rt.representation theory - Apocryphal Maschke theorem?

The result about bimodules is true, and standard. Here is one way to see it.

By Frobenius reciprocity, $Hom_G(V,k[G]) = Hom_k(V,k)$, since $k[G]$ is the induction (or coinduction, depending on your terminology) of the trivial representation of the trivial subgroup of G to G.$

Since Frobenius reciprocity is functorial, one easily sees that this canonical isomorphism
is an isomorphism of right $G$-representations, where the source has a right $G$-action coming from the right $G$-action on $k[G]$, and the target has a right $G$-action coming as the transpose of the left $G$-action on $V$.

Now if by Maschke's semisimplicity theorem, we know that
$k[G] = bigoplus_V V otimes_k Hom_G(V,k[G])$, where the sum is over all irreducible left $G$-representations. (Indeed, Mashke shows that this is true
for any left $G$-module in place of $k[G]$.) Again, this is a natural isomorphism, and so respects the right $G$-actions on source and target.

Combined with the preceding computation, we find that
$k[G] = bigoplus_V Votimes_k Hom_k(V,k) = bigoplus_V End(V),$ as both left and right $G$-modules,
as required.

[Edit:] Leonid's remark about $k$ being needing to be big enough in his answer below is correct. Each simple $V$ comes equipped with an associated division algebra of $G$-endomorphisms
$A_V := End_G(V)$. The representation $V$ is absolutely irreducible (i.e stays irred. after passing to any extension field) if and only if $A_V = k$. When we consider $Hom_k(V,W)$ for another left $G$-module $W$, this is naturally an $A_V$-module, and Maschke's theorem
will say that $W = bigoplus_V Hom_k(V,W)otimes_{A_V} V$. (I have written the factors in the tensor product in this order because $V$ is naturally a left $A_V$-module (if we think of endomorphisms acting on the left), and then $Hom_k(V,W)$ becomes a right $A_V$-module.)

So in the case of $W$ being the group algebra, we have
$$k[G] = bigoplus_V Hom_k(V,k)otimes_{A_V} V$$
(an isomorphism of $G$-bimodules).

If all the $V$ are absolutely irreducible, e.g. if $k$ is algebraically closed,
then all the $A_V$ just equal $k$, and the preceding direct sum reduces to what I wrote above, and what was written in the question.