# The Unapologetic Mathematician

## An Alternative Path

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: $\displaystyle\left(\begin{array}{c|c}X(g)&Y(g)\\\hline{0}&Z(g)\end{array}\right)$

We worked this out for the product of two group elements, finding $\displaystyle\left(\begin{array}{c|c}X(gh)&Y(gh)\\\hline{0}&Z(gh)\end{array}\right)=\left(\begin{array}{c|c}X(g)X(h)&X(g)Y(h)+Y(g)Z(h)\\\hline{0}&Z(g)Z(h)\end{array}\right)$

We focused before on the upper-left corner to see that $X$ was a subrepresentation, but we see that $Z(gh)=Z(g)Z(h)$ as well. The thing is, if the overall representation acts on $V$ and $X$ acts on the submodule $W$, then $Z$ acts on the quotient space $V/W$. That is, it’s generally not a submodule. However, it so happens that over a finite group we have $V\cong W\oplus V/W$ 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 $\displaystyle\left(\begin{array}{c|c}X(g)&0\\\hline{0}&Z(g)\end{array}\right)$

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: $\displaystyle\left(\begin{array}{c|c}I&D\\\hline{0}&I\end{array}\right)\left(\begin{array}{c|c}X(g)&Y(g)\\\hline{0}&Z(g)\end{array}\right)=\left(\begin{array}{c|c}X(g)&0\\\hline{0}&Z(g)\end{array}\right)\left(\begin{array}{c|c}I&D\\\hline{0}&I\end{array}\right)$

Multiplying this out, we find $\displaystyle\left(\begin{array}{c|c}X(g)&Y(g)+DZ(g)\\\hline{0}&Z(g)\end{array}\right)=\left(\begin{array}{c|c}X(g)&X(g)D\\\hline{0}&Z(g)\end{array}\right)$

which gives us the equation $Y(g)+DZ(g)=X(g)D$, which we rearrange to give $\displaystyle X(g)DZ\left(g^{-1}\right)=D-X(g)Y\left(g^{-1}\right)$

as well as $\displaystyle X(g)DZ\left(g^{-1}\right)=D+Y(g)Z\left(g^{-1}\right)$

That is, acting on the left of $D$ by $X(g)$ and on the right by $Z\left(g^{-1}\right)$ doesn’t leave $D$ 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 $\displaystyle X(g)Y\left(g^{-1}\right)+Y(g)Z\left(g^{-1}\right)=Y\left(gg^{-1}\right)=Y(e)$

which must clearly be zero, as desired.

Anyway, I say that things will work out if we choose $\displaystyle D=\frac{1}{\lvert G\rvert}\sum\limits_{h\in G}X\left(g^{-1}\right)Y(g)$

Indeed, we calculate \displaystyle\begin{aligned}X(g)DZ\left(g^{-1}\right)&=X(g)\left(\frac{1}{\lvert G\rvert}\sum\limits_{h\in G}X\left(h^{-1}\right)Y(h)\right)Z\left(g^{-1}\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{h\in G}X\left(gh^{-1}\right)Y(h)Z\left(g^{-1}\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}X\left(f^{-1}\right)Y(fg)Z\left(g^{-1}\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}X\left(f^{-1}\right)\left(X(f)Y(g)+Y(f)Z(g)\right)Z\left(g^{-1}\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}\left(Y(g)Z\left(g^{-1}\right)+X\left(f^{-1}\right)Y(f)\right)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}Y(g)Z\left(g^{-1}\right)+\frac{1}{\lvert G\rvert}\sum\limits_{f\in G}X\left(f^{-1}\right)Y(f)\\&=Y(g)Z\left(g^{-1}\right)+D\end{aligned}

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 $G$ is a finite group.

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October 28, 2010