Let P be a Sylow p- group of a finite group G and let H be a subgroup of G containing $$N_{G}(P)$$ . Prove that $$H = N_{G}(H)$$.

Solution

Let $$P \in Syl_{P}(G) \ and H \leq G \ such \ that \ N_{G}(P) \subset H$$

Claim : Frattinis Argument :

If G is a finite group with normal subgroup h and if $$P \in Syl_{P}(H)$$ then G is not great notation can be confirmed in $$\backslash[G ,E]$$ .

$$G’ = H N_{G}(P)$$

where $$N_{G}$$ is the normalizer of P in G’ .

& $$H N_{G}(P)$$ means the product of group subsets .

PROOF :

By Sylow’s 2nd theorem we know that any two Sylow subgroups are conjugate to each other and here we are considering P to be a Sylow P subgroup of H .

So they are conjugate in H . ………………………………….(1)

Now , for any $$g \in G \ or \ gHg^{-1} = H \ \ [as \ H \leq G \ given] \Rightarrow gPg^{-1} \subset H \ but \ gPg^{-1 } is \ also \ a \ group . \\ \\ and |gPg^{-1}| = |P|$$

So $$gPg^{-1}$$ is also a sylow – p -subgroup in H .

so $$\exists h \in H \ such \ that gPg^{-1} = hPh^{-1} \ \ [by \ (1)] \Rightarrow h^{-1}gPg^{-1}h = P \Rightarrow h^{-1}g \in N_{G} (P) \Rightarrow g \in HN_{G}(P) \\ \Rightarrow G \subset HN_{G}(P) = H$$

Now clearly $$HN_{G}(P) \subset G$$ .

So , we have $$N_{\epsilon} = HN_{G}(P) = H$$

hence the proof of the claim .

Now come back to our question ,

If we consider $$G’ = N_{G}(H)$$ So , $$H \leq G’$$ .

and $$P \leq H$$ where P is a Sylow p-group of G then it is na Sylow p- group of G’ as well .

So by Frattinis Argument we have that .