Let t=( 1 2) be a transposition. Then C_G(t)={g \in G| tg=gt } is isomorphic to Sym(2) × Sym(n-2).
Can you help me to prove this?
thanx
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.
Remember f is a permutation which leaves 1 and 2 fixed. Thus, if f is a permutation on {1,2,...,n} it means we can think of it as a permutation of {3,4,...,n} because 1 and 2 are unchanged. Let A={3,4,...,n}. Then f|A (this reads "f restricted to A") is a element of Sym(n-2). Conversely, any element of Sym(n-2) can be obtained by f|A in this manner.
We need to define a isomorphism :C(1,2) Sym(2)xSym(n-2). If is a permutation in C(1,2) then = f i.e. some permutation which fixes 1 and 2, or = (1,2)f. If =f then define ( ) = (id,f|A). If =(1,2)f then define ( ) = ((1,2),f|A).
The centraliser of (12) certainly contains (12), so it contains the subgroup generated by (12), which is isomorphic to Sym(2). It also contains any permutation that leaves 1 and 2 fixed. The group of all such permutations is isomorphic to Sym(n-2). Therefore C_G(t) contains a subgroup isomorphic to Sym(2) × Sym(n-2).
To show that this subgroup is the whole of C_G(t), you need to show that any permutation that takes 1 or 2 to a number other than 1 or 2 is not in C_G(t). Suppose for example that σ is a permutation that takes 1 to 3. Then the product σ(12) takes 2 to 3. If σ(12) = (12)σ then (12)σ must take 2 to 3. This implies that σ takes 2 to 3. But that contradicts the fact that σ takes 1 to 3. The contradiction shows that σ is not in the centraliser of (12).