Why is $Ind^G_H(M)=Ind^{G/H}_{\{e\}}$? I was looking at some representation theory notes and found the following statement:
$Ind^G_H(V)=\mathbb{C}[G]\otimes_{\mathbb{C}[H]}V=\mathbb{C}[G/H]\otimes_\mathbb{C} V$.
Now, this makes intuitive sense but I am struggling to see precidely why this works. Moreover, does this still work if we generalise the field $\mathbb{C}$ to some (possibly noncommutative) ring $R$?
 A: Let $H$ be a subgroup of the finite group $G$, and let $M$ be a $G\times H^{\mathrm{opp}}$-bimodule. Then there is a functor from left $H$-modules to left $G$-modules given by 
$$V\mapsto M\otimes_{\mathbb{C}[H]}V.$$ 
If you take $M=\mathbb{C}[G]$ (with the left $G$-translation and the right $H$-translation), then this is the usual induction functor, which explains the first equality.
The second equality, however, usually does not hold when viewing them as $G$-modules (it holds when $H$ acts on $V$ trivially). For a simple example, consider $G=\mathbb{Z}/2\times \mathbb{Z}/2$, and $H=\mathbb{Z}/2$ is the left component. Take $V=\mathbb{C}[H]$. Then $\mathbb{C}[G]\otimes_{\mathbb{C}[H]}V$ is simply the regular representation $\mathbb{C}[G]$, on which $G=\mathbb{Z}/2\times \mathbb{Z}/2$ acts by
$$(a,b)\colon (c,d)\mapsto (a+c,b+d).$$
But $\mathbb{C}[G/H]\otimes_{\mathbb{C}}V$ is isomorphic to $\mathbb{C}[\mathbb{Z}/2\times \mathbb{Z}/2]$ with the $G=\mathbb{Z}/2\times \mathbb{Z}/2$-action ($G$ acts on $G/H$ by translations)
$$(a,b)\colon (c,d)\mapsto (b+c,d).$$
