Let $D_{2n}$ be the dihedral group of order $2n$. Then, the subgroup generated by a reflection is not normal.
I think I've come up with a proof for a converse that if an element is not $\left\langle h\right\rangle$ for $H$ the subgroup consisting of $n$ rotations, then its square equals the identity thus it is a reflection.
However, I am struggling to prove why a reflection would not be normal in this case. Is this because a reflection $y\in Y$ for $g \in G$ would not give $ghg^{-1}\in Y$ because we have the property $y^2=e$?
Any help will be appreciated!
Note $\mathcal R_n\cap\langle h\rangle=\{e\}$ where $h$ is a reflection symmetry in the dihedral group $\mathcal D_n$ and where $\mathcal R_n$ denotes the $\mathcal D_n$ subgroup of all rotation symmetries. Since $\mathcal R_n\lhd\mathcal D_n$ then we have $$h\in\mathcal Z(\mathcal D_n)\iff\langle h\rangle\lhd\mathcal D_n\implies\mathcal D_n=\mathcal R_n\oplus\langle h\rangle$$ As $\mathcal D_n$ is non-abelian it is not the product of abelian groups and thus $\langle h\rangle$ is not normal in $\mathcal D_n$.