Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$H$ normal subgroup of a group $G$ with cardinality finite. $p$ prime number dividing $|H|$. $P$ a $p$-Sylow subgroup of $H$, how can I prove that then $G=HN_G(P)$ where $N_G(P)$ is the normalizer of $P$ in $G$?

share|improve this question
    
What have you tried? –  Alex Becker Jan 2 '12 at 6:04
5  

1 Answer 1

up vote 7 down vote accepted

If $g\in G$, then $gHg^{-1}=H$, and so $gPg^{-1}\subseteq H$. Since $gPg^{-1}$ is a $p$-Sylow subgroup of $H$, by Sylow's Theorems we know that $gPg^{-1}$ is conjugate to $P$ in $H$. That is, there exists $h\in H$ such that $hPh^{-1} = gPg^{-1}$. Therefore, $g^{-1}hPh^{-1}g = P$, so $h^{-1}g\in N_G(P)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.