Let $G = Sym(n)$, $n$ even. Let $H<G$ be the stabilizer of the partition $\{\{1,2\},\{3,4\},\dotsc,\{n-1,n\}\}$, or, what is the same, the centralizer of $(1\;2) \dotsc (n-1\; n)$.

By Stirling's formula, the number $|H|$ of elements of $H$ is somewhat smaller than $G$. While $H\cap g H g^{-1}$ is not trivial for any $g\in G$ (there is a simple argument showing as much in [1], as a colleague kindly pointed out), it does not seem hard to show that $H\cap g H g^{-1}$ is small (meaning: having $n^{O(1)}$ elements) for a proportion tending to $1$ of all elements $g\in G$. Call such elements "good".

My question is what is the structure of the set of good elements. In particular: if I have a set of generators $A$ of $G$, is there an element of $B(n^{100}) := (A\cup A^{-1}\cup \{e\})^{n^{100}}$ (say) that is good? (Here I write $S^k$ to mean the set of elements of the form $g_1 g_2 \dotsb g_k$, $g_i\in S$.)

An analogy with algebraic sets might be in order. Say we had a variety $V$ of positive codimension in $SL_n$ (or some other algebraic group). Let us call the elements of the complement $G\setminus V$ "good". Then we can in fact show that there is always an element of $B(r)=(A\cup A^{-1}\cup \{e\})^r$ that is good, where $r$ depends only on $n$ and the degree of $V$. The argument proceeds by "escape from subvarieties": since $A$ generates $G$, there exists a $g\in A$ such that $V\cap g V$ is of lower dimension than $g$; we can then iterate (in what can become a slightly complicated way) to find $g_1,\dotsc,g_r\in B(r)$ such that $V\cap g_1 V \cap \dotsc \cap g_r V = \emptyset$, with the result that at least one of $e, g_1^{-1},\dotsc, g_r^{-1}$ does not lie in $V$.

It is unclear to me whether one can proceed in quite this way here.

[1] James, J. P., Partition actions of symmetric groups and regular bipartite graphs, Bull. London Math. Soc. 38 (2006), no. 2, 224–232.

  • $\begingroup$ $n$ is even; typo fixed. $\endgroup$ – H A Helfgott Sep 24 '18 at 17:36
  • $\begingroup$ Do you expect the {algebraic-groups} tag to be relevant here? $\endgroup$ – LSpice Sep 24 '18 at 20:51
  • 1
    $\begingroup$ Yes, by analogy, as in the end if the post. (Relevant, you said; the analogy could be helpful or misleading!) $\endgroup$ – H A Helfgott Sep 24 '18 at 21:43

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