Corollary 3

For i = 1, 2 let $E_i$ be a subfield of N containing K and let $\Delta_i = g(E_i)$. For any $\sigma \in \Gamma$ the relations $\sigma(E_1) = E_2$ and $\sigma \Delta_1 \sigma^{-1} = \Delta_2$ are equivalent.

For we have $\tau \in g(\sigma(E_1))$ if and only if $\tau \sigma(x) = \sigma(x)$, that is, $\sigma^{-1} \tau \sigma(x) = x$, for all $x \in E_1$; this amounts to saying 'that $\sigma^{-1} \tau \sigma \in \Delta_1$, whence $g(\sigma(E_1)) = \sigma \Delta_1 \sigma^{-1}$.