Prove /Disprove: Existence of Symmetric Subgroup of Order $n^2-1$ of $S_n$. $G$ is a symmetric subgroup of symmetric group $S_n$ acting on $n$ objects where  order of $G$ is $n^2-1$.
Problem:
Prove (or disporve),  Such $G$ exists for $S_n$  for finite $n$.
For $n=5,11,71$,  $G$ exists.
Note:
Note, that if Brocard's problem has infinite solution, it does not imply, that there are infinite  $S_n$ with  symmetric subgroup of order of  $n^2-1$. 
 A: It can't exist in general, in fact if $n=p-1$ where p is prime, then there does not even exist a sub-group of order $n^2-1$ because the order of a subgroup must divide that of the group and $\frac{(p-1)!}{(p-1)^2-1} = \frac{(p-1)!}{p(p-2)}$ is not an integer. 
A: 
There are infinitely many positive integers $n$ such that $S_n$ has an abelian subgroup of order $n^2-1$. 

Suppose $n$ is odd.  Let $u\in\{-1,+1\}$ be such that $n\equiv u\pmod{4}$.  Then, $$n^2-1=2^{k+1}\,\left(\frac{n-u}{2^k}\right)\,\left(\frac{n+u}{2}\right)\,,$$
where $2^k$ is the largest power of $2$ that divides $n-u$ (whence $k\geq 2$).  If
$$2^{k+1}+\left(\frac{n-u}{2^k}\right)+\left(\frac{n+u}{2}\right)\leq n\,,$$
then there exists an abelian subgroup of $S_n$ of order $n^2-1$, which is the product of three cyclic groups generated by cycles of length $2^{k+1}$, $\frac{n-u}{2^k}$, and $\frac{n+u}{2}$.  Note that the inequality above is equivalent to
$$n-u\geq\frac{2^{k+2}}{1-\frac{1}{2^{k-1}}}\,.$$
Hence, if $n-u=2^kq$, where $q\geq 9$ is an odd integer, then $S_n$ has a subgroup of order $n^2-1$.  In other words, if $$n=2^k(2r+1)\pm1$$ where $k\geq 2$ and $r\geq 4$ are integers, then $S_n$ has a subgroup of order $n^2-1$.
P.S. I made a stupid calculation error in my earlier answer (now deleted).
