TITLE: fixed points of permutation groups
QUESTION [10 upvotes]: As is well-known (see, for example, a nice exposition by our own Qiaochu: https://qchu.wordpress.com/2012/11/07/fixed-points-of-random-permutations/) that the distribution of the number of fixed points of random permutations (so, uniformly chosen elements of $S_n$) is Poisson, for largish $n.$ The question is: what is known for proper subgroups of $S_n?$

REPLY [2 votes]: Given $G\leq S_n$ let $X_G$ be the random variable on $\mathbf{N} = \{0,1,\dots,\}$ defined by the number of fixed points of a random $g \in G$.

Claim: For every random variable $X$ on $\overline{\mathbf{N}} = \mathbf{N} \cup \{\infty\}$ there is a sequence of groups $G_n$ such that $X_{G_n} \to X$ in distribution.

In other words, the set $\{\mu_{X_G}: G\leq S_n\}$ is dense in the space of all probability measures on $\overline{\mathbf{N}}$. As established in the comments, there are multiple ways of interpretting the question. This answers one particularly concrete interpretation.
Lemma 1: If both $X$ and $Y$ can be obtained, then so can their (independent) sum $X+Y$ and product $XY$.
(In the event of $0\cdot\infty$, our convention is $0\cdot\infty = 0$.)
Proof: Suppose $X_G \approx X$ and $X_H \approx Y$, where $G \leq \text{Sym}(\Omega_1)$ and $H \leq \text{Sym}(\Omega_2)$. Then $G\times H$ acts on $\Omega_1 \sqcup \Omega_2$ via
$$(g,h) \omega = \begin{cases} g \omega : \omega \in \Omega_1, \\ h \omega : \omega\in \Omega_2.\end{cases}$$
For this action we have
$$\text{fix}_{\Omega_1 \sqcup \Omega_2}((g,h)) = \text{fix}_{\Omega_1}(g) \cup \text{fix}_{\Omega_2}(h),$$
so the random variable for this action is $X_G + X_H \approx X + Y$.
Alternatively, $G \times H$ acts on $\Omega_1 \times \Omega_2$ via
$$(g,h)(\omega_1,\omega_2) = (g\omega_1, h \omega_2).$$
For this action we have
$$\text{fix}_{\Omega_1 \times \Omega_2}((g,h)) = \text{fix}_{\Omega_1}(g) \times \text{fix}_{\Omega_2}(h),$$
so the random variable for this action is $X_G X_H \approx X Y$.
Lemma 2: For each $t\in[0,1]$ we can obtain a random variable $X_t$ such that $X_t = 0$ with probability $t$ and $X_t = 1$ with probability $1-t$.
Proof: For prime $p$ consider the "$ax+b$ group" $\mathbf{F}_p^\times \ltimes \mathbf{F}_p$ acting on $\mathbf{F}_p$ via $(a,b) x = ax+b$. Evidently
$$|\text{fix}_{\mathbf{F}_p}(a,b))| = \begin{cases} 1 & \text{if}~a\neq 1,\\ 0&\text{if}~a=1, b \neq 0,\\ p & \text{if}~a=1,b=0.\end{cases}$$
So we can obtain a random variable $Y$ such that $Y = 1$ with probability $1-1/(p-1)$, $Y=0$ with probability $1/(p-1) - 1/p(p-1)$, and $Y=p$ with probability $1/p(p-1)$. By Lemma 1 then we can obtain $Y^* = \prod_{i=1}^k Y_i$, where each $Y_i$ is an independent copy of $Y_i$. Note that $Y^*=1$ with probability $(1-1/(p-1))^k$ and $Y^*\geq 2$ with probability $\leq k/p(p-1)$. Put $k = \lfloor \log(1/(1-t)) p \rfloor$ and send $p\to\infty$.
Proof of main claim: It suffices to show that we can obtain any random variable supported on $\{0,\dots,N-1\}\cup\{\infty\}$ for each $N$. We will prove this by induction on $N$. For the case $N=0$ (i.e., $X=\infty$ with probability $1$), consider $X_{S_n}$ as $n\to\infty$. Now let $N\geq 1$ and let $X$ be an arbitrary random variable supported on $\{0,\dots,N-1\}\cup\{\infty\}$. Let $p_i = \mathbf{P}(X = i)$. Let $Y$ be the random variable supported on $\{0,\dots,N-2\}\cup\{\infty\}$ defined by
$$\mathbf{P}(Y = i) = p_{i+1} / (1 - p_0).$$
By induction we can obtain $Y$. Thus by the lemmas we can obtain
$$(X_0 + Y) X_{p_0} \sim X.$$