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?

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Could you make the question self-contained? –  Did Mar 16 '12 at 19:32

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.
Hi J.D. Thank you for your answer. I thought in the same way, but unfortunately I did not see how to prove $x_B(\phi_B)x_\beta(t)=x_\beta(t)[x_B(\phi_B),x_\beta(t)]$. Is there a simple trick that I did overlook? –  Bernhard Boehmler Mar 17 '12 at 7:47
I'm a bit confused with notation and can't understand $(3.5)$. What does this notation $[\cdot, \cdot]$ (e.g. $[x_B(\phi_B),x_\beta(t)]$) really mean? –  user2468 Mar 17 '12 at 15:39
I think $[\cdot,\cdot]$ should be the group commutator, i.e. $[x_\alpha(a),x_\beta(b)]={x_\alpha(a)}{x_\beta(b)}{x_\alpha(a)}^{-1}{x_\beta(b)‌​}^{-1}$, but this is exactly the thing that made me wonder why the equality $x_B(\phi_B)x_\beta(t)=x_\beta(t)[x_B(\phi_B),x_\beta(t)]$ should hold. –  Bernhard Boehmler Mar 17 '12 at 17:37