Sorry for my ignorance in advance, this should be a very naive question and I would be happy for a reference. 
Let $G$ be an arbitrary group (not necessary finite) acting on two (connected) manifolds $M$ and $N$. The action on $M$ is on the left and the action on $N$ is on the right. Suppose that the action  

$M\times G\rightarrow M$ is free. 
Is it true that 
 $$H\_{\ast}(M\times\_G N, k)= H\_{\ast}(M/G \times N,k )$$
where $k$ is a field or $\mathbb{Z}$ and the topology on $M/G$ is defined by the quotient map $M\rightarrow M/G$.



Homology of manifold with action of group
No, this is in general quite far from being true (and also somewhat ill-posed, since there is not a natural candidate for what "=" means since there is no map).


For example, take $M = S^1$ and $N=S^2$, both with the antipodal $C\_2$-action (which you can view as either right or left action).


Then $M/C\_2 \simeq S^1$ so $H\_\*(M/C\_2 \times N;k) \cong H\_\*(S^1 \times S^2;k)$ as graded groups.


But $C\_2$ acts as $-1$ on $H\_2(N)$ and trivially on $H\_\*(M)$ so $H\_\*(M \times\_{C\_2} N;{\mathbb Q}) \cong H\_\*(M \times N;{\mathbb Q})\_{C\_2} \cong H\_\*(M;{\mathbb Q}) \cong H\_\*(S^1;{\mathbb Q})$ as graded abelian groups, where $C\_2$ on the outside denotes coinvariants.


