g=\varphi\circ i_H$.
(More generally, as indicated by Primoz, you can define a $\mathfrak{V}$-product for any variety $\mathfrak{V}$, for groups $G_i\in\mathfrak{V}$ as $$\frac{\mathop{*}\limits_{i\in I} G_i}{\mathfrak{V}(\mathop{*}\limits_{i\in I}G_i)},$$ where $\mathfrak{V}(K)$ is the $\mathfrak{V}$-verbal subgroup of $K$. If you don't assume the $G_i$ all lie in $\mathfrak{V}$, then you mod out by the intersection of that verbal subgroup with the cartesian $[G_i]_{i\in I}$, which is the kernel of the canonical map from the free product of the $G_i$ to the direct product of the $G_i$).