The proof I've got starts like this ...
The symmetric group order can be considered as the set of all permutations of letters.
Let us choose a permutation , so that .
Since , we can find another permutation which interchanges j and k and leaves everything else where it is. So the inverse of does the same thing, and both and leave i alone.
If and were to commute, (right multiply by ). But they don't.
Whatever is, you can always find a such that .
So no non-identity elements of commute with all elements of .