Irreducible solvable equation of prime degree Theorem: if $p(x)=0$ is an irreducible solvable equation of prime degree (say, $q$) then its Galois group can be embedded in a group of order $q(q-1)$, which is isomorphic to semidirect product of $С_q$ and $C_{q-1}$. Where can i read about the proof this theorem? Is it complicated?
 A: Let $G$ be a solvable transitive subgroup of $S_q$ with $q$ prime. So $G$ has a nontrivial abelian normal subgroup $N$. If $O$ is an orbit of the action of $N$, then so is $gO$ for any $g \in G$, and $|gO|=|O|$, so transitivity of $G$ implies that all orbits of $N$ have the same length. (This is true for any normal subgroup of any transitive permutation group.) Now $q$ prime implies that there is a unique orbit, so $N$ is transitive.
Now the centralizer of any permutation permutes the fixed points of that permutation. So, since $N$ is transitive and abelian, no non-identity element of $N$ can have any fixed points, so in fact $N$ acts regularly. Hence $|N|=q$, and $N = \langle g \rangle$, where $g$ is a $q$-cycle.
Since $G$ normalizes $N$, we need to prove that the normalizer $N_{S_q}(N)$ in $S_q$ of $N$ has order $q(q-1)$. From  the centralizer arguments in the previous paragraph, we see that $C_{S_q}(N)=N$, so $|N_{S_q}(N)| \le |N||{\rm Aut}(N)| = q(q-1)$. On the other hand the permutation representation of the semidirect product $N \rtimes {\rm Aut}(N)$ on the cosets of ${\rm Aut}(N)$ embeds this semidirect product into $S_q$, so we have equality. (In fact the normalizer is sharply 2-transitive.)
