Intuition behind Maschke's theorem

Question

I'm an undergraduate learning about group representations and Young tableaux, and have came across Maschke's theorem stating;

If $G$ is a finite group and $F$ is a field whose characteristic does not divide the order of $G$, then every finite dimensional $G$-module over $F$ is completely reducible

The many proofs I have seen, show this by proving an equivalent statement that if $G$ and $F$ are as above, and $H$ is a submodule of a $G$-module $V$, then there exists a submodule $H'$ of $V$ such that $V$ is the direct product of $H$ and $H'$. This is done by showing $H'$ is the kernal of some homomorphism of $G$-modules that maps $H$ to $H$. However, the this approach doesn't seem well-motivated in the sense of; where did this homomorphism come from?

Does anyone have a suitable alternative proof for an undergraduate? Or can possibly shed some intuition on the theorem? Thank you.

Answer 1

Actually, I think that approach can be made pretty well-motivated. Here is how I like to see it: we have a $FG$-module $V$ and a submodule $H$, and we would like to know that there is a $G$-stable complement $H'$ (under certain circumstances, e.g., always if the characteristic of $F$ is prime to $|G|$). If such a complement exists, then $V=H \oplus H'$ and the projection onto $H$ with kernel $H'$ is a $G$-module map $V \rightarrow H$ left inverse to the $G$-module map including $H \hookrightarrow V$. Conversely, given a $G$-module map left inverse to this inclusion, its kernel will be a complement.

Now we are left with the problem of how to find such a homomorphism. Here is the idea: consider the vector space $\mathrm{Hom}_F(V,H)$. Inside this vector space live many projections $\pi$ onto $H$. We'd like to find one that is a $G$-module map, that is, such that $g \pi g^{-1}=\pi$ for all $g \in G$. If you stare at that equation for a second, it becomes clear that what you are really looking for is a $G$-fixed point in $\mathrm{Hom}_F(V,H)$, where $G$ acts by conjugation. Now the time-honored way of finding a fixed point for a group action is by averaging---start with any projection whatsoever and average it over all its images by elements of $G$. The condition on the characteristic is precisely what allows you to do this!

A nice thing about this point of view is that it tells you something even when the characteristic is not prime to the order of the group: it says you should really care about the functor of $G$-fixed points. This is a good way to motivate group cohomology to first year grad students, in my experience.

Answer 2

Stephen's answer briefly mentions group cohomology. I can't resist and point out that Maschke's theorem can be viewed (or rephrased) as a consequence of group cohomology.

To see this, fix a group $G$ and a field $K$. Denote by $K[G]$ the group ring of $G$ over $K.$ Let $K[G]\mathsf{-mod}$ be the category of $K[G]$-modules.

First, note that the following are equivalent:

1. Any $K[G]$-submodule $V$ of any $K[G]$-module has a $G$-invariant complement.
2. Every $K[G]$-module is projective.
3. Every $K[G]$-module is injective.
4. The functor $h^V:=\operatorname{Hom}_{K[G]}(-,V)\colon (K[G]\mathsf{-mod})^{opp}\rightarrow \mathsf{Ab}$ is exact for every $V\in K[G]\mathsf{-mod}$.

Next, verify that the functor $h^V$ is isomorphic to the composition $I\circ M$ of functors $$M:=\operatorname{Hom}_K(-,V)\colon (K[G]\mathsf{-mod})^{opp}\rightarrow K[G] \mathsf{-mod}$$ $$I:=(-)^G\colon K[G]\mathsf{-mod}\rightarrow \mathsf{Ab}$$ for every $V\in K[G]-\mathsf{mod}$. Here, for $W$ a $K[G]$-module, we endow the set of $K$-linear maps $\operatorname{Hom}_K(W,V)$ with the structure of a $K[G]$-module via the diagonal action $g.\phi=g.\phi(g^{-1}.-)$, where $g\in G$ and $\phi\in \operatorname{Hom}_K(W,V).$ The second functor sends a $K[G]$-module to its abelian group of $G$-fixed points, i.e. its $G$-invariants. Both functors are additive. Since every $K$-vector space has a basis, the functor $M$ is even exact.

As the composition of exact functors is exact, Maschke's theorem now follows from the following result.

Proposition. Assume that $G$ is finite and the characteristic of $K$ does not divide the group order $n:=\vert G \vert$. Then the functor $I=(-)^G$ is exact.
Proof. Condider $K$ with the trivial $G$-action. Recall that the functor $I$ is isomorphic to the functor $\operatorname{Hom}_{K[G]}(K,-)$. In particular, the functor $I$ is left exact. Thus, it is exact if and only if its first right derived functor $R^1I\cong \operatorname{Ext}_{K[G]}^1(K,-)\cong\operatorname{Ext}_{\mathbb{Z}[G]}^1(\mathbb{Z},-)$ vanishes on every $K[G]$-module. In other words, $I$ is exact if and only if the first cohomology $H^1(G,V)$ of the group $G$ vanishes for every $K[G]$-module $V$. To see that the cohomology vanishes, consider the $K[G]$-module homomorphism $V\rightarrow V,x\mapsto n\cdot x.$ Since the characteristic of $K$ does not divide $n$, this $K[G]$-module homomorphism is an isomorphism in $K[G]\mathsf{-mod}$ with inverse $V\rightarrow V,x\mapsto 1/n\cdot x.$ (I find this quite illuminating: In Maschke's theorem the assumptions on the characteristic and the group order $n$ are precisely the way they are so that the multiplication by $n$ is invertible.) Consequently, the induced map $m_\ast: H^1(G,V)\rightarrow H^1(G,V)$ is an isomorphism. By the additivity of the cohomology functor $H^1(G,-)$, the map $m_\ast$ is multiplication by $m$. Thus, we have $H^1(G,V)=mH^1(G,V)$. With $mH^1(G,V)=0$, the proposition follows, QED.

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For a proof that $mH^1(G,V)=0$ holds, see Corollary 16.4 in A Course in Homological Algebra by P.J. Hilton and U. Stammbach.