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September 30, 2019

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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