# Escaping from a centralizer

Let $$G = Sym(n)$$, $$n$$ even. Let $$H be the stabilizer of the partition $$\{\{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.

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