MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a group $G$ and a tuple $J = (g_1,g_2 ... g_n) \in G^k$ for $k$ some constant, define a parametrized word $w : G^k \rightarrow G$ to be a function which takes $J$ to some product of the elements in $J$. So $w(J) = g_1g_1g_2$ for $k \geq 2$ would be an example.

The structure of the space of all $w$ for a particular group modulo the equivalence relation of functional equality is not trivial. For instance, over $\mathbb{Z}_2$, $g_1g_1g_2 = g_2$ for all $J$ ,and for a finite abelian group the space of $w$ is clearly finite.

I don't know whether this topic has been covered before; It seems simple enough that someone might have done work on it, but I cannot find anything. Does anyone know what this area might be called?

share|cite|improve this question
In a vague sense it sounds like you're looking not just at the relations in a space, but relations among the relations. In many ways that's what group cohomology is about. – Ryan Budney Jan 25 '12 at 23:34
Your space (if I understand correctly) is the free group on k letters modulo all identities satisfied by G. Groups defined by certain sets of identities have been extensively studied, although it can be difficult to understand them in general: see for instance the theory of Burnside groups or Engel groups. – Colin Reid Jan 26 '12 at 0:31
If your group G has infinite exponent then yes, it matters whether you allow negative powers. (The positive words only form a monoid in general.) If the group you start with has finite exponent, then you have inverses for free and you are effectively asking about certain quotients of Burnside groups defined by identities. – Colin Reid Jan 26 '12 at 8:40
Syzygy? – Alain Valette Jan 26 '12 at 10:19
If you want to identify all positive words that agree on all k-tuples of group elements you get the free monoid in the variety generated by G on k generators. If you allow all group words you get the free group in the variety generated by G on k-generators. If G is finite these two guys are the same and the resulting group is finite. – Benjamin Steinberg Jan 26 '12 at 12:15
up vote 6 down vote accepted

I believe this is the subject of "word maps". See this link for a list of relevant authors (there are papers by Shalev and Larsen, e.g.), it is a big area.

share|cite|improve this answer

From comments it seems you want to know when your group/monoid of functions is finite for all k. These are equivalent and the answer is when G is locally finite of finite exponent. The group you are looking at is the free group of rank k in the variety generated by G. If G has infinite exponent the words $x^n$ are all distinct functions on G.

If G is of exponent n and is not locally finite then it has an infinite k-generated subgroup so the free group in the variety generated by G is infinite. If G is locally finite, then since varieties of groups are closed under direct limits, it follows G belongs to the variety of groups generated by finite groups of exponent n. This variety has finite free objects on finite generating sets by the solution to the restricted Burnside problem. Thus the free objects in the variety generated by G are finite as well.

If G is finite things are trivial since there are finitely many k-ary functions on G. In fact it is a classical result of Birkhoff shows the variety generated by a finite universal algebra is locally finite.

share|cite|improve this answer
I might recommend Chapter 4 Section 1 of "Algebras, Lattices, Varieties" by McKenzie, McNulty, and Taylor. It starts out with clone theory (which is related), and later in the chapter is the result of Birkhoff you mention. Gerhard "Ask Me About System Design" Paseman, 2012.01.26 – Gerhard Paseman Jan 26 '12 at 20:26
Gerhard, this question is almost exactly about clones is it not? – Benjamin Steinberg Jan 26 '12 at 22:17
The question could use some clarity. I bet if the poster did the suggested reading (I think it accessible, and am willing to help on MathOverflow if it is not), they would determine quickly if it was clones or identities they were interested in. Gerhard "Ask Me About General Algebra" Paseman, 2012.01.26 – Gerhard Paseman Jan 26 '12 at 22:35
Agreed. This is why I have 2 answers. – Benjamin Steinberg Jan 27 '12 at 0:53

I am not sure if this is exactly the same as what you want. A mapping $f\colon G^k\to G$ is a polynomial if there is an element $u$ in the free product of $G$ with a free group on $k$-generators such that $f$ is obtained by substituting the k-tuple of elements of G in for the free group elements and taking the product in $G$. Rhodes and Maurer proved that every $k$-ary function on a finite group $G$ is polynomial (for all k) if and only if G is simple nonabelian. This has applications in circuit complexity theory and was rediscovered by Barrington in that context.

share|cite|improve this answer

You raise a specific question the comments: Let $G$ be a finite group, let $k$ be an integer and let $\Gamma$ be the group on $k$ generators with relators given by the identities of $G$. Is $\Gamma$ finite for all $G$ and $k$?

I don't know the answer, but the question does at least reduce to the case when $G$ is a finite simple group. To keep things short I will use 'identity of $G$' to mean a word whose only value on $G$ is the identity element. Suppose that $G$ has minimal order among the finite groups for which $\Gamma$ is infinite (for $k$ large), and that $G$ is not simple. Let $K$ be a normal subgroup of $G$ such that $G/K$ is simple, and let $\Gamma_1$ be the subgroup of $\Gamma$ generated by all identities of $G/K$. Then $\Gamma/\Gamma_1$ is finite by the minimality of $G$, so $\Gamma_1$ is finitely generated; let $X$ be some finite generating set for $\Gamma_1$. Then by the minimality of $G$, $\Gamma_1/\Gamma_2$ is finite, where $\Gamma_2$ is generated by all the identities of $K$, written in the alphabet $X$. I claim that $\Gamma_2$ is actually trivial: if we write an identity of $K$ using letters that are themselves identities of $G/K$, we get an identity of $G$. Thus $\Gamma$ is finite, contradiction.

The answer is clearly 'yes' for the cyclic group of order $p$ (and hence for all finite soluble groups): in this case $\Gamma$ is elementary abelian of order $p^k$. For the non-abelian finite simple groups, I can't see a clever way of doing it - it would be nice to see a proof that doesn't involve a CFSG trawl, certainly.

share|cite|improve this answer
Colin, of course it is finite if the group is finite. There are only finitely many k-ary functions on the group. The fact that the finitely generated free object in the variety generated by a finite universal algebra goes back to Birkhoff. – Benjamin Steinberg Jan 26 '12 at 18:38
Ah yes, of course. – Colin Reid Jan 27 '12 at 0:44

Your Answer


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.