Trivial Group Representation is Projective implies Semisimple Let $G$ be a group and $\mathbb k$ a field. Then, let $\mathfrak 1$ be the trivial $\mathbb k G$-module.
According to Lorenz's A Tour of Representation Theory, he states "it turns out $\mathfrak 1$ has a great significance, if it is a projective as a $\mathbb kG$-module, then all $\mathbb kG$-module is projective" and goes on to reference Maschke's Theorem.
I can understand the following claim, that a ring $\mathbb kG$ is semisimple if and only if all its modules are projective, so Maschke's Theorem does indeed show that $\mathfrak 1$ is projective for a semisimple $\mathbb k G$. I can also see that if $\mathfrak 1$ is projective, then it is direct summand of some free module.
However, I don't understand how the converse implication (i.e. $\mathfrak 1$ projective implies $\mathbb kG$ semisimple) suggested by Lorenz's remark is shown. It seems to suggest that we want $\mathfrak 1$ to be the direct summand of $\mathbb kG$ as a $\mathbb kG$-module, but I don't quite see how this is done and even if it is, how the remark is proven.

I'll write $k$ for the trivial module. There's a natural map $\varepsilon : k[G] \to k$ of $k[G]$-modules given by sending every $g \in G$ to $1$, and $k$ is projective iff this map splits as a map of $k[G]$-modules. A splitting of this map must send $1 \in k$ to a $G$-invariant element of $k[G]$, which must have the form $c \sum_{g \in G} g$ for some constant $c$; moreover, we must have
$$\varepsilon \left( c \sum_{g \in G} \right) = c |G| = 1$$
so a splitting exists iff $|G|$ is invertible in $k$, and by Maschke's theorem this condition is necessary and sufficient for $k[G]$ to be semisimple. (Maschke's theorem isn't always stated as an if-and-only-if but it is.)
