Proposition 9

— Let $(U_\sigma)_{\sigma \in \Gamma}$ be a family of elements of $\mathbf{GL}(n, N)$. For $A$ to exist in $GL(n, N)$ such that $U_\sigma = A^{-1}\,a(A)$ for all $\sigma \in \Gamma$ it is necessary and sufficient that $U_{\sigma \tau} = U_\sigma \cdot \sigma(U_\tau)$ for $\sigma,\,\tau$ in $\Gamma$.

The condition is necessary: if $U_\sigma = A^{-1}\,a(A)$, then we have

The condition is sufficient: we identify the elements of $N^n$ with matrices of $n$ rows and one column with coefficients in $N$. We let the groups $\Gamma$ act on $N^n$ by

For each $a \in \Gamma$ we denote by $u_\sigma$ the mapping $x \mapsto U_\sigma \cdot \sigma(x)$ of $N^n$ into itself. The verification of Formulae (1) to (3) of V, p. 62 is immediate. Moreover we have $u_\varepsilon \circ u_\varepsilon = u_\varepsilon$ and since $u_\varepsilon$ is bijective, we have $u_\varepsilon = \mathrm{Id}_{N^n}$. Let $V_0$ be the set of vectors $x \in N^n$ such that $u_\sigma(x) = x$ for all $\sigma \in \Gamma$. By Prop. 7 (V, p. 63), $V_0$ is a K-structure on $N^n$; in particular there exist in $V_0$ vectors $b_1, \ldots, b_n$ forming a basis of $N^n$ over $N$. The matrix $B$ with columns $b_1, \ldots, b_n$ is therefore invertible and the relation $u_\sigma(b_i) = b_i$ for $1 \leq i \leq n$ is equivalent to $U_\sigma \cdot \sigma(B) = B$. Writing $A = B^{-1}$, we obtain $U_\sigma = A^{-1} \sigma(A)$ for all $a \in \Gamma$.