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?
Best Answer
It's because the automorphism group of an elementary abelian $2$-group $N$ of order $2^n$ with $n\ge 3$ acts 2-transitively but not 3-transitively on its non-identity elements. (When $n=2$, it does act 3-transitively, and that corresponds to the case when $k=4$ and $G=S_4$, which is excluded by the theorem statement.)
To see that, think of $N$ as a vector space over the field of order 2 (and use additive notation in $N$). Then any two distinct nonzero vectors are linearly independent, and so there is a linear map (i.e. an automorphism of the group $N$) that maps them to any other two such vectors.
However, when $n\ge 2$, if we let $x,y,z$ be linearly independent elements of $N$ then there is no linear map (i.e. no group automorphism of $N$) that maps $x \to x$, $y \to y$, and $z \to x+y$, so the automorphism group does not act 3-transitively. Hence $G$ is not 4-transitive.