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Jun 24, 2012 at 4:05 comment added Omar Antolín-Camarena 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 23, 2012 at 19:15 comment added Arturo Magidin 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 13:16 answer added user6976 timeline score: 4
Jun 23, 2012 at 11:15 vote accept Inna
Jun 23, 2012 at 10:53 answer added Derek Holt timeline score: 6
Jun 23, 2012 at 9:13 history asked Inna CC BY-SA 3.0