Frobenius groups Let $G$ be a 2-transitive Frobenius group. So we have $|G| = n(n-1)$. If we know that the kernel is an elementary abelian $p$-group,
is it true that the complement of such a group is always abelian?
 A: There are two deleted answers to this question, by AnalysisStudent0414 and Ken Lebensold, that both mention a frobenius group of order 72 with kernel isomorphic to $C_3\times C_3$ and nonabelian complement. Ken Lebensold mentioned that the complement is the quaternion group. 
This frobenius group is a real thing, so the answer to the question is no. 
One can construct it as follows: viewing $C_3\times C_3$ as the additive group of 2-dimensional column vectors over the field $\mathbb{F}_3$ of order 3, its automorphism group is $GL(2,3)$, which is a non-split $C_2$-central extension of the symmetric group $S_4$. (Indeed, $PGL(2,3)$ acts sharply 3-transitively on the 4 points of the projective line over $\mathbb{F}_3$; this identifies it with $S_4$; and $PGL(2,3)$ is the quotient of $GL(2,3)$ by its center, which is $\{\pm I\}$.) Take the double-transposition subgroup of $S_4$ and pull it back to $GL(2,3)$; you get a group isomorphic to the quaternion group, generated, for example, by
$$ \mathbf{i} :=\begin{pmatrix}1&-1\\-1&-1\end{pmatrix}\text{ and } \mathbf{j} := \begin{pmatrix}1&1\\1&-1\end{pmatrix}.$$
To see this is the quaternion group, note that
$$\mathbf{i}\mathbf{j}=\mathbf{k}:=\begin{pmatrix} &-1\\1& \end{pmatrix},$$
while $\mathbf{j}\mathbf{i}=-\mathbf{k}$; one can check that $\mathbf{i}^2=\mathbf{j}^2=\mathbf{k}^2=-I$. Call this group $H$. Its action on the nonzero points of $\mathbb{F}_3^2$ is free and transitive. If $T$ is the regular action of the additive group of $\mathbb{F}_3^2$ on itself, then $H$ normalizes $T$ (as linear transformations always normalize translations in a vector space), so $TH$ is a group.
It is the desired group: $T$ acts regularly on the 9 points of $\mathbb{F}_3^2$, so $H$ is the stabilizer of $\begin{pmatrix}0\\0\end{pmatrix}$ in $TH$, and all other point stabilizers are conjugate via elements of $T$. Since $H$ acts regularly on the nonzero points, it intersects each of the other point stabilizers trivially. So this is a frobenius group with kernel $C_3\times C_3$ and complement the quaternion group.
NB: This community wiki answer is based on insights obtained from deleted answers.  I wasn't sure about the etiquette of naming the sources, but, considering all factors, I decided to include them since it felt wrong not to give attribution to the sources of the key insight behind the post.
