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Here's a problem that's been going around my friend group. I have solved this problem, and am interested in a generalization of it.

Here's the original problem: Pick an element of $\sigma\in S_n$ uniformity at random, and let $X(n)$ be the random variable that is the number of fixed points of $\sigma$. Compute

$$\lim_{n\to\infty}P(X(n)=0)$$

Here's the extension: Suppose instead of $S_n$ we are interested in a class of subgroups, $\{H_n:n\in I\subseteq\mathbb{N}\}$. We can then ask: what is the probability the limit yields $$\lim_{n\to\infty}\frac{\mu(H_n)}{|H_n|}$$

where $\mu(G)$ is the biggest number such that every element of $G$ fixes $\mu(G)$ or more elements of $[n]$ when viewed as a subset of $S_n$. The notion I'm trying to formalize here is "the least number of fixed points that is actually achieved" as for many groups, every element has a fixed point.

This seems to be the right generalization of the problem, though other generalizations are welcome. I'm mostly asking this out of curiosity to see what can be proven. Results about transitive groups, doubly-transitive subgroups, $D_{2n}$, or any other reasonably interesting class of groups would be exciting for me.

This link seems relevant, but it's not clear to me that this problem can be solved simply by summing Recontre numbers. This MSE post also seems relevant, but not quite enough to solve the problem on its own.

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Stella Biderman
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  • If we pick $\sigma \in A_n$ uniformly at random, and let $X(n)$ be the number of fixed points, then $\lim_{n\to \infty} \Pr[X(n)=0] = \frac1e$ as well, but I'm not sure if that's what you're asking. – Misha Lavrov Mar 31 '17 at 15:00
  • @Misha That's was I was intending at first, but for general groups that number would be 0, so I tried to formalize the notion of "what's the odds that it equals the smallest value it can take on, whatever that smallest value is." – Stella Biderman Mar 31 '17 at 15:05
  • @Misha Oh, I see why that's the answer for $A_n$. Do you know how to solve the problem for other subgroups of $S_n$? – Stella Biderman Mar 31 '17 at 15:11
  • For $A_n$, as for $S_n$, the method of moments can prove that $X(n)$ approaches a $\operatorname{Poisson}(1)$ distribution as $n \to \infty$, but I don't see a good approach for cases where that's not true. – Misha Lavrov Mar 31 '17 at 15:16
  • @Misha Interesting. I had solved the problem combinatorially for $S_n$, and there's a combinatorial reason $A_n$ and $S_n$ have the same answer. – Stella Biderman Mar 31 '17 at 15:18
  • @JackD'Aurizio I am not sure what you mean. $1/e$ is a limit as $n \to \infty$. But there are transitive groups in which the proportion of fixed points is arbitrarily small. For example, there are groups of prime degree $p$ and order $p(p-1)$ in which only $p-1$ of the elements have no fixed points. – Derek Holt Mar 31 '17 at 17:00
  • @DerekHolt: I guess my "reasonable conjecture" was blatantly wrong :D – Jack D'Aurizio Mar 31 '17 at 17:27
  • @JackD'Aurizio for another counter example, $C_n$ has no fixed points, for any $\sigma\in \langle(1;2;\ldots; n)\rangle\leq S_n$. Thus for $\langle(1;2\ldots; n)\rangle$ the answer is $1$. – Stella Biderman Mar 31 '17 at 17:36
  • @StellaBiderman I would be interested to know what you mean when you say "for general groups that number would be $0$". What do you mean by a general group here. – Derek Holt Mar 31 '17 at 17:40
  • @DerekHolt I articulated the idea poorly. I only meant that in the general case, for some group $G$, you can't assume that there are any permutations with no fixed points. – Stella Biderman Mar 31 '17 at 17:44
  • I suspect that in a random subgroup of $S_n$ (i.e. chosen with uniform distribution over all subgroups), the probability that no element is fixed-point-free tends to $1$ for large $n$. It might even be provable, because I think there might be results saying that random subgroups have lots of short orbits. It is known that the proprtion of transitive subgroups approaches $0$. – Derek Holt Mar 31 '17 at 17:47
  • @DerekHolt That's my suspicion too. – Stella Biderman Mar 31 '17 at 17:52

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It is pretty well known that there are $$ D_n = n! \sum_{k=0}^{n}\frac{(-1)^k}{k!} $$ derangements (permutations without fixed points) in $S_n$, so the limit probability for the first problem is $\color{red}{\large\frac{1}{e}}$. It also is the limit probability for the second case (random even permutation), since the difference between the numbers of even and odd derangements is given by the determinant of a $n\times n$ matrix $M_n$ with zero elements on the diagonal and elements equal to $1$ outside the diagonal. $M_n$ is the difference between a rank-$1$ matrix (having eigenvalues $n,0,0,\ldots$) and the identity matrix, hence the difference between even and odd derangements is $n-1$ in absolute value and the limit probability is the same in both cases.

Jack D'Aurizio
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