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September 19th, 2017, 11:25 PM   #1
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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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