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.

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.

• Thanks, but could you (or someone else) please elaborate a little more on the answer. I'm definitely not used to these lines of reasoning. Oct 11 '11 at 20:53
• A thousand thanks, Derek! Do you think that $K$ is contained in $[G,G]$, even if $K$ is infinite but $G$ is solvable? Oct 12 '11 at 13:24
• I don't think this works for infinite solvable groups. We could take $G$ to be a restricted wreath product of two infinite cyclic groups, with $K$ being the base group and $H$ a complement. The action of $H$ on $K$ is fixed-point-free, so we get a Frobenius group. But $K \cap [G,G]$ is equal to the exponent sum 0 subgroup of $K$, so $K/(K \cap [G,G])$ is infinite cyclic. Oct 14 '11 at 13:23