# Main theory

let $\lambda(\mathbf E)$ be the subgroup $G(\mathbf {K/F})$ leaving fixed $\mathbf E$.

Let $\mathbf{K}$ be a normal extension (splitting and seperable) of $\mathbf F$, where $\mathbf {F \le E \le K}$.

1. $\lambda (\mathbf E)\,=\,\mathbf {\mathrm G(K/E)}$

2. $\mathbf {E\,=\,K_{\mathrm G(K/E)}}$

3. $\mathbf{\mathrm H \le \mathrm G(K/F)},\quad\lambda\mathbf{(E_{\mathrm H})=\mathrm H}$

($\mathbf{E_H}$ is a field of a set of all element fixed by $\mathrm{H}$.)

4. $\mathbf{[K:E]=|\lambda(E)|,\quad [E:F]=\mathrm G(K/F):\lambda(E)}$

($\mathbf{\mathrm G(K/F):\lambda(E)}$ is means a order of $\mathbf{\mathrm G(K/F)\,/\,\lambda(E)}$ )

5. $\mathbf{\mathrm G(E/F)\simeq \mathrm G(K/F)\,/\,\mathrm G(K/E)}$

6. above the diagrams are equavalent.

## Proof

### 1. $\lambda (\mathbf E)\,=\,\mathbf {\mathrm G(K/E)}$

this is just the definition of $\lambda$.

### 2. $\mathbf {E\,=\,K_{\mathrm G(K/E)}}$

by the definition of $\mathbf{K_{\mathrm G(K/E)}}$, $E$ is fixed in $K$. this show that $\mathbf{E \leq K_{\mathrm G(K/E)}}$.

we prove $\mathbf{K_{\mathrm G(K/E)}\leq E}$.
Let $\alpha \in \mathbf K, \text{where } \alpha \notin \mathbf E$. Since $\mathbf K$ is a normal extension of $\mathbf E$, we can find an automoriphsim of $\mathbf K$ leaving $\mathbf E$ fixed and mapping $\alpha$ onto different a zero of $irr(\alpha, \mathbf F)$. therefore, at least there exists a element in $G(\mathbf {K/E})$ that move $\alpha$. this implies that $\mathbf{K_{\mathrm G(K/E)}\leq E}$,
so $\mathbf {E\,=\,K_{\mathrm G(K/E)}}$.

### 3. $\mathbf{\mathrm H \le \mathrm G(K/F)},\quad\lambda\mathbf{(E_{\mathrm H})=\mathrm H}$

we think about $\mathbf{\mathrm H,\,\lambda(K_{\mathrm H}),\,G(K/F)}$. a relations of them is $\mathbf{\mathrm H \leq \lambda (K_{\mathrm H}) \leq G(K/F)}$. we shall suppose that

$H < \lambda(\mathbf K_H)$

and shall derive contradiction. As a finite seperable extension, $\mathbf K = \mathbf K_{H}(\alpha)$ for some $\alpha \in K$, Let

$n=\mathbf{[K : K_{\mathrm H}]=\{K : K_{\mathrm H}\}=|\mathrm G(K/K_{\mathrm H})|}$

then $|H| < |G(\mathbf K / \mathbf K_H)|=n$. Let the elements of $H$ be $\sigma_1, \ldots,\, \sigma_{|H|}$. and consider the polynomial

$f(x) = \prod_{i=1}^{|H|}(x-\sigma_i(\alpha))\text .$

this function is of degree $|H|< n$ and symmetric expressions in the $\sigma_i$. because H being a group, then $\sigma\sigma_1,\,\ldots,\, \sigma\sigma_{|H|} \in H$, these cofficeients are invariant under each isomporiphism $\sigma_i$. Hence $f(x)$ has coefficients in $K_H$. and some $\sigma_i$ is $1$, so $f(\alpha)=0$. Therefore, we have

$deg(\alpha, \mathbf K_H) = |H| < n = \mathbf {[K : K_{\mathrm H}]=[K_{\mathrm H}(\alpha) : K_{\mathrm H}]}$

### 4.$\mathbf{[K:E]=|\lambda(E)|,\quad [E:F]=\mathrm G(K/F):\lambda(E)}$

since $\mathbf K$ is a normal extension of $\mathbf E$, $\mathbf{[K:E]=\{K:E\}}$. and the defintion is $|\lambda(E)|= \{\mathbf{K:E}\}$. therefore, $\mathbf{[K:E]=|\lambda(E)|}$.

we prove $\mathbf{[E:F]=\mathrm G(K/F):\lambda(E)}$. Let $\sigma$ and $\tau$ are int the same left coset of $\lambda(\mathbf E)$. Let $\sigma = h\alpha_1, \tau=h\alpha_2 \text{ where } \alpha_1, \alpha_2 \in E$. $\sigma^{-1}\tau = \alpha_1^{-1}h^{-1}h\alpha_2=\alpha_1^{-1}\alpha_2 \in G(\mathbf {K/E})$. for $\alpha \in \mathbf E$, $\sigma^{-1}\tau(\alpha)=\alpha$ and then $\sigma(\alpha)=\tau(\alpha)$. Hence, the same left coset of $\lambda(\mathbf E)$ induce the same element in the $G(\mathbf{E/F})$ so $|G(\mathbf{E/F})|=\mathbf{\mathrm G(K/F):\lambda(E)}$.
therfore, $\mathbf{[E:F]=\{E:F\}=|G(\mathbf{E/F})|=\mathbf{\mathrm G(K/F):\lambda(E)}}$

### 5. $\mathbf{\mathrm G(E/F)\simeq \mathrm G(K/F)\,/\,\mathrm G(K/E)}$

we prove $\mathbf E$ is a normal extension of $\mathbf F$. every extension $\mathbf E$ of $\mathbf F$, $\mathbf{E \leq E \leq K}$, is separable over $\mathbf F$. $\mathbf E$ is separable over $\mathbf F$.[1] every isomorphism of $\mathbf E$ onto a subfield of $\overline \mathbf F$ leaving $\mathbf F$ fixed can be extended to an automorphism of $\mathbf K$, since $\mathbf K$ is normal over $\mathbf F$. thus the automorphisms of $\mathbf {G(K/F)}$ induce all possible isomorphisms of $\mathbf E$ onto a subfield of $\overline\mathbf F$ leaving $\mathbf F$ fixed. so then $\mathbf E$ is a splitting filed over $\mathbf F$.[2] and hence is nomrmal over $\mathbf F$.

by property 2, $\mathbf E$ is the fixed filed of $\mathbf{G(K/E)}$, so for $\sigma \in \mathbf{G(K/F)}$, $\sigma(\alpha)\in \mathbf E$ if and only if for all $\tau \in \mathbf{G(K/E)}$, $\tau(\sigma(\alpha))=\sigma(\alpha)$. so $\sigma^{-1}\tau\sigma(\alpha)=\alpha$. this means $\sigma \in \mathbf{G(K/F)}$ and $\tau \in \mathbf{G(K/E)}$, that is

$\sigma^{-1}\tau\sigma \in \mathbf{G(K/E)}\text .$

this is the condition that $\mathbf{G(K/E)}$ be a normal subgroup of $\mathbf{G(K/E)}$

when $\mathbf E$ is a normal extension of $\mathbf F$, there is a one to one correpsondence between left coset $\mathbf{G(K/E)}$ over $\mathbf{G(K/F)}$ and automorphism of $\mathbf E$. for $\sigma \in G(\mathbf {K/F})$, let $\sigma_E$ be the automorphism of $\mathbf E$ induced by $\sigma$. thus $\sigma_E \in \mathbf{G(E/F)}$. the map $\theta:\mathbf{G(K/F) \rightarrow G(K/F)}$ given by

$\theta(\sigma)=\sigma_E$

for $\sigma \in \mathbf{G(K/F)}$ is a homomorphism. and $\theta$ is onto $\mathbf{G(E/F)}$. the kernel of $\theta$ is $\mathbf{G(K/E)}$. by the fundamental isomorphism theorem, $\mathbf{\mathrm G(E/F)\simeq \mathrm G(K/F)\,/\,\mathrm G(K/E)}$

1. A First Course In Abstract Algebra-Jb Fraleigh, 7Ed(2003), th 51.9, p438 ↩︎

2. A First Course In Abstract Algebra-Jb Fraleigh, 7Ed(2003), th 50.3, p432 ↩︎