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September 19th, 2017, 10:25 PM  #1 
Member Joined: Jan 2015 From: usa Posts: 92 Thanks: 0  Proving isomorphism
Let $H<G$ an automorphism of $G$set of $G/H$ is an equivariant bijection from the set to itself. The automorphism of $G/H$ from a group under composition, and we denote this group by $Aut_G(G/H)$. Show that $Aut_G(G/H)$ is isomorphic to $N_G(H)/H$. This is what i wrote: We note that any $\phi\in \mathrm{Aut}(G/H)$ satisfies $\phi(xH)=x\phi(H)$. Let us suppose that $\phi(H)=gH$, and write $\phi=\phi_g$ accordingly. Now,we should prove the following: 1. $\phi_g\in \mathrm{Aut}(G/H)$ if and only if $g\in N_G(H)$. 2. The map $\Phi:N_G(H)\to \mathrm{Aut}(G/H)$, $g\mapsto\phi_g$ is a surjective homomorphism. 3. $\ker\Phi=H$. Note that any $\phi\in \mathrm{Aut}(G/H)$ satisfies $\phi(xH)=x\phi(H)$. Now, suppose that $\phi(H)=gH$, and write $\phi=\phi_g$ accordingly. Now, you should prove the following: 1. $\phi_g\in \mathrm{Aut}(G/H)$ if and only if $g\in N_G(H)$. 2. The map $\Phi:N_G(H)\to \mathrm{Aut}(G/H)$, $g\mapsto\phi_g$ is a surjective homomorphism. 3. $\ker\Phi=H$. Please help me to prove the three points 1), 2) and 3). I am realy blocked. Thanks. 

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isomorphism, prooving, proving 
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