Let $G$ be a group which acts primitively on $X$. When $G$ is solvable and $|X|=n$, then $n$ is a prime power.
Hello,
I have a question to this task. As a hint is given, to show, that every minimal normal subgroup is elemantary abelian. Therefore abelian and every non-trivial element has order $p$. Then use:
Let $G$ be a group which acts transitively on $X$. If the action is primitive and $N\unlhd G$, then $N$ is transitive or in the kernel of the action.
Showing that the minimal subgroup is abelian is easy:
Since $G$ is solvable, $N$ is solvable. Since $N$ is minimal $N/\{e\}\cong N$ is abelian.
Now I want to show, that every non-trivial element has order $p$ (where $p$ is prime of course). But I am stuck.
My try was to take $N\ni x\neq e$ and observe $\langle x\rangle$. But since $N$ is minimal it has to be $\langle x\rangle=\{e\}$.
I wanted to use the Lagrange theorem.
Can you give me a hint? Thanks in advance.