1. ## Normalizers

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} )-I've managed to prove this one...
Can't understand how to prove this one:
B. Prove that the normalizer of H is a 2-sylow group of S4.

TNX to all the helpers!

2. Originally Posted by WannaBe
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} )-I've managed to prove this one...
Can't understand how to prove this one:
B. Prove that the normalizer of H is a 2-sylow group of S4.

TNX to all the helpers!

Check that $(12)(34)\cdot(1234)\cdot(12)(34)=(1432)=(1234)^{-1}\Longrightarrow$ $(12)(34)\in N_{S_n}(H)\Longrightarrow 4\mid ord(C(H))\mid 24$ so...

Tonio

3. I didn't completely understand...
The equality you've wrote is obvious... By the definition, it means that (12)(34) is in N[Sn](H) ofcourse...
but why it means that 4|o(C(H))? And why from here we get that it's a sylow-p group? It can has an order 12 also...

TNX

4. Originally Posted by WannaBe
I didn't completely understand...
The equality you've wrote is obvious... By the definition, it means that (12)(34) is in N[Sn](H) ofcourse...
but why it means that 4|o(C(H))? And why from here we get that it's a sylow-p group? It can has an order 12 also...

TNX

That was a typo: it should have been $4\mid o(N_{S_4})$ , and it follows at once from Lagrange's theorem and the fact that for any subgroup H of any group G we always have $H\le N_G(H)$.
Finally, no: $N_{S_4}$ can't be all $A_4$ (which is the only subgroup of order 12 in $S_4$)

Tonio

5. There is an easy proof for the fact A4 is the only subgroup of order 12 in S4?

TNX a lot!

6. Originally Posted by WannaBe
There is an easy proof for the fact A4 is the only subgroup of order 12 in S4?

TNX a lot!

Well, this depends on what you consider "easy", but the following isn't, imo, hard:

Lemma: If $H\le S_n$ , then either $H\le A_n$ or else exactly half the elements of $H$ are even permutations (and thus, of course, $[H:H\capA_n]=2$)

Proof: (Hint) If $H\nleq A_n$ , define a map from the set of odd permutations of $H$, $H_o$ to $H_e=$the set of even permutations of H, by $h'\mapsto h'x\,,\,\,\forall\,x\in H_o$ , and check it is $1-1$ ...

From this lemma it follows at once what we want.

Tonio

7. Tnx a lot... If I'll have any questions I'll ask in this post...