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.
 A: 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
$$\begin{eqnarray*}
[xy=m] &=& [y=x^{-1}m] \\
&=& \sum_{r\in G}[x=r][y=r^{-1}m].
\end{eqnarray*}$$
A: Fix $x,y,m\in G$. The lefthand side, $[xy=m]$, is $1$ if $xy=m$ and $0$ otherwise. Suppose first that $xy=m$, so that the lefthand side is $1$. 
Now consider the righthand side. As $r$ ranges over $G$, one of the terms will be the $r=x$ term, since $x\in G$. For that term $[x=r]=1$, so we have
$$[x=r][y=r^{-1}m]=[y=r^{-1}m]=[y=x^{-1}m]\;.$$
But $y=x^{-1}m$ if and only if $xy=xx^{-1}m=m$, which is in fact the case, so this term of the sum is $1$. And of course all the other terms are $0$, because if $r\ne x$, then $[x=r]=0$, and the other factor doesn’t matter. Thus, the formula is correct when $xy$ is equal to $m$.
Now suppose that $xy\ne m$, so that $[xy=m]=0$. Again the only possible non-zero term in the summation is the $x=r$ term, which is
$$[x=r][y=r^{-1}m]=[y=x^{-1}m]\;.\tag{1}$$
But $xy\ne m$, so $y\ne x^{-1}m$, and the term $(1)$ is equal to $0$. In this case, therefore, every term of the summation is zero, and so is the sum — which is exactly what we want.
A: Note that $\lbrack r = x \rbrack\lbrack y = r^{-1}m\rbrack = 0$ for $r \neq x$. So this equation simplifies to $\lbrack xy = m\rbrack = \lbrack x=x\rbrack\lbrack y = x^{-1}m \rbrack$ or just
$$\lbrack xy = m\rbrack = \lbrack y = x^{-1}m \rbrack,$$
which is true by the definition of $x^{-1}$.
A: 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 $y = x^{-1}m = r^{-1}m$.
