Question about matrix multiplication I am reading a paper from Diaconis and Thiem about supercharacters and wanted to ask the following question:
In the proof of Lemma 4.1(a) I don't understand how to get from the third to the fifth line, i.e. why does the equality
$x_A(\phi_A)x_\beta(\phi(\beta))x_B(\phi_B)x_\beta(t)-x_\beta(t)=x_A(\phi_A)x_\beta(\phi(\beta)+t)[x_B(\phi_B),x_\beta(t)]-x_\beta(t)$
hold?
Thank you for your help.
 A: This is long for a comment
Under the assumption that we can cancel, and multiply by inverses, let's simplify
$$ \color{red}{x_A(\phi_A)}x_\beta(\phi(\beta))x_B(\phi_B)x_\beta(t)-\color{red}{x_\beta(t)}\stackrel{?}{=}
\color{red}{x_A(\phi_A)}x_\beta(\phi(\beta)+t)[x_B(\phi_B),x_\beta(t)]-\color{red}{x_\beta(t)} $$
We get
$$ x_\beta(\phi(\beta))x_B(\phi_B)x_\beta(t)\stackrel{?}{=}
x_\beta(\phi(\beta)+t)[x_B(\phi_B),x_‌​{\beta} (t)] $$ The proof say by "By relation (3.6)", and you have $$ x_{\alpha}(a) x_{\alpha}(b) =
 x_{\alpha}(a+b) \tag{3.6}$$ 
Here is a question for you. According to (3.5), does we have
$$ x_B(\phi_B) x_{\beta}(t) = x_{\beta}(t)[x_B(\phi_B),x_\beta(t)]??$$
I'm guessing so. Then by (3.6) we have
$$ x_\beta(\phi(\beta))x_B(\phi_B)x_\beta(t) \stackrel{?}{=}
 \color{red}{x_\beta(\phi(\beta))x_\beta(t)} [x_B(\phi_B),x_\beta(t)]\\ \stackrel{?}{=}
 \color{red}{x_{\beta}(\phi(\beta)+t)} [x_B(\phi_B),x_\beta(t)]$$
So you get what you want.
But again.. warning: I only skimmed few pages of the paper and all of this are just guessing.
