(1) Suppose that
\(G(E/K) = G(E/L) = G\text{.}\) Both
\(K\) and
\(L\) are fixed fields of
\(G\text{;}\) hence,
\(K=L\) and the map defined by
\(K \mapsto G(E/K)\) is one-to-one. To show that the map is onto, let
\(G\) be a subgroup of
\(G(E/F)\) and
\(K\) be the field fixed by
\(G\text{.}\) Then
\(F \subset K \subset E\text{;}\) consequently,
\(E\) is a normal extension of
\(K\text{.}\) Thus,
\(G(E/K) = G\) and the map
\(K \mapsto G(E/K)\) is a bijection.
(2) By Theorem
Theorem 23.7,
\(|G(E/K)| = [E:K]\text{;}\) therefore,
\begin{equation*}
|G(E/F)| = [G(E/F):G(E/K)] \cdot |G(E/K)| = [E:F] = [E:K][K:F]\text{.}
\end{equation*}
Thus, \([K:F] = [G(E/F):G(E/K)]\text{.}\)
Statement (3) is illustrated in
Figure 23.24. We leave the proof of this property as an exercise.
(4) This part takes a little more work. Let
\(K\) be a normal extension of
\(F\text{.}\) If
\(\sigma\) is in
\(G(E/F)\) and
\(\tau\) is in
\(G(E/K)\text{,}\) we need to show that
\(\sigma^{-1} \tau \sigma\) is in
\(G(E/K)\text{;}\) that is, we need to show that
\(\sigma^{-1} \tau \sigma( \alpha) = \alpha\) for all
\(\alpha \in K\text{.}\) Suppose that
\(f(x)\) is the minimal polynomial of
\(\alpha\) over
\(F\text{.}\) Then
\(\sigma( \alpha )\) is also a root of
\(f(x)\) lying in
\(K\text{,}\) since
\(K\) is a normal extension of
\(F\text{.}\) Hence,
\(\tau( \sigma( \alpha )) = \sigma( \alpha )\) or
\(\sigma^{-1} \tau \sigma( \alpha) = \alpha\text{.}\)
Conversely, let \(G(E/K)\) be a normal subgroup of \(G(E/F)\text{.}\) We need to show that \(F = K_{G(K/F)}\text{.}\) Let \(\tau \in G(E/K)\text{.}\) For all \(\sigma \in G(E/F)\) there exists a \(\overline{\tau} \in G(E/K)\) such that \(\tau \sigma = \sigma \overline{\tau}\text{.}\) Consequently, for all \(\alpha \in K\)
\begin{equation*}
\tau( \sigma( \alpha ) ) = \sigma( \overline{\tau}( \alpha ) ) = \sigma( \alpha );
\end{equation*}
hence, \(\sigma( \alpha )\) must be in the fixed field of \(G(E/K)\text{.}\) Let \(\overline{\sigma}\) be the restriction of \(\sigma\) to \(K\text{.}\) Then \(\overline{\sigma}\) is an automorphism of \(K\) fixing \(F\text{,}\) since \(\sigma( \alpha ) \in K\) for all \(\alpha \in K\text{;}\) hence, \(\overline{\sigma} \in G(K/F)\text{.}\) Next, we will show that the fixed field of \(G(K/F)\) is \(F\text{.}\) Let \(\beta\) be an element in \(K\) that is fixed by all automorphisms in \(G(K/F)\text{.}\) In particular, \(\overline{\sigma}(\beta) = \beta\) for all \(\sigma \in G(E/F)\text{.}\) Therefore, \(\beta\) belongs to the fixed field \(F\) of \(G(E/F)\text{.}\)
Finally, we must show that when \(K\) is a normal extension of \(F\text{,}\)
\begin{equation*}
G(K/F) \cong G(E/F) / G(E/K)\text{.}
\end{equation*}
For \(\sigma \in G(E/F)\text{,}\) let \(\sigma_K\) be the automorphism of \(K\) obtained by restricting \(\sigma\) to \(K\text{.}\) Since \(K\) is a normal extension, the argument in the preceding paragraph shows that \(\sigma_K \in G( K/F)\text{.}\) Consequently, we have a map \(\phi:G(E/F) \rightarrow G(K/F)\) defined by \(\sigma \mapsto \sigma_K\text{.}\) This map is a group homomorphism since
\begin{equation*}
\phi( \sigma \tau ) = (\sigma \tau)_K = \sigma_K \tau_K = \phi( \sigma) \phi( \tau )\text{.}
\end{equation*}
The kernel of \(\phi\) is \(G(E/K)\text{.}\) By (2),
\begin{equation*}
|G(E/F)| / |G(E/K)| = [K:F] = |G(K/F)|\text{.}
\end{equation*}
Hence, the image of \(\phi\) is \(G(K/F)\) and \(\phi\) is onto. Applying the First Isomorphism Theorem, we have
\begin{equation*}
G(K/F) \cong G(E/F) / G( E/K )\text{.}
\end{equation*}
