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Assume that $G$ is a finite group. Let $k$ be a field. Let $\varepsilon$ be the augmentation $kG\rightarrow k$. Consider the following map

$\varepsilon\otimes id:k[G]\otimes_k k[G]\rightarrow k[G]$

since $k[G]\otimes_kk[G]$ is isomorphic to $k[G\times G]$ this allows us to view $k[G]$ as a $k[G\times G]$-module. My question is: is this module a projective $k[G\times G]$-module?

It is clear when $char\;k\not\mid|G|$ since in this case $k[G\times G]$ is semisimple so everything is projective, but what happens in general? Is that module still projective?

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The answer is no. Take for example $A=\mathbb{F}_2[C_2\times C_2]$, where $C_2=\langle c\mid c^2=1\rangle$ is the cyclic group of order 2. I claim that $A$ is indecomposable. Indeed, it is easy to see that the left regular module contains precisely one 1-dimensional submodule $W$ spanned by $$w=(1,1)+(1,c)+(c,1)+(c,c).$$ (Indeed, suppose $x=\alpha(1,1)+\beta(1,c)+\delta(c,1)+\gamma(c,c)$ spans a 1-dimensional submodule. Acting on $x$ with $(1,c)$ shows that $\alpha=\beta$ and $\delta=\gamma$, and acting with $(c,1)$ shows $\alpha=\delta$ and $\beta=\gamma$. Hence, $x=w$.)

On the other hand, take $u=(1,1)+(1,c)$ and $v=(1,1)+(c,1)$. Then $V=\mathrm{span}\{u,v,w\}$ is a 3-dimensional invariant subspace of $_AA$ containing $W$. If it had a compliment, it would be a 1-dimensional submodule different from $W$, which is impossible.

Now, $\mathbb{F}_2[C_2]$ is 2-dimensional, so it cannot be a direct summand of a free $A$-module, as $\dim A=4$ and $A$ is indecomposable.

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That module is projective precisely when the group algebra kG is separable over the field k (this is, essentially, the definition of separability) and separability implies semisimplicity. It follows that it is projective if and only if the characteristic of the field does not divide the order of the group.

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