Understanding the step in the proof about additive energy I'm reading a paper on arXiv on additive combinatorics, and I have trouble understanding a step in the proof on page 16. Suppose $\Gamma \subseteq F^\times_p $ is a multiplicative subgroup of integers modulo prime $p$.
Define quantities
$$ r_{k\Gamma}(x) = |\{g_1, ..., g_k \in \Gamma: g_1+...+g_k = x\}| $$
(i.e. the number of possible representations of an element $a$ as a sum of $k$ elements of $\Gamma$), and
$$ T_{2k} (\Gamma) = |\{g_1, g_2, ..., g_{4k} \in \Gamma: g_1 + ...+g_{2k} = g_{2k+1} + ... +g_{4k} \}| = \\= \sum_{x,y,z} r_{k\Gamma}(x)r_{k\Gamma}(y)r_{k\Gamma}(x+z)r_{k\Gamma}(y+z)$$
The author then says that from the last equation it follows that
$$T_{2s}(\Gamma) = \frac{1}{|\Gamma| ^ 2} \sum_{\gamma_1, \gamma_2 \in \Gamma} \sum_{\substack{a,b,c,d\\a+\gamma_1 b = c+\gamma_2 d}} r_{k\Gamma}(a)r_{k\Gamma}(b)r_{k\Gamma}(c)r_{k\Gamma}(d) + E $$
where $a,b,c,d$ are nonzero and $E$ corresponds to the term where at least one of $a,b,c,d$ is zero, but I don't understand why. I take it that somehow follows from
$$ T_{2s}(\Gamma) = \sum_{\substack{a,b,c,d\\a+ b = c+ d}} r_{k\Gamma}(a)r_{k\Gamma}(b)r_{k\Gamma}(c)r_{k\Gamma}(d) + E$$
but I'm not able to work out the exact steps the author took to obtain it.
Any and all help is appreciated.
 A: For any set $\Gamma\subset \mathbb{F}_p^\times$ and function $f:\mathbb{F}_p^\times \to \mathbb{R}$, we have
$$\sum_{x\in \mathbb{F}_p^\times}f(x) = \frac{1}{\lvert \Gamma\rvert}\sum_{\gamma\in \Gamma}\sum_{y\in \mathbb{F}_p^\times}f(\gamma y),$$
since as $x$ ranges over $\mathbb{F}_p^\times$ so does $\gamma y$ for any fixed $\gamma\neq 0$. 
In particular, if $r(x) = r_{k\Gamma}(x)$ then (using the above identity twice, once on the $b$ variable then again on the $d$ variable)
$$\sum_{\substack{a,b,c,d\\ a+b=c+d\\ abcd\neq 0}}r(a)r(b)r(c)r(d)= \frac{1}{\lvert \Gamma\rvert^2}\sum_{\gamma_1,\gamma_2\in \Gamma}\sum_{\substack{a,b,c,d\\ a+\gamma_1b=c+\gamma_2d\\ abcd\neq 0}}r(a)r(\gamma_1b)r(c)r(\gamma_2d).$$
The identity you're after comes from the fact that if $\gamma\in \Gamma$ then, for any $x$,
$$ r_{k\Gamma}(\gamma x) = r_{k\Gamma}(x).$$
This is true because the left-hand side counts $(\gamma_1,\ldots,\gamma_k)\in \Gamma^k$ such that $\gamma_1+\cdots+\gamma_k=x$, and the right-hand side counts (after dilating by $\gamma^{-1}$) $(\gamma_1,\ldots,\gamma_k)\in (\gamma^{-1}\Gamma)^k$ such that $\gamma_1+\cdots+\gamma_k=x$. But since $\Gamma$ is a multiplicative subgroup, $\gamma^{-1}\Gamma=\Gamma$, and hence both sides are counting the same thing.
