Let $K=\langle b,c,d\mid b^{2}=c^{2}=d^{2}=bcd=1\rangle $. Now we consider $$D=K*\mathbb Z/2\mathbb Z=\left\{a,b,c,d\mid a^{2}=b^{2}=c^{2}=d^{2}=bcd=1\right\}$$ where $*$ is the free product. Then we can construct a group algebra $$\mathbb C(D)=\left\{h_{0}g_{0}+\Sigma_{i=1}^{n}h_{k}g_{k}\mid h_{i}\in \mathbb C,g_{i}\in D,g_{0}=1,g_{i}\neq 1 \ when\ i\neq 0\right\}$$ where ${\rm tr}(h_{0}g_{0}+\Sigma_{i=1}^{n}h_{k}g_{k})=h_{0}$ and $(h_{0}g_{0}+\Sigma_{i=1}^{n}h_{k}g_{k})^{*}=\overline{h_{0}}g_{0}^{-1}+\Sigma_{i=1}^{n}\overline{h_{k}}g_{k}^{-1}$. We can define an inner product on $C(D)$ by $\langle f,g\rangle ={\rm tr}(g^{*}f)$. By taking completion, we get a Hilbert space $H_{D}$.
Now we consider $u=\frac{b+c+d-1}{2}$, it's easy to check that $u^{2}=1$. We have a subspace $H$ of $H_{D}$ spanned by $\left\{(au)^{k}, u(au)^{l}|k,l\in \mathbb Z\right\}$. This $H$ is very similar to the Hilbert space $H_{D_{\infty}}$ where $D_{\infty}$ is infinite dihedral group. But their inner products are different. $(au)^{k}$'s are not orthogonal. It's natural to consider the ${\rm tr}(au)^{k}$ in order to dig deeper for this $H$. $(au)^{k}=a(\frac{b+c+d-1}{2})a(\frac{b+c+d-1}{2})...$, the trace is the coefficient of 1.When $k$ is odd , it's obvious that ${\rm tr}(au)^{k}=0$ since we don't have $1$ term after expand $(au)^{k}$.
Things become complicated when $k$ is even because we have plenty of "$1$" terms. For instance, after expanding $(au)^{k}$, a prototype for "$1$" terms is $$aa...\text{even numbers}... aabaa...\text{even numbers}...aabaaa...$$ Since there are tons of such kind terms, I don't know how to estimate the ${\rm tr}(au)^{k}$. Since I am not so familiar with group theorem, I don't know if there is some tools in group theory to deal with such kind problem.
