# If $M$ is a $G$-module, and $H\leq G$, is $kG\otimes_{kH}\otimes kG\otimes_{kG} M\simeq M$ as $G$-modules?

Say you're given a group $G$ with a subgroup $H$. Suppose $M$ is a $G$-module, and $k$ is some coefficient field. If we restrict to $H$ and then induct back to $G$, do we get a module isomorphic to $M$? That is, viewing $kG$ as a $(kG,kH)$-bimodule or a $(kH,kG)$-bimodule, respectively below, where appropriate, is $$kG\otimes_{kH} kG\otimes_{kG} M\simeq M?$$

I believe it to be true, since I think $kG\otimes_{kG} M$ is basically just viewing $M$ as a left $H$-module, since $G$ is an $(H,G)$-bimodule here, and then left tensoring by $kG\otimes_{kH}$ extends the scalars back to $kG$, but I'm not sure how to write down an explicit isomorphism.

Maybe more simply, is $kG\otimes_{kH} kG\simeq kG$ as a $(G,G)$-bimodule? Because then I know the usual multiplication isomorphism gives $kG\otimes_{kG} M\simeq M$.

If $H=\{e\}$ then the question becomes: is $kG\otimes_k kG\cong kG$. But the former has dimension $|G|^2$ over $k$ and the latter has dimension $|G|$.