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.
Best Answer
The thing you are looking for is called the exponent of a group. The answer you want (when the group is $S_n$) and some details of why it is true, can be found here.
The answer given in the link is $$ lcm \{ k \mid 1 \le k \le n \}$$ So in your case, your answer of $5 \cdot 7 \cdot 8 \cdot 9$ is correct.