I have tried to compute the de Rham cohomology and the homology over the integers of the space $S^3/D^*_k$, where $D^*_k$ is the binary dihedral group of order $4k$ and I would like to know if what I have done is correct.
If we think of $S^3\subset \mathbb{C}^2$, $S^3=\{(z^1,z^2):|z^1|^2+|z^2|^2=1\}$, the action of $D^*_k$ that I want to consider is generated by \begin{equation} \begin{split} g_1&:(z^1,z^2)\mapsto \mathrm{e}^{\frac{i\,\pi}{k} }(z^1,z^2),\\ g_2&:(z^1,z^2)\mapsto i\,(\bar z ^2,-\bar z ^1). \end{split} \end{equation}
Let us start with de Rham cohomology. Since $H^*(S^3/D^*_k)=H^{*\,D^*_k}(S^3)$ and $H^*(S^3)$ is non-vanishing only in degree 0 and 3, we immediately have $H^1(S^3/D^*_k)=H^2(S^3/D^*_k)=0$. Since $S^3/D^*_k$ is connected, $H^0(S^3/D^*_k)=\mathbb{R}$ and by Poincare duality $H^3(S^3/D^*_k)=\mathbb{R}$, hence
\begin{equation} H^p_{\mathrm{dR}}(S^3/D^*_k)=\begin{cases} \mathbb{R}&\text{if $p=0,3$,}\\ 0&\text{otherwise}. \end{cases} \end{equation}
Next I have tried to calculate homology with integer coefficients. $H_0(S^3/D^*_k)=\mathbb{Z}$ as the space is connected. To calculate $H_1$ I thought of looking at the abelianization of $\pi_1$. Since $S^3$ is the universal cover, $\pi_1(S^3/D^*_k)=D^*k$ which has the presentation \begin{equation} D^*_k=\langle x,a:a^{2k}=e,x^2=a^k,xax^{-1}=a^{-1} \rangle. \end{equation} Now I am not sure if this is the correct way to calculate the ablianization of a group, but I thought that I can treat the product in the relations as commutative thus getting $a=a^{-1}$, hence $a$ has order 2. Then if $k$ is even $x$ is also of order 2 and the abelianization is $\mathbb{Z}_2\times\mathbb{Z}_2$, while if $k$ is odd we get $\mathbb{Z}_4$. Is this correct?
By Poincare duality $H_3(S^3/D^*_k)=H^0(S^3/D^*_k)=\mathbb{Z}$. Hence
\begin{equation} H_p(S^3/D^*_k,\mathbb{Z})=\begin{cases} \mathbb{Z}&\text{if $p=0,3$},\\ \mathbb{Z}_2\times\mathbb{Z}_2&\text{if $p=1$ and $k$ is even,}\\ \mathbb{Z}_4&\text{if $p=1$ and $k$ is odd,}\\ 0 &\text{otherwise}. \end{cases} \end{equation}
I do not know how to compute $H_2$.
EDIT: I think I have found how to compute $H_2$: Since $S^3/D^*_k$ is compact and orientable $H_2$ has no torsion. Its rank can be found using the fact that $0=\chi(S^3)=|D^*_k|\cdot\chi(S^3/D^*_k)\Rightarrow \chi(S^3/D^*_k)=0$. Since we know the ranks of all the other groups it follows that the rank of $H_2$ is zero.