Counting Frobenius groups with abelian Frobenius complement

Question

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).

Answer 1

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.