If $M$ is a $kG$-module, $H\leq G$, why is the module of coinvariants isomorphic to $k(G/H)\otimes_{kG} M$? Suppose $k$ is a commutative unital ring, and $M$ is a left $kG$-module where $G$ is a finite group. If $H$ is a subgroup of $G$, why is the module of coinvariants $M_H=M/\langle m-h\cdot m:m\in M, h\in H\rangle$ isomorphic to $k(G/H)\otimes_{kG}M$?
I know that the module of coinvariants of $M$ under $H$ can also be interpreted as $k\otimes_{kH} M$ where $H$ acts trivially on $k$ on the right, the isomorphism given by 
$$
k\otimes_{kH}M\to M_H:a\otimes m\mapsto a\bar{m}.
$$
Why is either of these isomorphic to $k(G/H)\otimes_{kG}M$? How does $kG$ act on $k(G/H)$ on the right, since $H$ need not be normal? I was trying to construct an isomorphism in a similar way assuming a trivial right action, but I don't think it works. 
 A: The right action of $G$ on $G/H$ is given by inverse left mutliplication, that is, $gH \cdot g' := g'^{-1}gH$. Given that action, the desired isomorphism is given by $$k(G/H) \otimes_{kG} M \to M_H,\quad gH \otimes m \mapsto g^{-1}m +  \langle m' - hm' \:|\: m' \in M, h \in H\rangle.$$
Of course, some things have to be checked here. First of all, the map is well-defined, that is, different representatives for the coset $gH$ give the same element in $M_H$ and also $gH \otimes g'm$ maps to the same element as $g'^{-1}gH \otimes m$. Both things can be easily verfied.
To show that the given map is an isomorphism, we can consider the map 
$$ M_H \to k(G/H) \otimes_{kG} M, \quad m +  \langle m' - hm' \:|\: m' \in M, h \in H\rangle \mapsto 1H \otimes m$$
This map can be shown to be well-defined two. Now, all that is left to do is to show that these two maps are inverses of one another.
Also note that instead of working with $G/H$ you could have considered the set of left cosets $H\backslash \! G$ with the usual right action. That might be a bit more intuitive since you immediately know the right action. Also note that the isomorphisms above are very similar to and generalize the isomorphism $kG \otimes_{kG} M \cong M$ (the case where $H = \{1\}$).
