Subgroups $N$ and $H$ of $D_{4}$ such that $N\trianglelefteq H$, $H\trianglelefteq D_{4}$ but $N$ is not normal in $D_{4}$

Question

Problem: Find subgroups $N$ and $H$ of $D_{4}(D_{8})$ such that $N \trianglelefteq H$ and $H \trianglelefteq D_{4}$ but $N$ is not normal in $D_{4}$

Question: I read that $\{1\}$ is not normal in $D_{4}$, and if so I think this is an answer for $N$, along with $\{1,r^{2}\}=H$ . But isn't the identity always normal?

Is it true?

Answer 1

Use $N=\{1,r\}$ and $H=\{1,r,r^2,r^3\}$ . $|H:N|=2$ and $|D_4:H|=2$ ,so both of them normal subgroups . But $H$ is not normal in $G$ . As $sHs^{-1}=\{1,r^3\} \neq H$ .