I am currently working through an exercise in Dummit & Foote, in which I try to prove that every automorphism of $S_n$ with $n \geq 3, n \neq 6$ is an inner automorphism ($S_n$ being the permutation group of the set $\{1,\ldots, n\}$). I showed that the number of permutations equals $n(n-1)/2$ and the number of elements of order $2$ which are the product of $m$ disjoint transpositions equals $$\frac{n!}{(n - 2m)!m!2^m}.$$ I tried to find solutions for $m$, but did not succeed and then stumbled on this question. I understand all of the accepted answer, but the last step: is it trivial to see that there could never be any solutions for $m > 3$?
any help would be appreciated.
In the linked question, the claim is that $\frac{m}{(2m-3)!!}$ is never an integer for $m>3$. This is true because $2m-3>m$, so $$0<\frac{m}{(2m-3)!!}\leq \frac{m}{2m-3}<1,$$ and of course there can be no integer between $0$ and $1$...