If G is commutative, then centralizer, normalizer of any subset and center of G are all three equal to G.

The center of S_n is trivial if n > 2. It only consists of the identity. Indeed, if c is an element of the center, it has to commute with the transposition (1,2). Hence

c(2) = (c(1,2)) (1) = (1,2)(c(1)).

Since c(2) and c(1) are distinct, we find that c(1) is in the support of (1,2). The same reasoning with (i,j) instead of (1,2) yields that c(i) is in the support of (i,j). Varying j implies then that c(i) = i. How about the case n = 2?

The center of the general linear group GL_n(R), with n > 1, equals the set of diagonal matrices with nonzero determinant. The proof of this is similar to the above case.

Fix a basis B consisting of b₁, ..., b_n. Let P_i,j be the matrix of the linear map that permutates the basis vectors b_i and b_j. If C is an element in the center, then it commutes with all P_i,j. Suppose n > 2 and let k be different from i,j. Then we have

C(b_k) = CP_i,j(b_k) = P_i,jC(b_k).

Thus C(b_k) is contained in the 1-eigenspace of P_i,j.

Similarly we obtain that C(b_k) is contained in the 1-eigenspace of Q_i,j, the linear map that fixes all b_k except for b_i and b_j. On these two elements Q_i,j acts as follows: Q_i,j(b_j) = -b_i and Q_i,j(b_i) = b_j.

Thus C(b_k) is contained in the space generated by B\{b_i, b_j}.

Varying the i and j, we easily find that C(b_k) is a multiple of b_k.

Again the case n = 2 is left to the reader.