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$\begingroup$

$\phi(H) = H$ 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?

  • 1
    The subgroups $H$ that satisfy your hypothesis are called [characteristic subgroups](http://en.wikipedia.org/wiki/Characteristic_subgroup), 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.2012-06-03

2 Answers 2

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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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$\phi (H)=H$ for all automorphism
We wanted to show that H is normal subgroup that is for every $x \in G$ $xHx^{-1}=H$ Now as we have $\phi_x(g)=xgx^{-1}$ as inner automorphism .$\phi_x(H)=xHx^{-1}=H$ Hence Done