# Probability of words summing to $1$ in $S_n$ or $\mathrm{PGL}_2(n)$

$$\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Conj{Conj}$$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 $$G$$.

Question:

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

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

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

Known:

If there's a variable occurring 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 $$\PGL_2(n)$$.

• What is $\varepsilon$? It doesn't seem to appear anywhere else except in the clause "for each $\varepsilon >0$". Jun 8, 2019 at 13:02
• I have edited the question: There should be no $ε$. Jun 8, 2019 at 13:03
• 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 $N^{-N/k} \exp(N/k+o(N))$. Jun 8, 2019 at 13:42
• 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 $e^{c\sqrt{n}}$. Jun 9, 2019 at 13:15
• 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 $X=x_1^{a_1}\cdots x_m^{a_m}$. Jun 9, 2019 at 22:22

$$\DeclareMathOperator\PGL{PGL}$$I believe the best result in this direction is due to M. Larsen & A. Shalev (2012); see this paper. I'll summarize their results here. This doesn't answer the questions whether the limit exists or what its precise value is, though, so this is not a full answer.

Denote the length of the word $$X$$ by $$\ell$$ (while $$k$$ is the size of the alphabet).

As for $$S_n$$, Proposition 2.3 gives $$P < n!^{-\eta}$$ for large enough $$n$$, if $$\eta < \frac{2k-1}{4\left((2k)^{2\ell+1}-1\right)}$$, so $$\log_{|S_n|}P < -\eta.$$

As for $$\PGL_n(\mathbb{F}_q)$$, Proposition 3.3 gives $$\log_{|G|}P \le -\eta$$ for large enough $$n$$ (and every $$q$$), if $$\eta < \frac{1}{1800\ell^2}$$. (This holds for all classical groups of Lie type).

Shalev and Larsen handle all finite simple groups in their paper. They also refer to this paper, which gives a bound for simple groups $$G$$ of Lie type of bounded rank $$r$$: $$P \le \frac{C}{q}$$ where $$C = C(r, X)$$ is a constant, and $$q \ge G^{\epsilon}$$ for some $$\epsilon = \epsilon(r)$$, hence $$\log_{|G|}P \le -\eta$$ for large enough $$G$$ if $$\eta < \epsilon$$. EDITED: this last paragraph answers the $$n=2, q\to\infty$$ case.

• OP asked the asymptotics for $\mathrm{PGL}_2(q)$ for large $q$, nor $\mathrm{PGL}_n(q)$ for fixed $q$ and large $n$ (but agree that the analogy with $S_n$ makes it more natural to consider $\mathrm{PGL}_n(q)$ for $n\to\infty$).
– YCor
Mar 5, 2022 at 11:07
• The last paragraph handles this case (bounded rank). I admit I wasn't sure which group OP meant, so I tried to answer both possible meanings. Mar 5, 2022 at 12:08