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.

Thus:

So .

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 .