Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
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. –  Arturo Magidin Jun 23 '12 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. –  Omar Antolín-Camarena Jun 24 '12 at 4:05
add comment

2 Answers 2

up vote 6 down vote accepted

In the free class 2 nilpotent group with generators $x,y$, we have, for example, $xy^2x=yx^2y$.

More generally, since nilpotent groups have polynomial growth, they cannot contain free semigroups of rank greater than 1, so such equations must exist for all nilpotent groups.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.