Enumeration and structure of abelian 2-subgroups of a symmetric group I am struggling with a group theoretic problem arising in my research. Given a symmetric group $\Sigma_{n}$, let's consider all its abelian 2-subgroups up to conjugation. Is it possible to give a classification of all these subgroups and count how many of them for arbitrary $n$?
 A: As Derek pointed out in his comment, asymptotically there are $2^{\frac{n^2}{16}+o(n^2)}$ abelian $2$-subgroups of $S_n$. Indeed, let $H:=\langle (1,2),(3,4),\dots,(n-1,n)\rangle$ if $n$ is even, and $H:=\langle (1,2),(3,4),\dots,(n-2,n-1)\rangle$ if $n$ is odd. Then $H$ is elementary abelian of order $2^{\lfloor\frac{n}{2}\rfloor}$. The number of $r$-dimensional subspaces of a vector space of dimension $k$ over a field of order $q$ is the Guassian binomial coefficient $\binom{k}{r}_q$, and this is maximised at $r:=\lfloor\frac{k}{2}\rfloor$. Summing over $r$ we get that, asymptotically, the total number of subspaces is $2^{\frac{k^2}{4}+o(k^2)}$. So $H$ has $2^{\frac{n^2}{16}+o(n^2)}$ subgroups.
For an upper bound, note that an abelian transitive permutation group of degree $n$ is regular. Now, it is no loss to fix a partition $n=n_1+\dots +n_t$ of $n$ and count those abelian $2$-groups with orbits of size $n_1$, $\dots$, $n_t$, since the number of such partitions is at most $2^{O(\sqrt{n})}$. Clearly we may assume that each $n_i$ is a power of $2$: say $n_i=2^{e_i}$. Now, each abelian $2$-subgroup of $S_n$ with these orbit sizes is conjugate to a subgroup of $T(A_1,A_2,\dots,A_t):=A_1\times A_2\times\dots\times A_t\le S_n$, where $A_i$ is a maximal abelian transitive $2$-subgroup of $S_{n_i}$. Since the abelian maximal abelian transitive $2$-subgroup of $S_{n_i}$ act regularly, there is a one to one correspondence between such groups and the isomorphism classes of finite groups of order $n_i=2^{e_i}$. The number $\prod_{i=1}^{t}f(n_i)$, where $f(n_i)$ denotes the number of isomorphism classes of finite groups of order $n_i=2^{e_i}$, is at most $2^{O(n)}$, so we may assume that each group $A_i$ is fixed.   
So we now just need to find an upper bound for the number of subgroups of $T:=T(A_1,A_2,\dots,A_t)$. Let $e:=\sum_{i=1}^t e_i$, so that $T$ has order $2^e$. By a result of Shalev, the number of subgroups of $T$ is at most $4\times 2^{\frac{e^2}{4}}$ (See Lemma 4.2 of {A. Shalev, Growth functions, p-adic analytic groups, and groups of finite coclass, J. London Math. Soc. (2) 46 (1992), 111–-122). Since $e\le \frac{n}{2}$ in this case, we deduce that the number of subgroups of $T$ is at most $4\times 2^{\frac{n^2}{16}}$.  
Combining both the lower and upper bounds above, we see that the number of abelian $2$-subgroups of $S_n$ is at most $2^{\frac{n^2}{16}+o(n^2)}$.
One can go through and get more precise estimates for the `little o' part above, if it is needed. 
