Counting Frobenius groups with abelian Frobenius complement In R. Brown, D. K. Harrison Abelian Frobenius kernels and modules over number rings. J. Pure Appl. Algebra 126 (1998), no. 1-3, 51–86, Remark 11.13 (A), the authors show that the number of isomorphism classes of metabelian Frobenius groups of order $\leq 10^3$ is 569 and of order $\leq 10^6$ is 568220. I am interested in similar statistics if we change some of the restrictions on the Frobenius kernel and keep the restriction on the Frobenius complement.
More specifically, for $n=10^3$ or $10^6$ how many isomorphism classes of Frobenius groups of order at most $n$ are there subject to the following restrictions: (a) the Frobenius complement is abelian, (b) the Frobenius complement is abelian and the Frobenius kernel is of $p$-power order for any prime $p$?
I suspect one should be able to find the answers to (a) and (b) when $n=10^3$ by using GAP or Magma and the small groups database, but I've been unable to do this because I don't see an easy way to identify Frobenius groups (and their corresponding Frobenius kernels and complements). When $n=10^6$ I would guess this isn't going to work, but maybe there is some other approach (an approximate number would also be interesting if it's not possible to find the exact answer).
 A: I'll make a few theoretical remarks which might be useful in computation.
As I said in comments, an Abeliean Frobenius complement is necessarily cyclic.
Given a cyclic group $A$ of order $d$, here's a strategy for determining Frobenius groups with complement $A$ and elementary Abelian kernel $V$ which is a $p$-group for some prime $p$. It is enough to consider the case that $A$ acts irreducibly on $V$ (I will return to this point later).
It is of course necessary that $p$ does not divide $d$. Let $e$ be the smallest positive integer such that $d|p^{e}-1$. Then we may construct a Frobenius group with kernel $V$ of order $p^{e}$ and complement isomorphic to $A$ as follows:
The polynomial $\Phi_{d}(x) \in \mathbb{Z}[x]$ reduces (mod $p$) as a product of 
$\frac{\phi(d)}{e}$ irreducible polynomials of degree $e$. Given any one of these irreducible factors, say $m(x)$, we may let $A$ act on $V$ as a linear transformation whose matrix is the companion matrix of $m(x)$. In this way, we obtain $n(d) = \frac{\phi(d)}{e}$ non-isomorphic Frobenius groups with complement isomorphic to $A$ and kernel elementary Abelian of order $p^{e}$.
Then it is easy to check that there are $\frac{n(d)^{2}+n(d)}{2}$ non-isomorphic Frobenius groups with complement isomorphic to $A$ and kernel elementary Abelian of order $p^{2e}$, and similar results hold for elementary Abelian kernels of order $p^{me}$ for any positive integer $m$.
This strategy suffices to construct all non-isomorphic Frobenius groups with cyclic complement and Abelian kernel of squarefree exponent up to any specified order.
Dealing with other kernels (eg non-Abelian or Abelian, but not of squarefree exponent) requires more work, as indicated in comments, but for groups of relatively small order this should still be manageable.
