I am not understanding a proof in Cameron's Permutation Groups.
Theorem 4.1. let G be k-transitive, but not $S_k $, with k>1. Then a non-trivial normal subgroup N of G is (k-1)-transitive, except possibly when k=3, when N may be an elementary abelian 2-group.
The proof given in the book is by induction on k. The base case of k=2 is clear: 2-transitive groups are primitive, and primitive groups' normal subgroups are transitive. Now we assume the result holds for k-1 and endeavour to prove it for k. Choose some $\alpha \in \Omega$ (where $\Omega$ is the set G acts on). Then the stabiliser $N_\alpha$ of $\alpha$ in N is a normal subgroup of the (k-1)-transitive stabiliser $G_\alpha$ of $\alpha$ in G (by second isom. thm.).
Now the bit I struggle with: by the induction hypothesis, one of three possibilities occurs:
1) $N_\alpha$=1. Then N is regular, so N is an elementary abelian 2-group, and G is not 4-transitive.
I won't list the other two possibilities, because I just want an explanation for why G is not 4-transitive. A previous theorem showed that if G is k-transitive for $k\ge 3$ then a regular normal subgroup is an elementary abelian 2-group. But that theorem says nothing about G not being 4-transitive, and the proof of the theorem in question implies that the previous theorem shows this.
Why is G not 4-transitive?