# 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).

• i) If a Frobenius complement is Abelian, then it is cyclic. ii) A finite cyclic group $A$ can act as a group of automorphisms of a (necessarily nilpotent group $G$) with ${\rm gcd}(|G|,|A|) = 1$ in such a way that $GA$ is a Frobenius group if and only if $PA$ is a Frobenius group with complement $A$ for each Sylow subgroup $P$ of $G$. – Geoff Robinson Jan 18 '16 at 10:22
• It is easy to determine for cyclic $A$ which elementary Abelian $p$-groups $P$ can have $PA$ a Frobenius group. The difficulty seems to be whether one can extend one such group by another (with the same $p$) and admit the same action of $A$ on each factor (and whether (or how far) the process can be iterated). – Geoff Robinson Jan 18 '16 at 10:40
• There are $578$ Frobenius groups of order $\leq10^3$ whose complements are abelian. – M. Farrokhi D. G. Jan 19 '16 at 0:35
• @M. Farrokhi D. G. Thanks for this. Can you explain how you arrived at this number? – Henri Johnston Jan 19 '16 at 10:58
• A finite group is Frobenius if its Fitting subgroup is nontrivial and contains the centralizer of its nontrivial elements. Using this equivalent from, one can simply check groups of small orders by GAP. – M. Farrokhi D. G. Jan 20 '16 at 1:13

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 $m$.