Can u help me to prove that the Z(Sym(n))={1}, centralizer of the symmetric group is equal to 1.
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 .
In this context, is the set of all permutations of objects (which we can call 1, 2, 3, ..., without loss of generality).
A permutation of objects is a rearrangement of them. For example, a rearrangement of (1, 2, 3) is (1, 3, 2), and another one is (2, 1, 3), and so on.
is an element of , and so it is a rearrangement of objects.
So applied to one of the objects of an n-element set will either move it to somewhere else or leave it where it is.
For example, if is the permutation that changes (1, 2, 3) to (1, 3, 2), then and .
In this context, is any permutation you care to think of, but not the identity permutation (which leaves everything where it is). So there's bound to be at least one element of this n-element set which is not left where it is.
Call this element .
What I mean by is the result of applying this arbitrarily selected permutation to an equally arbitrarily selected element , and calling it .
Don't confuse (the set of permutations on obects) with the actual objects that is acting on. They are not the same thing.