Little groups compute the irreducible characters of a dihedral group

For each integer $k$ the formula ${\theta }_k(r^j):={\zeta }^{jk}$ is a well-defined homomorphism $A\to {\mathbb{C}}^{\times }$, because ${\zeta }^n=1$ makes it independent of the representative $j$ modulo $n$, and ${\theta }_k(r^{j+l})={\zeta }^{(j+l)k}={\theta }_k(r^j){\theta }_k(r^l)$. By [F3] every irreducible complex representation of the abelian group $A$ is one-dimensional, hence of this form, and by [F4] the $n$ functions ${\theta }_0,\dots ,{\theta }_{n-1}$ are distinct (they take the distinct values ${\zeta }^k$ at $r$); so $\widehat{A}=\{{\theta }_0,\dots ,{\theta }_{n-1}\}$ with ${\theta }_k={\theta }_{k^{\mathrm{\prime }}}$ exactly when $k\equiv k^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}n)$.

The generator $s$ acts on $\widehat{A}$ by ${}^s{\theta }_k={\theta }_{-k}$: for all $j$, ${}^s{\theta }_k(r^j)={\theta }_k(s^{-1}{r}^{j}s)={\theta }_k(r^{-j})={\zeta }^{-jk}={\theta }_{-k}(r^j)$, using $s=s^{-1}$ and [F5]. Hence the $H$-orbit of ${\theta }_k$ is $\{{\theta }_k,{\theta }_{-k}\}$, of size one exactly when $k\equiv -k(\mathrm{m}\mathrm{o}\mathrm{d}n)$, i.e. $2k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$, and of size two otherwise; correspondingly $H_{{\theta }_k}=H$ in the first case and $H_{{\theta }_k}=1$ in the second.

Fixed case: if $2k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ then $H_{{\theta }_k}=C_2$ and $I_{{\theta }_k}=A⋊C_2=G$ by step 2.1, so [F2] lists the representations over ${\theta }_k$ as ${\tilde{\theta }}_k\otimes \mathrm{Infl}\sigma$ with $\sigma \in \mathrm{Irr}(C_2)$; the group $C_2$ has exactly its two linear characters, of degree $1$, and $[H:H_{{\theta }_k}]=1$, so this orbit contributes two linear characters of $D_{2n}$.

Non-fixed case: if $2k\equiv ̸0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ then $H_{{\theta }_k}=1$ and $I_{{\theta }_k}=A=⟨r⟩$ by step 2.1, so the only $\sigma$ is the trivial character of the trivial group and [F2] gives the single representation ${\mathrm{Ind}}_A^G{\theta }_k$, of degree $[H:1]\cdot 1=2$, induced from $C_n$. The two members of the orbit induce isomorphic representations, since ${\theta }_k$ and ${\theta }_{-k}$ are conjugate under $s$ and [F2] uses one orbit representative; so each orbit of size two contributes exactly one irreducible of degree two.

Counting: by [A1] exactly $f=\gcd (2,n)$ indices $k$ modulo $n$ satisfy $2k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$, so there are $2f$ linear characters by step 3.1 and the remaining $n-f$ characters of $\widehat{A}$ form $(n-f)\mathrm{/}2$ orbits of size two, contributing $(n-f)\mathrm{/}2$ irreducibles of degree two by step 3.2; the sum of squares of the degrees is $2f\cdot 1^2+\frac{n-f}{2}\cdot 2^2=2f+2(n-f)=2n=\mathrm{\mid }{D}_{2n}\mathrm{\mid }$ by [F1], and by [F2] this list is exactly $\mathrm{Irr}(D_{2n})$ with no repetitions.

The degenerate cases are included. For $n=1$ one has $f=1$ by [A1] and $D_2\cong C_2$ by [F1], and the list consists of the two linear characters of $C_2$, with no degree-two character. For $n=2$ one has $f=2$ and $D_4\cong C_2\times C_2$ by [F1], and the list consists of the four linear characters, with no degree-two character; both agree with the classification of step 4.1 since $(n-f)\mathrm{/}2=0$ in these cases.

The example is verified: the characters ${\theta }_k(r^j)=\exp (2\pi ijk\mathrm{/}n)$ of $C_n$ have $H$-orbits $\{k,-k\}$ by step 2.1; the fixed indices with $2k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ contribute two linear characters each by step 3.1; every other orbit contributes the single degree-two representation induced from $C_n$ by step 3.2; and by steps 4.1 and 5.1 these exhaust $\mathrm{Irr}(D_{2n})$ with the stated degree count, including $n=1$ and $n=2$.