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# 4 questions from Sylow’s theorem: Qn 2

I have come up with the Question no. 2 of four problems related to Sylow's theorem with high difficulty level. Let's take a look at the problem and understand the concept.

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.

Make sure to visit the Question no. 1 to have the your concepts clear.

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