2-regular group action implies elementary Abelian regular normal subgroup

Statement
Suppose $$G$$ is a finite group with a 2-regular group action on a set $$S$$. Then, there exists a normal subgroup $$N$$ of $$G$$ such that the induced action of $$N$$ on $$S$$ is a regular group action, and a permutable complement $$M$$ to $$N$$ in $$G$$. More precisely, $$G$$ is a Frobenius group with Frobenius kernel $$N$$ and a Frobenius complement $$M$$ whose action by $$N$$ is a regular group action on the non-identity elements of $$N$$.

Facts used

 * 1) uses::Frobenius' theorem
 * 2) uses::Finite and automorphism group is transitive on non-identity elements implies elementary Abelian

Proof
Given: A finite group $$G$$ with a 2-regular group action on a set $$S$$.

To prove: $$G$$ has an elementary Abelian regular normal subgroup $$N$$ with a permutable complement $$M$$ whose induced action on $$M$$ by conjugation is a regular group action on the non-identity elements. Further, if $$S$$ has order $$n$$, then $$n = p^k$$ for some prime $$p$$ and positive integer $$k$$, and the order of $$G$$ is $$p^k(p^k - 1)$$.

Proof: Let $$M$$ be the isotropy subgroup of any element $$s$$ of $$S$$. The conjugate subgroups of $$M$$ in $$G$$ are precisely the isotropy subgroups of different elements of $$S$$.


 * 1) For any $$g \notin M$$, $$M \cap gMg^{-1}$$ is trivial: Since $$g \notin M$$, we have $$g \cdot s \ne s$$. Consider the ordered pair $$(s, g \cdot s)$$. Since the action of $$G$$ on $$S$$ is 2-regular, we know that the only element of $$G$$ that fix the ordered pair $$(s, g \cdot s)$$ is the identity element. Since $$gMg^{-1}$$ and $$M$$ are the isotropy subgroups of $$g \cdot s$$ and $$s$$ respectively, their intersection must be trivial.
 * 2) $$M$$ is a Frobenius subgroup of $$G$$: This follows directly from step (1).
 * 3) The complement of the union of conjugates of $$M$$ in $$G$$ forms then non-identity elements of a normal subgroup $$N$$ of $$G$$: This follows from fact (1).
 * 4) The induced action of $$N$$ on $$S$$ is semiregular: Note that $$N$$ intersects every conjugate of $$M$$ trivially, so $$N$$ acts semiregularly on $$S$$.
 * 5) The induced action of $$N$$ on $$G$$ is regular: Suppose $$S$$ has size $$n$$. Then, $$G$$ has size $$n(n-1)$$, and each of the subgroups $$M$$ has size $$n-1$$. There are $$n$$ of these, each with the non-identity elements distinct. Thus, we get a total of $$n(n-2)$$ non-identity elements in the union of conjugates of $$M$$, leaving $$n$$ elements for $$N$$. Thus, $$N$$ acts semiregularly on $$S$$ and has the same size as $$S$$, so $$N$$ acts regularly on $$S$$.
 * 6) The induced action of $$M$$ on $$N$$ by conjugation is a semiregular group action on the non-identity elements: Recall that $$M$$ is the isotropy subgroup of the element $$s \in S$$. For non-identity elements $$n \in N, m \in M$$, we want to show that $$mnm^{-1} \ne m$$. Suppose, for a contradiction, that $$mnm^{-1} = n$$. Then, we have $$m \cdot (n \cdot s) = (mn) \cdot s = (nm) \cdot s = n \cdot s$$. Thus, $$m$$ fixes both $$s$$ and $$n \cdot s$$, a contradiction to 2-regularlity.
 * 7) The induced action of $$M$$ on $$N$$ by conjugation is a regular group action on the non-identity elements: The action is semiregular, and the size of $$M$$ equals the number of non-identity elements of $$N$$, so the action is regular.
 * 8) The automorphism group of $$N$$ is transitive on its non-identity elements: Since $$M$$ acts regularly on the non-identity elements of $$N$$ by conjugation, and conjugation by elements of $$M$$ gives automorphisms of $$N$$, the automorphism group of $$N$$ is transitive on the non-identity elements.
 * 9) $$N$$ is elementary Abelian: This follows from fact (2).
 * 10) Conclusion: $$N$$ is an elementary Abelian regular normal subgroup, and $$M$$ acts regularly on the non-identity elements of $$N$$ by conjugation.
 * 11) It further follows that since $$N$$ is elementary Abelian, $$N$$ has order $$p^k$$. So, $$n = p^k$$ and the order of $$G$$ is $$p^k(p^k - 1)$$.