$\phi(H) = H$ 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?
$\endgroup$ 22 Answers
$\begingroup$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 $H\trianglelefteq G$.
$\endgroup$ 5 $\begingroup$$\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
