The conjugation relation $a^{b}=a^{{-1}}$ allows us to place every element in the normal form $a^{i}b^{j}$. If $a^{i}b^{j}=a^{k}b^{l}$ then $a^{{i-k}}=b^{{l-j}}$. Yet $\langle a\rangle\cap\langle b\rangle=1$ so $i-k\equiv 0\;\;(\mathop{{\rm mod}}n)$ and $l-j\equiv 0\;\;(\mathop{{\rm mod}}2)$. Thus we have and irredundant list as required.

For the conjugacy classes note that $(a^{i})^{{a^{j}b}}=(a^{i})^{b}=a^{{-i}}$ so that these conjugacy classes are established. Next

for all $i\in\mathbb{Z}$. When $2\nmid n$ we have $(2,n)=1$ so $2$ is invertible modulo $n$. We let $j=2^{{-1}}(k-i)$ for any $k\in\mathbb{Z}$ and we see that $a^{i}b$ is conjugate to any $a^{k}b$. However, when $2|n$ we have a parity constraint that so far creates the two classes. We need to also verify conjugation by $a^{j}b$ does not fuse the two classes. Indeed

thus we retain two conjugacy classes amongst the reflections.

Finally, the order of the elements $(a^{i}b)$ is 2 – a fact used already. Thus the only cyclic subgroup of order $n$, when $n>2$, is $C_{n}$ and thus by its uniqueness it is characteristic. ∎

If $H$ is normal and contains an element of the form $a^{i}b$, then it contains the entire conjugacy class of $a^{i}b$. If $n$ is odd then all reflections are conjugate to $a^{i}b$ so $H$ contains all reflections of $D_{{2n}}$ and so $H$ is $D_{{2n}}$ as the relfections generate $D_{{2n}}$.

If instead $n$ is even then $H$ is forced only to contain one of the two conjugacy classes of reflections. If $i$ is even then $H$ contains $b$ and $a^{2}b$ so it contains $a^{2}$. If $i$ is odd then $H$ contains $ab$ and $a^{3}b$ so it contains $a^{2}=aba^{3}b$ (note $n>3$ as $n>2$ and $2|n$).

The two maximal subgroups of index $2$ which can exist when $n$ is even can be interchanged by an outer automorphism which maps $a\mapsto a^{{-1}}$ and $b\mapsto ab$ so these two are not characterisitic. The subgroups of a characterisitic cyclic group are necessarily characteristic. ∎

The homomorphic image of a dihedral group has two generators $\hat{a}$ and $\hat{b}$ which satisfy the conditions $\hat{a}^{{\hat{b}}}=\hat{a}^{{-1}}$ and $\hat{a}^{n}=1$ and $\hat{b}^{2}=1$, therefore the image is a dihedral group.

For subgroups we proceed by induction. When $n=1$ the result is clear. Now suppose that $D_{{2n}}$ has some proper subgroup $H$ that is not dihedral or cyclic. $H$ is contained in some maximal subgroup $M$ of $D_{{2n}}$. However the maximal subgroups of $D_{{2n}}$ are cyclic or dihedral so $H$ falls to the induction step for $M$ – together with the fact that subgroups of cyclic groups are cyclic. Thus $H$ must actually be dihedral or cyclic to avoid contradictions. ∎

When $n=1$, $D_{{2n}}\cong\mathbb{Z}_{2}$ which is nilpotent and so also solvable. Now let $n>1$. Then $D_{{2n}}/\langle a\rangle\cong\mathbb{Z}_{2}$ and $\langle a\rangle\cong\mathbb{Z}_{n}$. Both $\mathbb{Z}_{n}$ and $\mathbb{Z}_{2}$ are nilpotent and so they are both solvable. As extensions of solvable groups are solvable, $D_{{2n}}$ is solvable for all $n>0$. ∎