# Prove that the character induced from $\langle x \rangle$ to $D_n$ is irreducible

If we take the dihedral group of order $$2n$$ for odd $$n$$, $$D_n=\langle x, y | \, x^n=y^2=1, yxy=x^{-1} \rangle$$, then we have $$\frac{n-1}{2}$$ complex characters, that are induced from $$\langle x \rangle$$ to $$D_n$$. For $$s=1,2, \dots, \frac{n-1}{2}$$, these characters are 2-dimensional and take non-zero values on the conjugacy classes containing $$1, x, x^2, \dots, x^{\frac{n-1}{2}}$$. If we denote the $$s$$th such character by $$\chi_{\zeta_n^s}\uparrow_{\langle x \rangle }^{D_n}$$, then $$\chi_{\zeta_n^s}\uparrow_{\langle x \rangle }^{D_n}(x^k)=\zeta_{n}^{sk}+\zeta_{n}^{-sk}$$.

Now, I've been struggling to show that $$\langle \chi_{\zeta_n^s}\uparrow_{\langle x \rangle }^{D_n}, \chi_{\zeta_n^s}\uparrow_{\langle x \rangle }^{D_n} \rangle=1$$.

Let $$\zeta_n^s=\cos(\frac{2s \pi}{n})+i\sin(\frac{2s \pi}{n})$$, so for all $$k$$

$$\zeta_{n}^{sk}+\zeta_{n}^{-sk}=2 \cos(\frac{2ks \pi}{n})$$, hence

$$\langle \chi_{\zeta_n^s}\uparrow_{\langle x \rangle }^{D_n}, \chi_{\zeta_n^s}\uparrow_{\langle x \rangle }^{D_n} \rangle=\frac{1}{2n}(4+2\sum_{k=1}^{\frac{n-1}{2}}4\cos^2(\frac{2ks\pi}{n}))$$ (we multiplied the sum by two, because $$x^i$$ and $$x^{-i}$$ form a conjugacy class in $$D_n$$)

I rewrote this to this form:

$$\frac{1}{2n}(2n+2+4\sum_{k=1}^{\frac{n-1}{2}} \cos(\frac{4ks\pi}{n}))$$

and it should be equal to one.

So basically (if I not yet have made a mistake) what I have left to prove is this: For odd $$n$$

$$\sum_{k=1}^{\frac{n-1}{2}} \cos(\frac{4ks\pi}{n})=-\frac{1}{2}$$

I have no idea, how to calculate this, or if I'm even on the right track still. I would appreciate any help!

• I've fixed the presentation of $D_n$, hopefully nothing got wrong. – lisyarus Mar 4 at 13:14