Let permutation group $G$ contains a minimal normal subgroup $\neq 1$ which is transitive and Abelian. Show that $G$ is primitive.
My attempts:
Because of Proposition 4.4. of Wielandt's book $N_\alpha = 1$ and $C_{S^\Omega}(N) = N$, hence $C_G(N) = N$.
For every normal subgroup $1 \neq L$ of $G$, since $ L \cap N \trianglelefteq G$ and $N$ is minimal normal subgroup of $G$, so $N \subseteq L$ or $ L \cap N = 1 $. But if $ L \cap N = 1 $, so $ L \subseteq C_G(N) = N $, a contradiction! So $N \subseteq L$. Say, $N$ is a subgroup of every non-trivial normal subgroup of $G$.
For every maximal subgroup $M$ containing $G_\alpha$ properly, It's easy to see that $\alpha^M$ is a non-trivial block. $$ |M| / |G_\alpha| = |M| / |M_\alpha| = |\alpha^M| < |\Omega| = |N| $$ and since $|N \cap G_\alpha| = 1$, $$ |M| < |N| |G_\alpha| = |N G_\alpha| $$ So $N G_\alpha = G$.
If we show that $G_\alpha$ is maximal subgroup of $G$ then we done!
Note that $N \cap M$ is normal in both $M$ and in $N$ (since $N$ is abelian). So it is normal in $MN$, and since $G_\alpha < M$ and $G=NG_\alpha$, we have $N \cap M \unlhd G$.
If $G_\alpha < M < G$, then this contradicts $N$ being a minimal normal subgroup of $G$. This is because $N \cap M = 1 \Rightarrow M=G_\alpha$ and $N \le M \Rightarrow M=G$.