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.
 A: 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.)
A: One can see in the following way that every $G$-representation (over $\mathbb{k}$) is projective.
Step 1.
For every two representations $V$ and $W$ of $G$, the vector space $\operatorname{Hom}_{\mathbb{k}}(V,W)$ is again a representation of $G$ via
$$
  (g \cdot f)(v) = g \cdot f( g^{-1} \cdot v )
$$
for all $g \in G$, $f \in \operatorname{Hom}_{\mathbb{k}}(V, W)$ and $v \in V$.
A linear map $f$ from $V$ to $W$ is $G$-invariant if and only if it is a homomorphism of representations, i.e.
$$
  \operatorname{Hom}_{\mathbb{k}}(V,W)^G
  =
  \operatorname{Hom}_G(V,W) \,.
$$
Step 2.
We have for every $\mathbb{k}$-vector space $V$ the natural isomorphism of vector spaces
$$
  \operatorname{Hom}_{\mathbb{k}}(\mathbb{k}, V) \longrightarrow V \,,
  \quad
  f \longmapsto f(1) \,.
$$
Suppose now that $V$ is a representation of $G$.
The above isomorphism of vector spaces is then an isomorphism of $G$-representations.
It therefore restricts to a natural isomorphism of vector spaces
$$
  \operatorname{Hom}_G(\mathbb{k}, V)
  \longrightarrow
  V^G \,.
$$
Step 3.
Suppose now that the trivial representations $\mathbb{k}$ is projective.
This means that the functor
$$
  \operatorname{Hom}_G(\mathbb{k}, -)
  \colon
  \mathbf{Rep}(G)
  \longrightarrow
  \mathbf{Vect}(\mathbb{k})
$$
is exact.
It follows from the (natural) isomorphism $\operatorname{Hom}_G(\mathbb{k}, -) \cong (-)^G$ that the functor
$$
  (-)^G
  \colon
  \mathbf{Rep}(G)
  \longrightarrow
  \mathbf{Vect}(\mathbb{k})
$$is exact.
Step 4.
Let now $V$ be an arbitrary representations of $G$.
The functor $\operatorname{Hom}_G(V, -)$ equals the composition
$$
  \mathbf{Rep}(G)
  \xrightarrow{ \enspace \operatorname{Hom}_{\mathbb{k}}(V, -) \enspace }
  \mathbf{Rep}(G)
  \xrightarrow{ \enspace (-)^G \enspace }
  \mathbf{Vect}(\mathbb{k}) \,.
$$
Both of these functors are exact, whence $\operatorname{Hom}_G(V, -)$ is exact.
This shows that $V$ is projective.

There is also another way to look at this problem:
The group structure of $G$ gives its group algebra $\mathbb{k}G$ the structure of a Hopf algebra.

Partial Definition.
A Hopf algebra over $\mathbb{k}$ is a $\mathbb{k}$-algebra $H$ together with homomorphisms of $\mathbb{k}$-algebras
$$
    \Delta
    \colon
    H
    \to
    H \otimes H
    \,,
    \quad
    \varepsilon
    \colon
    H
    \to
    \mathbb{k}
  $$
such that certain conditions hold (which I won’t mention here).
The map $\varepsilon$ is the counit of $H$, and the map $\Delta$ (which we won’t care about in the following) is the comultiplication of $H$.

Given a Hopf algebra $H$, the ground field $\mathbb{k}$ becomes an $H$-module via
$$
  x \cdot y
  =
  \varepsilon(x) y
$$
for all $x \in H$ and $y \in \mathbb{k}$.
The counit map $\varepsilon$ becomes in this way a homomorphism of $H$-modules from $H$ to $\mathbb{k}$.

Example.
In the case of $H = \mathbb{k}G$ the maps $\Delta$ and $\varepsilon$ are given on the basis $G$ of $\mathbb{k}G$ by
$$
    \Delta(g)
    =
    g \otimes g \,,
    \quad
    \varepsilon(g)
    =
    1
  $$
for every $g \in G$.
The resulting $\mathbb{k}G$-module structure on $\mathbb{k}$ is given by
$$
    g \cdot y
    =
    \varepsilon(g) y
    =
    1 \cdot y
    =
    y
  $$
for all $g \in G$ and $y \in \mathbb{k}$.
This is precisely the $\mathbb{k}G$-module structure on $\mathbb{k}$ that corresponds to the trivial action of $G$ on $\mathbb{k}$.

There is now a generalization of Maschke’s theorem to Hopf algebras.

Theorem (Maschke’s theorem for Hopf algebras).
Let $H$ be a Hopf algebra.
The following conditions on $H$ are equivalent.

*

*$H$ is semisimple.

*There exists an element $a$ of $H$ such that

*

*$x \cdot a = \varepsilon(x) a$ for all $x \in H$ and $a \in A$ (invariance), and

*$\varepsilon(a) = 1$ (normalization).



*There exists an $H$-submodule $A$ of $H$ with $H = A \oplus \ker(\varepsilon)$.

*The kernel of $\varepsilon$ (which is an $H$-submodule of $H$ because $\varepsilon$ is a homomorphism of $H$-modules) is a direct summand of $H$.

*The $H$-module $\mathbb{k}$ is projective.


One shows this by proving the implications
$$
  1 \implies 5 \implies 4 \implies 3 \implies 2 \implies 1 \,.
$$

Example.
Consider again $H = \mathbb{k}G$.
Let $a$ be an element of $H$ as in characterization 2 of the above theorem.
This element is given by a linear combination $a = \sum_{g \in G} a_g g$.
The condition $x \cdot a = \varepsilon(x) a$ holds for every $x \in \mathbb{k}G$ if and only if $g \cdot a = \varepsilon(g) a$ for every group element $g$ of $G$.
We have $\varepsilon(g) = 1$, so this condition is furthermore equivalent to $g \cdot a = a$ for every $g \in G$.
This means that all coefficients $a_g$ of $a$ have to be equal.
If $G$ is infinite, then this means that $a = 0$ (because $a$ can’t have infinitely many nonvanishing coefficients).
But then $\varepsilon(a) = 0$, contradicting $\varepsilon(a) = 1$.
So if $G$ is infinite, then no such element $a$ exist.
If $G$ is finite, then $a$ can be any element of the form $a = C \sum_{g \in G} g$ for some scalar $C$ in $\mathbb{k}$.
But we have
$$
    1
    =
    \varepsilon(a)
    =
    C \sum_{g \in G} \varepsilon(g)
    =
    C \cdot \lvert G \rvert.
  $$
We hence find that no such element $a$ exists if the characteristic of $\mathbb{k}$ divides $\lvert G \rvert$, and otherwise $C = 1 / \lvert G \rvert$ and thus
$$
    a
    =
    \frac{1}{\lvert G \rvert} \sum_{g \in G} g \,.
  $$
This is the “averaging element” (hence the letter $a$) used in the usual proof of Maschke’s theorem for groups.

By actually going through the full proof of the above theorem, one sees that when $\mathbb{k}$ is projective as an $H$-module, then the surjective homomorphism of $H$-modules $\varepsilon$ from $H$ to $\mathbb{k}$ splits.
Such a split gives us a homomorphism of $H$-modules from $\mathbb{k}$ to $H$, which in turn corresponds to an element of $H$.
This element is then the “averaging element” $a$.
