Proof of Order of Galois Group equals Degree of Extension I am struggling to understand this proof that for a Galois extension, the order of the Galois group equals the degree of extension. Here's a summary:


*

*Let $f(x)$ be a polynomial in $F[x]$ with no repeated roots and suppose that $E$ is the splitting field for $f(x)$ over $F$.

*This result is trivial for $n=1$, so do induction on $n$. Assume the result holds true for all $f(x)$ such that $0 \leq k < n$.

*Let $p(x)$ be an irreducible factor of $f(x)$ with $r$ and one of the roots being $\alpha$ so that $F \subset F(\alpha) \subset E$. Thus, $[E : F(\alpha)]=n/r$ and $[F(\alpha) : F]=r$.

*From the induction hypothesis, $[E : F(\alpha)]=\lvert G(E/F(\alpha)) \rvert$.

*There are exactly $r$ isomorphisms from that maps $F(\alpha)$ to a $F(\beta)$ for some $\beta$ is some root of $p(x)$.

*Consequently, there are $[E : F]=[E : F(\alpha)][F(\alpha) : F]=n$ possible automorphisms of $E$ that fix $F$, or $\lvert G(E/F) \rvert=[E : F]$.


I'm having trouble understanding how 4 and 5 come together to imply 6. How are we supposed to combine an isomorphism from $F(\alpha)$ to $F(\beta)$ with an automorphism of $E$ that fixes $F(\alpha)$? If someone could please explain to me how these two automorphisms can be combined to get a new automorphism of $E$ that fixes $F$, that would be great. Thanks!
 A: First just to point out that this is when $E$ is a finite separable normal extension of $F$. At the steps you ask about $\alpha$ and $\beta$ have the same min poly over $F$ and so an earlier result shows that $F(\alpha)$ is isomorphic to $F(\beta)$ and there is an isomorphism $\psi_{j}$ say that acts as the identity on $F$ and maps $\alpha$ to $\beta$. But now the splitting field $E$ sits on top of both these fields and (essentially) using the result that proves splitting fields are isomorphic you can construct an isomorphism $\phi_{j}$ say from $E$ to $E$  such that its restriction to $F(\alpha )$ is $\psi_{j}$. Now there is a $\phi_{j}$ for each root and there are $r$ of these and $\phi_{j}$ is in Gal($E$/$F$). Let $\theta_{s}$ be the elements of Gal($E$/$F(\alpha))$ and so there are $n/r$ such $\theta_{s}$. Now look at the maps $\phi_{j}\theta_{s}$. They are distinct, there are $n$ of them and you show they exhaust Gal($E$/$F$).
Hope this helps  
A: This is a rather old post now, but I think it's worth it for me to fill in the gaps in this book's proof that the order of Galois group equals the degree of extension now that I have a better understanding of Galois theory.
Let $f(x)$ be a polynomial in $F[x]$ with no repeated roots and suppose that $E$ is the splitting field for $f(x)$ over $F$. Let $n=[E : F]$.
Clearly, if $[E : F]=1$, then $E=F$, so $G(E/F)=\{id\}$. Thus, $|G(E/F)|=[E : F]=1$.
Now, do induction on $n$: Assume that for all fields $F$, for any splitting field $E$ of some polynomial $f(x)\in F[x]$ such that $[E : F] < n$, $|G(E/F)|=[E : F]$.
At this point, I'm going to restate two theorems from the book below:

Theorem 21.32: Let $\phi: E \rightarrow F$ be an isomorphism of fields. Let $K$ be an extension field of $E$ and $\alpha \in K$ be algebraic over $E$ with minimal polynomial $p(x)$. Suppose that $L$ is an extension field of $F$ such that $\beta \in L$ is the root of the polynomial in $F[x]$ obtained from $p(x)$ under the image of $\phi$. Then, $\phi$ extends to a unique isomorphism $\bar \phi : E(\alpha) \rightarrow F(\beta)$ such that $\bar \phi(\alpha)=\beta$ and $\bar \phi$ agrees with $\phi$ on $E$.
Theorem 21.33: Let $\phi: E \rightarrow F$ be an isomorphism of fields and let $p(x)$ be a nonconstant polynomial in $E[x]$ and $q(x)$ be the corresponding polynomial under the isomorphism $\phi$. If $K$ is a splitting field of $p(x)$ and $L$ is a splitting field of $q(x)$, then $\phi$ extends to an isomorphism $\psi: K \rightarrow L$.

Let $p(x)$ be an irreducible factor of $f(x)$ and let $r$ be the degree of $p(x)$. Since $p(x)$ splits in $E$, there are $r$ roots of $p(x)$ in $E$. Label these roots $\alpha_1, \alpha_2, ..., \alpha_r$. By Theorem 21.32, for any $1 \leq i \leq r$, there is a unique isomorphism $\phi_i : F(\alpha_1) \rightarrow F(\alpha_i)$ such that $\phi_i$ fixes $F$ and $\phi_i(\alpha_1)=\alpha_i$. Moreover, by Theorem 21.33, there exists an isomorphism $\psi_i : E \rightarrow E$ such that $\psi_i$ agrees with $\phi_i$ on $F(\alpha_1)$.
Now, take some arbitrary automorphism $\sigma \in Gal(E/F)$. I will now restate another theorem from the book:

Proposition 23.5: Let $E$ be a field extension of $F$ and $f(x)$ be a polynomial in $F[x]$. Then, any automorphism in $G(E/F)$ defines a permutation of the roots of $f(x)$ which lie in $E$.

By Proposition 23.5, $\sigma(\alpha_1)=\alpha_i$ for some $1\leq i \leq r$. Moreover, $\sigma(F)=F$ since $\sigma$ must fix $F$ by the definition of $Gal(E/F)$. Therefore, $\sigma(F(\alpha_1))$ contains both $F$ and $\alpha_i$, so $\sigma(F(\alpha_1)) \subset F(\alpha_i)$.
Now, consider $\sigma^{-1}(F(\alpha_i))$. Clearly, $\sigma^{-1}(F)=F$, since $\sigma$ fixes $F$. Also, because $\sigma(\alpha_1)=\alpha_i$, $\sigma^{-1}(\alpha_i)=\alpha_1$. Thus, $\sigma^{-1}(F(\alpha_i))$ contains both $F$ and $\alpha_1$, so $\sigma^{-1}(F(\alpha_i)) \subset F(\alpha_1)$. In other words, $F(\alpha_i)\subset \sigma(F(\alpha_1))$. This suffices to show that $\sigma(F(\alpha_1))=F(\alpha_i)$.
Thus, $\sigma$ restricted to $F(\alpha_1)$ is an isomorphism from $F(\alpha_1)$ to $F(\alpha_i)$ such that $\sigma$ fixes $F$ and $\sigma(\alpha_1)=\alpha_i$. However, by Theorem 21.32, there is only one isomorphism which satisfies these conditions, and $\phi_i$, as defined above, is also an isomorphism such that $\phi_i$ fixes $F$ and $\phi_i(\alpha_1)=\alpha_i$. This means that $\sigma$ must agree with $\phi_i$ on $F(\alpha_1)$.
Now, consider $\psi_i^{-1}\sigma$. As shown above, both $\sigma$ and $\psi_i$ agree with $\phi_i$ on $F(\alpha_1)$. Thus, for any $x\in F(\alpha_1)$:
$$\psi_i^{-1}\sigma(x)=\psi_i^{-1}(\phi_i(x))=\phi_i^{-1}(\phi_i(x))=x$$
Thus, $\psi_i^{-1}\sigma$ is an automorphism of $E$ which fixes $F(\alpha_1)$. In other words, $\psi_i^{-1}\sigma$ is a member of $G(E/F(\alpha_1))$.
We now restate another theorem:

Theorem 21.17: If $E$ is a finite extension of $F$ and $K$ is a finite extension of $E$, then $K$ is a finite extension of $F$ and $[K : F]=[K : E][E : F]$.

Thus:
$$[E : F]=[E : F(\alpha_1)][F(\alpha_1) : F]\implies [E : F(\alpha_1)]=\frac{[E : F]}{[F(\alpha_1) : F]}=\frac n r$$
Thus, $[E : F(\alpha_1)] < n$, so the induction hypothesis applies: $|G(E/F(\alpha_1))|=[E : F(\alpha_1)]=n/r$. We can thus label the elements of $|G(E/F(\alpha_1))|$ as $\theta_1, \theta_2, ..., \theta_{n/r}$.
As discussed earlier, $\psi_i^{-1}\sigma$ is a member of $G(E/F(\alpha_1))$. Thus, $\psi_i^{-1}\sigma=\theta_j$ for some $1 \leq j \leq n/r$. This means that, for any $\sigma \in G(E/F)$, $\sigma=\psi_i\theta_j$ for some $1\leq i\leq r$ and some $1 \leq j \leq n/r$. Since there are $r$ possibilities for $\psi_i$ and $n/r$ possibilities for $\theta_j$, there are $r\cdot(n/r)=n$ possibilities for $\sigma$. Therefore, $|G(E/F)|=n=[E : F]$. Q.E.D.
Now, here's a fun exercise which is relevant to the end of the proof:

Let $1 \leq i,i_2 \leq r$ and $1 \leq j,j_2 \leq n/r$. Prove that if $\psi_i\theta_j=\psi_{i_2}\theta_{j_2}$, then $i=i_2$ and $j=j_2$.

