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.



Find the trace for some elements in group algebra
You essentially want to count the number of 'words' of length 2k in the 'alphabet' $a$, $(ab)$, $(ac)$ and $(ad)$ that evaluate to 1. A quick computational check for $k=0,1, ..., 6$ suggests that this number corresponds to [sequence A194724 on the OEIS](http://oeis.org/A194724). The reason for this eludes me at the moment though...


