Thread: Center of ....

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

Originally Posted by dimuk

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?

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.

Thanx for ur reply. But I have some doubts,

What do u mean by
"Map f into (id,f) and (1,2). Map (1,2)f into ((1,2),f). "

Pls explain me.

Originally Posted by dimuk

Thanx for ur reply. But I have some doubts,

What do u mean by
"Map f into (id,f) and (1,2). Map (1,2)f into ((1,2),f). "

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).

Originally Posted by dimuk

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?

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).

Thanx for help.

Need more help.

If \alpha \in Aut (S_n) then \alpha(S_(n-2))=S_(n-2). Can u help me to prove this.

Originally Posted by dimuk

If \alpha \in Aut (S_n) then \alpha(S_(n-2))=S_(n-2). Can u help me to prove this.

This is not true. Given an embedding of S_{n-2} in S_n, there will be automorphisms of S_n that do not preserve S_{n-2}.

Can u give me a example for this.