Let $G$ be the symmetric group $S_n$ or the projective general linear group $PGL_2(n)$.

Let $X$ be a cyclically reduced word in the abstract variables $x_1, x_2, \ldots,x_k$, i.e. $X$ is a product containing $x_1, x_2, \ldots,x_k$ and their inverses, without any element appearing next to its own inverse in any cyclic permutation. (Only words with length $4$, $6$, $8$ are needed in my research.)

Consider the probability $P$ that the word sums to $1$, with each $x_i$ chosen uniformly and independently from $G$.


What are the upper bounds of $\log_{|G|}P$?

If $\log_{|G|}P$ converges when $n→\infty$, what's the value?

Answers are acceptable for either $G=S_n$ or $G=PGL_2(n)$.


If there's a variable occuring only once in $X$, then $P$ is exactly $1/|G|$.

If $X=x_1^k$, then the limit is $-1/k$ for symmetric groups by David E Speyer's argument.

As Richard Stanley pointed out, if $X=x_1x_2x_1^{-1}x_2^{-1}$, then $P=|Conj(G)|/|G|$. ($|Conj(G)|$ is the number of conjugacy classes of $G$)

The formula $P=|Conj(G)|/|G|$ holds for the words $x_1x_1x_2x_2$ and $x_1x_2x_1x_2^{-1}$ if all the characters of $G$ are real, and that's exactly the case for $S_n$ and $PGL_2(n)$.

  • $\begingroup$ What is $\varepsilon$? It doesn't seem to appear anywhere else except in the clause "for each $\varepsilon >0$". $\endgroup$ – Noam D. Elkies Jun 8 '19 at 13:02
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    $\begingroup$ I have edited the question: There should be no $ε$. $\endgroup$ – LeechLattice Jun 8 '19 at 13:03
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    $\begingroup$ If $X = A^k$ then the limit is $-1/k$. The generating function for the probability that an element of $S_n$ obeys $g^k=1$ is $f_k(x) := \exp \left( \sum_{d|k} t^d/d \right)$. Taking the contour integral $\oint f_k(x) x^{-N-1} dx$ around a circle of radius $N^{1/k}$ gives asymptotics of the form $N^{-N/k} \exp(N/k+o(N))$. $\endgroup$ – David E Speyer Jun 8 '19 at 13:42
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    $\begingroup$ A paper that is somewhat related is A. Nica, On the number of cycles of given length of a free word in several random permutations, Random Structures & Algorithms 5 (1994), 703-730. An unrelated comment: for $X=xyx^{-1}y^{-1}$, the limit is 1, since the number of commuting pairs in any finite group $G$ is $|G|\cdot k(G)$, where $k(G)$ is the number of conjugacy classes, and the number of conjugacy classes in $S_n$ is around $e^{c\sqrt{n}}$. $\endgroup$ – Richard Stanley Jun 9 '19 at 13:15
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    $\begingroup$ As for $X=x_1x_1x_2x_2$, this is part of Exercise 7.69(h) of Enumerative Combinatorics, vol. 2. Possibly part (i) can solve the problem for $xy^kxy^{-k}$ and $xy^kx^{-1}y^k$, but I have not tried to do this. See also (f) for a possible approach to $X=x_1^{a_1}\cdots x_m^{a_m}$. $\endgroup$ – Richard Stanley Jun 9 '19 at 22:22

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