The question is stated exactly like I put it in the title. Note that $S_9$ is the symmetric group on $9$ elements and that $e$ here denotes the identity element of the group.
My initial thought was that the answer needs to be $9!$ because we need to account for all possible cycle lengths that appear in the elements of the group.
However, I think that this can be bounded further. The final answer I came to is $5 \cdot 7 \cdot 8 \cdot 9$. My logic is that this number $m$ needs to be a multiple of all the prime factors that appear in numbers that could possibly be the order of any element $g \in S_9$.
So for example, we can imagine having an element $g \in S_9$ that is of order 20, because it contains one cycle of length 4 and one cycle of length 5 (we know that the order of some permutation $\sigma$ is going to be the least common multiple of the cycle lengths of cycles that appear in its cycle decomposition). In order for our $m$ to work here, in its prime factorization it needs to contain 2 twice and 5 once.
The first question I have is obviously whether this result I got is correct or not. If yes, I would like to further ask whether this is a good way to think about this and whether there exists a nicer, and more concrete way to be sure about one's reasoning when tackling problems like this.
As always, any and all help is much appreciated.
