Question:
Consider $D_{16}$ and $H =\{ ε,p^4,p^8,p^{12}\}$. Show that $H$ is a normal subgroup of $D_{16}$.
My Reasoning
I know I need to show that $gHg^{-1} \subseteq H$ for all $g \in D_{16}$. I also know that $D_{16} = \{u^ip^j: 0 \leq i \leq 1 , 0 \leq j \leq 15\}$. Does that mean that I just need to show this: $(u^ip^j)H(u^ip^j)^{-1} \subseteq H$? Is this how I'm supposed to do this? This is the first time I encounter this sort of problem so I'm not sure how to answer these types of questions.
For example: $(u^ip^j)p^4 (u^ip^j)^{-1} = (u^ip^j)p^4 (p^{-j}u^{-i}) = u^ip^{j+4-j}u^{-i} = u^ip^4u^{-i} = u^iu^{-1}p^{-4} = p^{-4} = p^{12} \in H$
First, show that $H$ is a subgroup.
Test:
i) $1_{D_{16}} = \varepsilon \in H$.
ii) For any $p^{4n}, p^{4m}\in H$, we have that $p^{4n}p^{4m}=p^{4n+4m}=p^{4(n+m)}\in H$.
iii) For any $p^{4n}\in H$, we have that $p^{16-4n}p^{4n}=p^{4n}p^{16-4n}=\varepsilon$, so $p^{16-4n}=(p^{4n})^{-1}$. Furthermore, for any $n\in \{0, 1, 2, 3\}$, we have that $p^{16-4n}\in H$, so $H$ is closed under inverses.
To show that $H$ is normal, your proof is fine! Just replace $p^4$ with $p^{4n}$ for an arbitrary $n\in \{0, 1, 2, 3\}$.