TITLE: Decomposing a matrix representation
QUESTION [5 upvotes]: I am currently working on the following problem:

Assume that $X$ is a reducible matrix representation of the form
\begin{equation} X(g)=\left( \begin{array}{c|c}   A(g) & B(g)\\ \hline
  0 & C(g) \\  \end{array}\right). \end{equation}
Use part (a) to show that
  $$TX(g)T^{-1} = \left( \begin{array}{c|c}   A(g) & 0\\ \hline
  0 & C(g) \\  \end{array}\right),$$
  where $T = \left( \begin{array}{c|c}   I & D\\ \hline
  0 & I \\  \end{array}\right)$ and $D = \frac{1}{|G|}\sum_{g \in G}A(g^{-1})B(g).$
In part (a) we found that $A(gh) = A(g)A(h)$, $C(gh) = C(g)C(h)$, by equating blocks in $X(gh) = X(g)X(h)$.

This is what I have got so far:
Clearly $T^{-1} = \left( \begin{array}{c|c}   I & -D\\ \hline
  0 & I \\  \end{array}\right)$.
Then $$TX(g)T^{-1} = \left( \begin{array}{c|c}   I & D\\ \hline
  0 & I \\  \end{array}\right)\left( \begin{array}{c|c}   A(g) & B(g)\\ \hline
  0 & C(g) \\  \end{array}\right)\left( \begin{array}{c|c}   I & -D\\ \hline
  0 & I \\  \end{array}\right)= \left( \begin{array}{c|c}   A(g) & B(g) + DC(g)\\ \hline
  0 & C(g) \\  \end{array}\right)
\left( \begin{array}{c|c}   I & -D\\ \hline
  0 & I \\  \end{array}\right) = \left( \begin{array}{c|c}   A(g) & -A(g)D + B(g) + DC(g)\\ \hline
  0 & C(g) \\  \end{array}\right).$$
Thus, we are required to show that $-A(g)D + B(g) + DC(g) = 0$.
But $-A(g)D = -A(g)\frac{1}{|G|}\sum_{g \in G}A(g^{-1})B(g)$ and $DC(g) = \frac{1}{|G|}(\sum_{g \in G}A(g^{-1})B(g))C(g)$.
Here I do not see any way to continue. Do you know what I could do?
Thank you very much for your help!

REPLY [3 votes]: This is rather simple actually. As noted in the OP, it suffices
to show that $-A(g)D+B(g)+DC(g)=0$.
Now, equating yet another block in $X(gh)=X(g)X(h)$, we obtain
$A(g)B(h)+B(g)C(h)=B(gh)$, or (switching variable names) 
$A(h)B(g)+B(h)C(g)=B(hg)$. But then
$$
\begin{array}{lcl}
DC(g)&=&\frac{1}{|G|}\sum_{h\in G}A(h^{-1})B(h)C(g) \\
&=& \frac{1}{|G|}\sum_{h\in G}A(h^{-1})\big(B(hg)-A(h)B(g)\big) \\
&=& \bigg(\frac{1}{|G|}\sum_{h\in G}A(h^{-1})B(hg)\bigg)-B(g) \\
&=& \bigg(\frac{1}{|G|}\sum_{w\in G}A(gw^{-1})B(w)\bigg)-B(g) \ (\text{putting } w=hg)\\
&=& A(g)D-B(g).
\end{array}
$$