# Sylow group

• Dec 6th 2008, 05:10 AM
Prometheus
Sylow group
I was asked to prove that given a finite group G, a normal group N in G, and a p-sylow supgroup P in G, then
$\displaystyle (N_{G}(P)N)/{N}\cong N_{G/N}(PN/N)$ where $\displaystyle N_{G} (P)$ is the normalizer of P in G (same with $\displaystyle N_{G/N}(PN/N)$ )
I tried to solve it using the homomorphism
$\displaystyle \varphi:N_{G}(P)N\rightarrow G/N$ defined by $\displaystyle \varphi(gn)=gnN$
I can show that $\displaystyle Im(\varphi)\subseteq N_{G/N}(PN/N)$ and also that $\displaystyle ker(\varphi)=N$.
all I need now is to show that $\displaystyle N_{G/N}(PN/N)\subseteq Im(\varphi)$ in order to use the first isomorphism theorem, but for some reason I can manage it.

any help would be appreciated
• Dec 6th 2008, 06:28 AM
NonCommAlg
Quote:

Originally Posted by Prometheus
I was asked to prove that given a finite group G, a normal group N in G, and a p-sylow supgroup P in G, then
$\displaystyle (N_{G}(P)N)/{N}\cong N_{G/N}(PN/N)$ where $\displaystyle N_{G} (P)$ is the normalizer of P in G (same with $\displaystyle N_{G/N}(PN/N)$ )
I tried to solve it using the homomorphism
$\displaystyle \varphi:N_{G}(P)N\rightarrow G/N$ defined by $\displaystyle \varphi(gn)=gnN$

note that $\displaystyle \color{red}gnN=gN$ because $\displaystyle \color{red}n \in N.$ so we define $\displaystyle \color{red}\varphi(gn)=gN.$ you also need to show that $\displaystyle \color{red}\varphi$ is well-defined and a homomorphism.

I can show that $\displaystyle Im(\varphi)\subseteq N_{G/N}(PN/N)$ and also that $\displaystyle ker(\varphi)=N$.
all I need now is to show that $\displaystyle N_{G/N}(PN/N)\subseteq Im(\varphi)$ in order to use the first isomorphism theorem, but for some reason I can manage it.

you need a little trick to show that $\displaystyle N_{G/N}(PN/N)\subseteq Im(\varphi).$ so let $\displaystyle gN \in N_{G/N}(PN/N).$ then we'll get $\displaystyle g^{-1}Pg \subseteq PN.$ thus $\displaystyle g^{-1}Pg$ is a Sylow p-subgroup of $\displaystyle PN.$ but $\displaystyle P$ is a Sylow p-subgroup of

$\displaystyle PN$ too. thus $\displaystyle P$ and $\displaystyle g^{-1}Pg$ are conjugates in $\displaystyle PN,$ i.e there exists $\displaystyle n \in N: \ g^{-1}Pg=n^{-1}Pn.$ therefore $\displaystyle g_0=gn^{-1} \in N_G(P).$ hence: $\displaystyle \varphi(g_0)=g_0N=gN. \ \Box$
• Dec 6th 2008, 12:08 PM
NonCommAlg
we can actually prove a better result, which is: $\displaystyle (N_G(P)N)/N=N_{G/N}(PN/N).$ then you even don't need to define an isomorphism. here's how to prove the equality:

first if $\displaystyle gN \in (N_G(P)N)/N,$ with $\displaystyle g \in N_G(P),$ then $\displaystyle g^{-1}Pg=P,$ which will quickly give us: $\displaystyle gN \in N_{G/N}(PN/N).$

conversely, suppose $\displaystyle gN \in N_{G/N}(PN/N).$ then as i showed in my previous post $\displaystyle gn^{-1} \in N_G(P),$ for some $\displaystyle n \in N.$ thus: $\displaystyle g \in N_G(P)N$ and hence: $\displaystyle gN \in (N_G(P)N)/N.$