# Analyzing words in a "free" group of nilpotency class 2

Suppose that we have a group $G$ generated by a set $S$ of elements with the only family of relations being that all commutators are central. Equivalently, $G$ is the largest group of nilpotency class $2$ generated by $S$. Now suppose that we have two equivalent words $x_1x_2\cdots x_n$ and $y_1y_2\cdots y_n$, where each $x_i$ and $y_j$ is a generator. Does it follow that $x_i = y_i$ for all $i$?

Intuitively, the answer should be "no", since that's the usual answer to these kinds of questions, but I am having trouble constructing a counterexample.

• In the relatively free nilpotent group of class $c$, you can perform Hall's collection process (with respect to a particular ordering of the generators) to obtain a normal form for any particular word. If $c_1,c_2,\ldots,c_t$ is a sequence of basic commutators up to weight $c$, then every word can be written uniquely in the form $c_1^{a_1} c_2^{a_2}\cdots c_t^{a_t}$. I think that's as good a uniqueness in words as you'll get. Jun 23, 2012 at 19:15
• You know there are relations and you're just wondering if there is any relation "with no denominators". Well, I'd try finding some word $W$ in $x$ and $y$ so that you can "clear denominators" from the relation $xyx^{-1}y^{-1} W = W xyx^{-1}y^{-1}$. It seems reasonable to guess $W = yx$ to clear denominators from the LHS, and this indeed works and leads to the example Derek Holt gave. Jun 24, 2012 at 4:05

In the free class 2 nilpotent group with generators $x,y$, we have, for example, $xy^2x=yx^2y$.
Malcev characterized nilpotent groups of class $c$ by semigroup laws. Let $u_0=x$, $v_0=y$. By induction let $u_{n+1}=u_nz_{n+1}v_n$, $v_{n+1}=v_nz_{n+1}u_n$ where $z_i$ are different letters. Then a group is nilpotent of class $c$ if and only if it satisfies the law $u_c=v_c$. The proof is easy, see Malʹcev, A. I. Nilpotent semigroups. Ivanov. Gos. Ped. Inst. Uč. Zap. Fiz.-Mat. Nauki 4 (1953), 107–111.