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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 $k_i,l_i\in {\mathbb Z}$)?

Update: Following the discussion below (especially questions of Sergey Ivanov, here is a group theory problem closely related to the one before.

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 $G$.

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.

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.

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The number 665 in the comment below, no doubt, arises from the negative solution of Burnside's problem. Can you, please, edit the question and indicate, as a motivation, what you know about the connection? –  Victor Protsak Sep 1 '10 at 8:21
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It does not apply to the bounded Burnside problem. Together with the solution of Burnside problem it applies to some other group theory problem. Unlike the bounded Burnside problem, this is about finite groups. In fact group theory motivation is misleading here because I do not see how to apply any group theory to this problem. It may be just a combinatorial Olympiad-type problem about graphs or it may belong to a completely different area of mathematics. So in this case, I think, the less you know about Burnside problems the better. –  Mark Sapir Sep 1 '10 at 9:30
    
Does the hypothetical construction have to work for all large enough values of $n$, or would it suffice to give a sequence $n_i \to \infty$ for which the construction works? –  Keivan Karai Sep 1 '10 at 11:58
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The condition that $a,b$ act transitively on a set with at least $n$ elements is equivalent to the condition that the group they generate has order at least $n.$ One direction is trivial and the other direction is given by the Cayley construction (action on itself by left multiplication). Thus the question can be reformulated as follows: does there exist a 2-generated group of large order $n>2^{2^p}$ in which all short words (length at most $2^p$) have order dividing $p$? –  Victor Protsak Sep 6 '10 at 6:39
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@Victor: I just noticed your claim that "The condition that a,b act transitively on a set with at least n elements is equivalent to the condition that the group they generate has order at least n. " That is not true because $S_{n-1}$ is a non-transitive subgroup of $S_n$. –  Mark Sapir Sep 21 '10 at 1:46

1 Answer 1

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.

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The question is for all $p$'s. The first "interesting" p is $>665$. The expected answer is "no" for all $p$. –  Mark Sapir Sep 1 '10 at 3:38
    
To be more precise: the case $p\le 3$ is easy. I do not know how to do the case $p=5$. The first case interesting for applications is $p=673$. –  Mark Sapir Sep 1 '10 at 4:31

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