Frobenius kernel is regular normal elementary abelian p-subgroup?

I'm attempting Exercise 3.4.6 in Dixon & Mortimer's book on Permutation Groups:

Let $G$ be a finite primitive permutation group with abelian point stabilisers. Show that $G$ has a regular normal elementary abelian $p$-subgroup for some prime $p$.

I know that if $G$ is a primitive permutation group with abelian point stabilisers and $G$ does not have prime order, then it is a Frobenius Group. Is it also a Frobenius group if $G$ has finite order?

If so, I know that the Frobenius kernel $K$ is a normal regular subgroup from the Structure Theorem for Finite Frobenius Groups. I'm not sure how to go about proving that $K$ is also an elementary abelian $p$-subgroup though.

From the Frattini Argument I know that if $P$ is a Sylow $p$-subgroup of $K$ then the direct product of $K$ and the normaliser of $P$ is equal to $G$. But if $G$ is Frobenius it is also equal to the semi-direct product of the Frobenius kernel and complement - perhaps I can use this to show that $K$ itself is a $p$-subgroup? And therefore an elementary group since a direct product of a $p$-subgroup and the identity (which is finite and cyclic or order $1$).

I'm a bit stuck as to where to start on proving that $K$ is abelian, and this is all assuming that $G$ is a Frobenius group and $K$ is the subgroup in question!

Any help would be much appreciated. Thanks!

• Thank you @DerekHolt. I understand that the point stabilizers can't fix more than one point, but how do we know that the point stabilizers aren't all trivial? If G is regular then it can't be Frobenius... – M.Hope Apr 16 '15 at 14:24

You can reduce to the case when $G$ is Frobenius as follows. In a primitive permutation group which is not cyclic of prime order, the point stabilizers must fix a unique point, or else the sets of points fixed by the point stabilizers would form a system of imprimitivity for $G$.
Suppose that $G$ is not Frobenius and let $g \in G$ fix more than one point. Then, since point stabilizers are abelian, the centralizer of $G$ contains the full point stabilizers of each of the fixed points of $t$, which are all distinct, so $C_G(t)$ properly contains a point stabilizer, contradicting the result that these are maximal subgroup of $G$.