Complement of a projective two sided ideal is two sided again? If a finite dimensional algebra $A$ over a field $\mathbb{k}$ is semisimple then any two sided ideal of $A$ is generated (as a left module) by a central idempotent, so its (unique) complement is a two sided ideal again. In general, this observation is wrong as e.g., in the $\mathbb{k}$-algebra of upper triangular ($2 \times 2$)-matrices the (non-central) matrix unit $e_{2,2}$ generates a two sided ideal $Ae_{2,2}$ having a complement $Ae_{1,1}$ which is not two sided.
However, I am not aware of such an example if $A$ is a group algebra.
Question: Let $\mathbb{k}$ be a field, and let $G$ be a finite group such that the algebra $\mathbb{k} G$ is not semisimple. Let $e \in \mathbb{k} G$ be an idempotent generating a two sided ideal $\mathbb{k} G e$. Does this imply that $e$ is central?
Note that since group algebras are self-injective, any left ideal $L$ isomorphic to a two sided ideal $I$ must be equal to $I$. So if $\mathbb{k} G e$ has any two sided complement then it has to be unique, and $e$ must be central.

If $A$ is any finite dimensional algebra and $e$ any idempotent, then $A=Ae\oplus A(1-e)$ as a direct sum of left ideals. The endomorphism ring of $A$ as a left module is $A^{op}$ acting by right multiplication, so the condition that $Ae$ is a two sided ideal is equivalent to $\text{Hom}_A\left(Ae,A(1-e)\right)=0$.
But for a finite group algebra this is equivalent to $\text{Hom}_A\left(A(1-e),Ae\right)=0$, and hence to $A(1-e)$ also being a two sided ideal, since finite group algebras are examples of symmetric algebras, for which $\text{Hom}_A(P,Q)$ is naturally dual to $\text{Hom}_A(Q,P)$ for finitely generated projective modules $P$ and $Q$.
