Have any publications been made in this area of group theory? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T15:44:06Z http://mathoverflow.net/feeds/question/86667 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86667/have-any-publications-been-made-in-this-area-of-group-theory Have any publications been made in this area of group theory? unknown (google) 2012-01-25T23:32:06Z 2012-04-27T08:23:20Z <p>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. </p> <p>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.</p> <p>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?</p> http://mathoverflow.net/questions/86667/have-any-publications-been-made-in-this-area-of-group-theory/86676#86676 Answer by Igor Rivin for Have any publications been made in this area of group theory? Igor Rivin 2012-01-26T00:35:44Z 2012-01-26T00:35:44Z <p>I believe this is the subject of "word maps". See <a href="http://www.mpim-bonn.mpg.de/node/3421" rel="nofollow">this link</a> for a list of relevant authors (there are papers by Shalev and Larsen, e.g.), it is a big area.</p> http://mathoverflow.net/questions/86667/have-any-publications-been-made-in-this-area-of-group-theory/86677#86677 Answer by Benjamin Steinberg for Have any publications been made in this area of group theory? Benjamin Steinberg 2012-01-26T00:38:50Z 2012-01-26T00:38:50Z <p>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. </p> http://mathoverflow.net/questions/86667/have-any-publications-been-made-in-this-area-of-group-theory/86733#86733 Answer by Colin Reid for Have any publications been made in this area of group theory? Colin Reid 2012-01-26T16:32:26Z 2012-01-26T16:32:26Z <p>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 $k$?</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/86667/have-any-publications-been-made-in-this-area-of-group-theory/86748#86748 Answer by Benjamin Steinberg for Have any publications been made in this area of group theory? Benjamin Steinberg 2012-01-26T19:04:23Z 2012-01-26T19:04:23Z <p>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. </p> <p>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. </p> <p>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. </p>