An identical relation on a group G is a word w in Fr, the free group on r elements (for some r), such that evaluating w on any r-tuple of elements of G yields the identity (this just means substituting elements of g for the variables in w, and evaluating the product).

Does the complete set of identical relations characterize a finite group? That is, are there two finite groups with precisely the same set of identical relations?

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    $\begingroup$ Your "identical relations" are often called "laws". $\endgroup$ – HJRW Dec 28 '09 at 17:04

Z/2 and Z/2 x Z/2 have the same identical relations: those words w such that every variable in w appears an even number of times. (This is obviously sufficient for w to be an identical relation. If w has a variable occurring an odd number of times, assign it a nonzero value and the other variables zero to obtain a nonzero value for w.)

  • $\begingroup$ More generally, this will be true for any group G and any product G^n. $\endgroup$ – Eric Wofsey Oct 15 '09 at 19:24
  • $\begingroup$ Of course - thanks. How about if the order of the group is fixed? If G and H are non-isomorphic of the same order, can they still have the same identical relations? Thanks again - if nobody notices the comment I'll post this as a separate question. $\endgroup$ – Alon Amit Oct 15 '09 at 21:33
  • $\begingroup$ Yes, they can. If G is abelian, then the set of identical relations of G only remembers its exponent (gcd of the orders of all elements) by similar reasoning. So we can take Z/4 x Z/4, Z/4 x Z/2 x Z/2 for instance. $\endgroup$ – Reid Barton Oct 15 '09 at 22:45
  • $\begingroup$ <bangs head on keyboard> of course. I need to think about what my question really ought to be. $\endgroup$ – Alon Amit Oct 15 '09 at 22:50

Two finite-dimensional prime algebras (in a certain extended sense, what includes the usual algebras with binary multiplication over a field) with the same identities are isomorphic over the algebraic closure of the ground field (Yu.P. Razmyslov, Identities of Algebras and Their Representations, AMS, 1994, p. 30 onwards). This suggests that the proper condition one should impose on finite groups to guarantee that the same identities imply isomorphism, would be something related to primeness.

Unfortunately, Razmyslov's reasonings seem not be extendible to a broader class of algebraic systems, in particular, to groups: the linear structure is crucial there (a relatively free algebra in the corresponding variety is enlarged, via the action of a suitable extension of the ground field, to an algebra which is isomorphic to an extension of the initial algebra). Also, it is not clear what the group-theoretic analog of primeness should be in this context.

However, there is an old result saying that two finite simple groups with the same identities are isomorphic (H. Neumann, Varieties of Groups, p. 166, Corollary 53.35). I believe that the machinery developed in that book (critical groups, etc.) would allow to answer this question for any reasonably defined class of finite groups.


Are you assuming the two groups are quotients of the same free group?

  • $\begingroup$ Er... no, why? Of course any two finite groups are quotients of the same free group if you choose it to have enough generators. Oh, I think I see what you're confused by: identical relations aren't the same thing as "relations" as in "generators and relations". Those are not relations satisfied by particular generators, but rather relations that are satisfied by all the elements of the group. $\endgroup$ – Alon Amit Oct 15 '09 at 22:49
  • $\begingroup$ Yes, that's why I was confused. It sounded (to me) like you might have been asking something like if two finitely presented groups are given in different alphabets but with the same number of generators and "identical" relations. $\endgroup$ – Josh Oct 16 '09 at 2:16

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