Trivial Group Representation is Projective implies Semisimple

Question

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.

Answer 1

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.

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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.

1. $H$ is semisimple.
2. 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).
3. There exists an $H$-submodule $A$ of $H$ with $H = A \oplus \ker(\varepsilon)$.
4. The kernel of $\varepsilon$ (which is an $H$-submodule of $H$ because $\varepsilon$ is a homomorphism of $H$-modules) is a direct summand of $H$.
5. 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” $a$.

Answer 2

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.)