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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!

Thanks in advance!

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    $\begingroup$ I've fixed the presentation of $D_n$, hopefully nothing got wrong. $\endgroup$ – lisyarus Mar 4 at 13:14
  • $\begingroup$ thanks, that was a typo, didn't realize! $\endgroup$ – Máté Kadlicskó Mar 4 at 13:17

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