# P-Sylow subgroups, Normalizers, Indexes

• Aug 26th 2010, 06:31 AM
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):={ $g\in G: gH=Hg$ }

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 $m*p^n$ , 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 :)
• Aug 26th 2010, 07:12 AM
Swlabr
Quote:

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):={ $g\in G: gH=Hg$ }

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 $m*p^n$ , 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!
• Aug 26th 2010, 07:38 AM
Quote:

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 $[G:N(H)]=n_p$, 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 :) ?
• Aug 26th 2010, 07:41 AM
Swlabr
Quote:

Oh, I see now that if H is a P-Sylow group of G, then $[G:N(H)]=n_p$, 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!
• Aug 26th 2010, 07:46 AM
Quote:

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.
• Aug 26th 2010, 08:04 AM
Swlabr
Quote:

Well, $a \equiv b \text{ mod }p \Rightarrow a=pn+b$ for some $n \in \mathbb{Z}$. Thus, $2=pn+1 \Rightarrow pn=1 \Rightarrow p|1 \Rightarrow p= \pm 1 \Rightarrow$ is not prime, a contradiction.