What are the properties of cyclic and normalizer groups in S4?

[TheForumLord]

Homework Statement

In the permutations group S4, let H be the cyclic group the is generated by the cycle
(1 2 3 4).
A. Prove that the centralizer C(H) of H is excatly H ( C(H)={g in G|gh=hg for every h in H} )
B. Prove that the normalizer of H is a 2-sylow group of S4.

Homework Equations

The Attempt at a Solution

Well, there's a quick calculation of H:
(1234)(1234)=(13)(24)
(13)(24)(1234)=(1 4 3 2)
(1432)(1234)=(1)(2)(3)(4)
Hence: H={(1234),(13)(24),(1432),1}

The order of the group is 4. Hence [S4]=6...
Since it's a cyclic group, it's abelian. Which means H is a sub group of C(H)...
How can I prove that C(H) is in H? Or how excatly can I prove the 1st statement?

About 2, I actually have no clue... My abilities in sylow&homomorphzm are really lame and I need to send a lot of excercices 'till next week so I really need your help :(

TNX to all the helpers!

 

[rasmhop]

I haven't really thought a lot about it, but a fairly straightforward way I see to solve it (without the Sylow theorems; don't know if that's a problem) would be to use the theorem that for any subset H of a group G, the number of conjugates of H equals the index of the normalizer of H in G. So count the number of conjugates of H (there should be 3) and you will get the order of the normalizer of H in G, which will prove (B). As for (A) use the special case that says that for some element x of a group G, the number of conjugates of x equals the index of the centralizer of x in G. So count the number of conjugates of (1 2 3 4) (there should be 6 in total; all 4-cycles) and then you get the order of the centralizer of (1 2 3 4) which will convince you that you have found all elements of the centralizer of H.

I assume you algebra book has these theorems (in Dummit-Foote they are proposition 6 in section 4.3).

 

[TheForumLord]

Hey there, I know these theorems of course, but the way you've suggested is very "not elegant" ...There might be an easier way?

TNX!