# Math Help - Sylow group

1. ## 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
$(N_{G}(P)N)/{N}\cong N_{G/N}(PN/N)
$
where $N_{G} (P)$ is the normalizer of P in G (same with $N_{G/N}(PN/N)$ )
I tried to solve it using the homomorphism
$\varphi:N_{G}(P)N\rightarrow G/N$ defined by $\varphi(gn)=gnN$
I can show that $Im(\varphi)\subseteq N_{G/N}(PN/N)$ and also that $ker(\varphi)=N$.
all I need now is to show that $N_{G/N}(PN/N)\subseteq Im(\varphi)$ in order to use the first isomorphism theorem, but for some reason I can manage it.

any help would be appreciated

2. 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
$(N_{G}(P)N)/{N}\cong N_{G/N}(PN/N)
$
where $N_{G} (P)$ is the normalizer of P in G (same with $N_{G/N}(PN/N)$ )
I tried to solve it using the homomorphism
$\varphi:N_{G}(P)N\rightarrow G/N$ defined by $\varphi(gn)=gnN$

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

I can show that $Im(\varphi)\subseteq N_{G/N}(PN/N)$ and also that $ker(\varphi)=N$.
all I need now is to show that $N_{G/N}(PN/N)\subseteq Im(\varphi)$ in order to use the first isomorphism theorem, but for some reason I can manage it.
you need a little trick to show that $N_{G/N}(PN/N)\subseteq Im(\varphi).$ so let $gN \in N_{G/N}(PN/N).$ then we'll get $g^{-1}Pg \subseteq PN.$ thus $g^{-1}Pg$ is a Sylow p-subgroup of $PN.$ but $P$ is a Sylow p-subgroup of

$PN$ too. thus $P$ and $g^{-1}Pg$ are conjugates in $PN,$ i.e there exists $n \in N: \ g^{-1}Pg=n^{-1}Pn.$ therefore $g_0=gn^{-1} \in N_G(P).$ hence: $\varphi(g_0)=g_0N=gN. \ \Box$

3. we can actually prove a better result, which is: $(N_G(P)N)/N=N_{G/N}(PN/N).$ then you even don't need to define an isomorphism. here's how to prove the equality:

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

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