# Thread: P-Sylow subgroups, Normalizers, Indexes

1. ## P-Sylow subgroups, Normalizers, Indexes

Let G be a group.

Prove that G has no P-Sylow subgroup 'H' for which [G:N(H)]=2.

For every subgroup H in G, the Normalizer is defined as N(H):={ $\displaystyle g\in G: gH=Hg$ }

I need to find a way to prove this. If |G| is infinity, then for every prime p and natural m, we get that |G| is never $\displaystyle m*p^n$ , therefore there is no p-Sylow subgroups anyway (Is this right? I can't think of any other way to 'take care' of the infinite case).

Thanks 2. Originally Posted by adam63 Let G be a group.

Prove that G has no P-Sylow subgroup 'H' for which [G:N(H)]=2.

For every subgroup H in G, the Normalizer is defined as N(H):={ $\displaystyle g\in G: gH=Hg$ }

I need to find a way to prove this. If |G| is infinity, then for every prime p and natural m, we get that |G| is never $\displaystyle m*p^n$ , therefore there is no p-Sylow subgroups anyway (Is this right? I can't think of any other way to 'take care' of the infinite case).

Thanks The finite case is an application of Sylow's theorem. Look up the part of the theorem which tells you how many Sylow p-subgroups there are and work out which bit is of interest here. Then think about why there can't just be 2 Sylow p-subgroups.

For the infinite case you need to remember that, well, there is no infinite case! Essentially, you were right with what you said; you are working with Sylow subgroups, which are only applicable to finite groups!

3. Originally Posted by Swlabr The finite case is an application of Sylow's theorem. Look up the part of the theorem which tells you how many Sylow p-subgroups there are and work out which bit is of interest here. Then think about why there can't just be 2 Sylow p-subgroups.

For the infinite case you need to remember that, well, there is no infinite case! Essentially, you were right with what you said; you are working with Sylow subgroups, which are only applicable to finite groups!
Oh, I see now that if H is a P-Sylow group of G, then $\displaystyle [G:N(H)]=n_p$, therefore if [G:N(H)]=2, we get that 2=1mod(p), which is impossible (considering that p has to be natural, since p is prime).

That's it ?

4. Originally Posted by adam63 Oh, I see now that if H is a P-Sylow group of G, then $\displaystyle [G:N(H)]=n_p$, therefore if [G:N(H)]=2, we get that 2=1mod(p), which is impossible (considering that p has to be natural, since p is prime).

That's it ?
Yup, that's all there is to it!

5. Originally Posted by Swlabr Yup, that's all there is to it!
Thanks Just to be 'formal', is a=b mod(p) , what does it mean (about the existance of an integer c for which...)? I want to know it, and also to complete the proof.

6. Originally Posted by adam63 Thanks Just to be 'formal', is a=b mod(p) , what does it mean (about the existance of an integer c for which...)? I want to know it, and also to complete the proof.
Well, $\displaystyle a \equiv b \text{ mod }p \Rightarrow a=pn+b$ for some $\displaystyle n \in \mathbb{Z}$. Thus, $\displaystyle 2=pn+1 \Rightarrow pn=1 \Rightarrow p|1 \Rightarrow p= \pm 1 \Rightarrow$ is not prime, a contradiction.

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