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. Frobenius' theorem
2. 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 $g\cdot s\neq s$. 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 $mnm^{-1}\neq m$. 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 $p^{k}(p^{k}-1)$.