Morita equivalence between $k$ and $kG$ I would like to show that $k$ is Morita equivalent to $kG$ iff $G$ is the trivial group.
clearly, if $G$ is the trivial group, $kG\cong k$ and so $k$ is Morita equivalent to $kG$
However for the other way, I have no idea. I tried to find some contradictions in supposing that they where Morita equivalent but have found none.
Any hints?
 A: Morita equivalent rings have the same center (exercise). The center of a group ring $k[G]$ has a basis given by sums over the conjugacy classes of $G$, and in particular has dimension given by the number of conjugacy classes of $G$ (exercise). A finite group of size at least $2$ has at least $2$ conjugacy classes. 
(Edit: As pointed out by Eric Wofsey in the comments, this argument only shows that the centers are not isomorphic as $k$-algebras, so only addresses Morita equivalence over $k$.)
If $G$ is infinite then the center of $k[G]$ involves only sums over the finite conjugacy classes, and so can be trivial if every nontrivial conjugacy class of $G$ is infinite. So we can instead argue as follows. I assume $k$ is a field, in which case the only rings Morita equivalent to $k$ are $M_n(k)$, which are in particular all finite-dimensional over $k$. You can also use this to finish: there's an obvious homomorphism $k[G] \to k$ given by sending all $g \in G$ to $1$, and $M_n(k)$ doesn't admit such a homomorphism unless $n = 1$.
