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.

  • $\begingroup$ I like to think of this as the statement that "$G$-modules work as one might expect them to, except possibly if $|G|$ divides $\operatorname{char}(F)$." $\endgroup$ – Alexander Gruber Mar 31 '13 at 15:16

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.

  • $\begingroup$ After taking some time to ponder the response, I believe this is much more motivated than was presented to me in class. As such I thank you for your response; however I still am slightly disturbed at having to construct H' as the kernel of some G-module homomorphism.. It seems more indirect than I'd like.. $\endgroup$ – gone Apr 3 '13 at 0:11
  • 2
    $\begingroup$ @Zak, well, here is an alternative approach, if you are working over the reals or complexes: construct the complement using a positive definite inner product! Of course, for the complement to be $G$-invariant, the inner product must be compatible with the $G$ action, and the proof that such and inner product exists is almost the same. But it has a more "geometric" feel to it. $\endgroup$ – Stephen Apr 3 '13 at 0:20

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