Commutator subgroup of Frobenius groups If $K$ is any additive abelian group of odd order and $H$ has order 2 and acts on $K$ by inversion, then I know that the commutator subgroup of the dihedral group $G=K.H$ is $2K$ (multiplication by 2).
My question: are there formulas for the commutator subgroup of a general Frobenius group with Frobenius kernel $K$ and Frobenius subgroup $H$?
Thanks.
EDIT: In case the question is too general, I'm in fact interested in solvable groups. It would be nice to know if something can be said when $K$ and/or $H$ is moreover abelian.
 A: Assuming that you are talking about finite Frobenius groups, the commutator subgroup of $G = K.H$ is $K.[H,H]$.
To prove that, it is sufficient to show that $K = [K,H]$. Fix $1 \ne h \in H$. Then $C_K(h) = 1$, so the elements $\{ k^{-1}h^{-1}kh \mid k \in K\}$ are all distinct elements of $K$, and hence must comprise all of $K$.
Added material follows. OK, here is a bit more detail Firstly, the quotient group $G/K[H,H] = KH/[H,H]$ is isomorphic to $H/[H,H]$, which is abelian. One of the principal properties of the commutator subgroup $[X,X]$ of a group $X$ is that, for an arbitrary normal subgroup $N$ of $X$, $X/N$ is abelian if and only if $[X,X] \le N$. So $G/K[H.H]$ abelian implies that $[G,G] \le K[H,H]$.
So it remain to prove that $K[H,H] \le [G,G]$. If we can prove that $K = [K,H]$, then we will have $K[H,H] = [K,H][H,H] \le [G,G]$ and we will be done.
By definition of a Frobenius group, $H \ne 1$, so we can choose a non-identity element $h \in H$. Let $k,l \in K$. Suppose that $k^{-1}h^{-1}kh = l^{-1}h^{-1}lh$. Then rearranging this equation gives
$lk^{-1} = h^{-1}lhh^{-1}k^{-1}h = h^{-1}(lk^{-1})h$,
so $h$ and $lk^{-1}$ commute. But in Frobenius group, no non-identity element of $H$ can commute with any non-identity element of $K$. So, since we assumed that $1 \ne h$, we must have $1 = lk^{-1}$; i.e. $k=l$.
This proves that, as $k$ ranges over $K$, but with a fixed $h$, the $|K|$ elements $h^{-1}k^{-1}hk$ are all distinct. Since $K$ is a normal subgroup of $G$, they are distinct elements of $K$, and so every element of $K$ must be a commutator of the form $h^{-1}k^{-1}hk$. This proves that $K = [K,H]$, which completes the proof.
