most recent 30 from http://mathoverflow.net 2013-05-20T06:53:36Z http://mathoverflow.net/feeds/question/37344 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37344/orders-of-products-of-permutations Mark Sapir 2010-09-01T02:36:47Z 2011-10-20T15:22:12Z

<p>Let $p$ be a prime, $n\gg p$ not divisible by $p$ (say, $n>2^{2^p}$). Are there two permutations $a, b$ of the set $\{1,...,n\}$ which together act transitively on $\{1,2,...,n\}$ and such that all products $w(a,b)=a^{k_1}b^{l_1}a^{k_2}...$ of length at most $n$ satisfy $w(a,b)^p=1$ (here $k_i,l_i\in {\mathbb Z}$)? </p> <p>Update: Following the discussion below (especially questions of Sergey Ivanov, here is a group theory problem closely related to the one before.</p> <p>Is there a torsion residually finite infinite finitely generated group $G$ such that $G/FC(G)$ is bounded torsion? Here $FC(G)$ is the FC-radical of $G$, that the (normal) subgroup of $G$ which is the union of all finite conjugacy classes of $G$.</p> <p>For explanations of relevance of this question see below (keep in mind that the direct product of finite groups coincides with its FC-radical). Note that if we would ask $G$ to be bounded torsion itself, the question would be equivalent to the restricted Burnside problem and would have negative answer by Zelmanov. </p> <p>If the answer to any of the two questions above is negative for some $p>665$, then there exists a non-residually finite hyperbolic group. </p>

http://mathoverflow.net/questions/37344/orders-of-products-of-permutations/37346#37346 Robert Bell 2010-09-01T02:58:16Z 2010-09-01T02:58:16Z

<p>If $p=2$, then your condition on words of length at most $2^p$ implies that $a$ and $b$ have order two and commute. Therefore they generate a group of order at most four. So, they together cannot act transitively on a large set.</p>