Subgroup of Order $n^2-1$ in Symmetric Group $S_n$ when $n=5, 11, 71$

Question

$G$ 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:

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

2. How can I compute such calculation ? Is there a online system/ website I can use? in this case, some introductory infromation will help.

Answer 1

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.

Answer 2

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.

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

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Update 2 Direct computation with GAP show that $S_{9}$ has seven conjugacy classes of subgroups of order $80$.

Answer 3

$S_n$ 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 $5$.