Iverson bracket help How can I show that if $G$ is a group and $x,y,m$ are in $G$ then,
$$[xy=m]=\sum_{r\in G}[x=r][y=r^{-1}m]$$

Where $[P]$ is the Iverson bracket.
Let $S$ be a set and suppose $x\in S$. 
Then 
$$\sum_{r\in S}[x=r]F(r) = F(x),$$
where $F(x)$ is any expression involving $x$. 
(Roughly, the Iverson bracket acts here like a delta function.) 

We have immediately that
$$\begin{eqnarray*}
[xy=m] &amp;=&amp; [y=x^{-1}m] \\
&amp;=&amp; \sum_{r\in G}[x=r][y=r^{-1}m].
\end{eqnarray*}$$
Assuming that the bracket on the left has an implied summation over all $x, y \in G$, then you are trying to show that for any fixed $m \in G$,
$$
\sum_{x,y\in G} [xy = m] = \sum_{r\in G} [x = r][y = r^{-1}m].
$$

In other words, you need to convince yourself that there's a bijection between the sets
$$
\{(x, y) \in G \times G \;|\; xy = m \}
\quad \longleftrightarrow \quad
\{r \in G \} = G.
$$ 

Once you identify $x = r$, then the condition on the left forces $y = x^{-1}m = r^{-1}m$.