It seems the group G= S_n. The set C(1,2) is called the 'centerzlier' not center. But anyway note that if f is a permutation in S_n which fixes 1 and 2 i.e. f(1) = 1 and f(2) = 2 then f is in C(1,2). Also, (1,2) is in C(1,2). Now argue that this is a complete discription of all elements in the centralizer. Therefore we have (n-2)! permutations which leave 1,2 fixed and a product of those (n-2)! permutations with the cycle (1,2) i.e. (1,2)f is also in C(1,2). Thus, we see that |C(1,2)| = |Sym(2) x Sym(n-2)|. To go furthermore and actually prove isomorphism we should realize that the permutations f which leave 1,2 fixed behave like ordinary permuations on {3,4,...,n}. With that the isomorphism is clear. Map f into (id,f) and (1,2). Map (1,2)f into ((1,2),f). Now show this is an isomorphism.