# Normal subgroups of k-transitive groups.

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?

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.