Corollary 2

Let $(a_\sigma)_{\sigma \in \Gamma}$ be a family of elements of $N$. For $b$ to exist in $N$ such that $a_\sigma = \sigma(b) - b$ for all $\sigma \in \Gamma$ it is necessary and sufficient that $a_{\sigma \tau} = a_\sigma + \sigma(a_\tau)$ for $\sigma, \tau$ in $\Gamma$.

We have $\sigma \tau(b) - b = [\sigma(b) - b] + \sigma[\tau(b) - b]$ for all $b$ in $N$ and $\sigma, \tau$ in $\Gamma$, whence the necessity.

Conversely suppose that $a_{\sigma \tau} = a_\sigma + \sigma(a_\tau)$ for any $\sigma$ and $\tau$ in $\Gamma$. Put $U_\sigma = \begin{pmatrix} 1 & a_\sigma \\ 0 & 1 \end{pmatrix}$ for $\sigma \in \Gamma$; then we have $U_{\sigma \tau} = U_\sigma \cdot \sigma(U_\tau)$ for $\sigma, \tau$ in $\Gamma$; by Prop. 9, there exists thus a matrix $A = \begin{pmatrix} x & y \\ z & t \end{pmatrix}$ with non-zero determinant such that $\sigma(A) = A U_\sigma$ for all $a \in \Gamma$; writing down the relation $\sigma(A) = A U_\sigma$ we find

In particular, $x$ and $z$ belong to $K$ and we have If $x \neq 0$ we have $a, = \sigma(b) - b$ with $b = x^{-1} y$; if $z \neq 0$, we have the same relation with $b = z^{-1} t$. Now $x$ and $z$ cannot both be zero because