Subgroup of Order $n^2-1$ in Symmetric Group $S_n$ when $n=5, 11, 71$ $G$ is a symmetric subgroup of symmetric group $S_n$ acting on $n$ objects where $n=5, 11, 71$ and order of $G$ is $n^2-1$.
Question: 


*

*Does $G$ (as defined above for $n=5, 11, 71$) exist?

*How can I compute such calculation ? Is there a online system/ website I can use? in this case, some introductory infromation will help.
 A: For $m\geq n$, the group $S_m$ contains a copy of $S_n$. For the given numbers if you calculate $n^2-1$ you get $4!,5!$ and $7!$ respectively. Thus they all have such subgroups.
You can compute these subgroups very easily. For example $$S_4=\{1, (12), (13), (14), (23), (24), (34)
(12)(34), (13)(24), (14)(23)
(123), (124), (132), (134), (142), (143), (234), (243) (1234), (1243), (1324), (1342), (1423), (1432)\}$$ and it can be seen as a subset of $S_5$ since $S_5$ also contains very similar permutations. Similarly, $S_5\subseteq S_{11}$ and $S_7\subseteq S_{71}$.
Edit : Of course the statement is not true in general. In some cases, $n^2-1$ does not divide $n!$ and even if it does there may not be such a subgroup. 
A: This is only a partial attempt at tackling a more general underlying question.
Note that if $n \ge 5$ is odd, then $n^{2} - 1$ divides $n!$.
In fact $n^{2} = (n - 1) (n +1)$. Clearly $n-1$ divides $n!$. As to $n+1$, since $n$ is odd, $n+1$ is even, so
$$
n+1 = 2 \cdot \frac{n+1}{2}.
$$
Clearly also $2$ and $\frac{n+1}{2}$ divide $n!$,
Now if $n \ge 5$, we have that
$$
n - 1,  2, \frac{n+1}{2}
$$
are distinct.

Update 1
Direct computation with GAP show that $S_{7}$ has three conjugacy classes of subgroups of order $48$.

Update 2
Direct computation with GAP show that $S_{9}$ has seven conjugacy classes of subgroups of order $80$.
A: $S_n$ admits a subgroup (not necessarily symmetric) of order $n^2-1$ if and only if neither of $n\pm 1$ is prime. In particular, it admits such a subgroup if $n$ is odd and at least $5$.
