It turns out that our efforts last time were somewhat unnecessary, although they were instructive. Actually, we already had a matrix representation in our hands that would have done the trick.
The secret is to look at the block diagonal form from when we defined reducibility:
We worked this out for the product of two group elements, finding
We focused before on the upper-left corner to see that was a subrepresentation, but we see that as well. The thing is, if the overall representation acts on and acts on the submodule , then acts on the quotient space . That is, it’s generally not a submodule. However, it so happens that over a finite group we have as modules. That is, if we have a matrix representation that looks like the one above, then we can find a different basis that makes it look like
This is more than Maschke’s theorem tells us — not only do we have a decomposition, but we have one that uses the exact same matrix representation in the lower right as the original one. Proving this will reprove Maschke’s theorem, and in a way that works over any field!
So, let’s look for a change-of-basis matrix that’s partitioned the same way:
Multiplying this out, we find
which gives us the equation , which we rearrange to give
as well as
That is, acting on the left of by and on the right by doesn’t leave unchanged, but instead adds a certain offset. We’re not looking for an invariant of these actions, but something close. Incidentally, why are these two offsets the same? Well, if we put them together we find
which must clearly be zero, as desired.
Anyway, I say that things will work out if we choose
Indeed, we calculate
Just as we wanted.
Notice that just like our original proof of Maschke’s theorem, this depends on a sum that is only finite if is a finite group.