# 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?

• It is sufficient to show that any nontrivial group has got more representations than just the trivial one. – vap Jan 5 '18 at 14:13
• because ${\bf{Rep}}_G(k)\cong {}_{kG}{\bf{Mod}}\cong {}_{k}{\bf{Mod}}\cong{\bf{Rep}}_e(k)$? however does this induce a "bijection" between the objects of ${\bf{Rep}}_e(k)$ and the objects of ${\bf{Rep}}_G(k)$? – tomak Jan 5 '18 at 14:33
• What is $k$? A field? – Eric Wofsey Jan 5 '18 at 20:45

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 $n = 1$.
• Your first argument only shows they cannot be Morita equivalent over $k$, since you're using the $k$-algebra structure on the center. – Eric Wofsey Jan 5 '18 at 20:49
• You're right. The last argument works over $\mathbb{Z}$ and doesn't assume that $k$ is a field either. – Qiaochu Yuan Jan 5 '18 at 20:51