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$\phi(H) = H$ for $\phi$ any automorphism in $G$.

I tried to find a homomorphism for which $H$ is the kernel, which shows that $H$ is normal. However, I tried to have it maps $g$ to $\phi(gH) = \phi(g)H$, but cannot show it preserves the operation because we don't know that $\phi(a)\phi(b) = \phi(a)H\phi(b)H = \phi(a)\phi(b)H$ as we don't have $H$ is normal.

Is this the right way to go? Or should I try something else?

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The subgroups $H$ that satisfy your hypothesis are called characteristic subgroups, and the condition is more restrictive than being a normal subgroup. Using $\phi$ to define a homomorphism is not a good idea because $\phi$ stands for any automorphism, not any one of them in particular. In fact the idea of defining a homomorphism, though it can be done, is not very helpful here: justifying that whatever you define is a homomorphism will probably amount to proving that $H$ normal. –  Marc van Leeuwen Jun 3 '12 at 9:27

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up vote 9 down vote accepted

let $\phi$ be inner automorphism for example $\phi_g(x)=g^{-1}xg$ for any $g\in G$. Now you can apply this automorphism.

$\phi_g(H)=H$ $\Longrightarrow$ for any $h\in H$ then we have $\phi_g(h)\in H$. Hence $H^g=H$ for any $g\in G$. So $H\trianglelefteq G$.

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