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This is a crosspost of my (slightly longer) question on MSE since I'm not getting any responses there.


Definition. Let $G$ be a finite group and $F_1=\text{Fit}\,G$ and $F_2/F=\text{Fit}\left(G/F_1\right)$. If $F_2$ is a Frobenius group with kernel $F_1$ and $G/F_1$ is a Frobenius group with kernel $F_2/F_1$, we say that $G$ is $2$-Frobenius.

I have read about the characters of Frobenius groups in Isaacs and Huppert's books, but I have never seen $2$-Frobenius groups mentioned. Can anyone point me to some literature on the character theory of $2$-Frobenius groups?

Alternatively, does anyone know any theorems about Frobenius groups that could be adapted to $2$-Frobenius groups? I am especially interested in $2$-Frobenius groups where $F_1$ and $G/F_2$ are $p$-groups and $F_2/F_1$ is a $q$-group (for distinct primes $p$,$q$), but I would appreciate any representation theory at all which may help me better understand this class of groups.

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I wonder if you have written exactly what you meant? Surely you mean that $F_{2}$ is a Frobenius group with kernel $F_{1}$ and $G/F_{1}$ is a Frobenius group group with kernel $F_{2}/F_{1}?

Assuming that is what you meant,there are lots of $2$-Frobenius groups, as you are probably aware. One family of examples, which is in a sense typical, is when you have a Frobenius group $H$ with Abelian Frobenius kernel $A$, and you take a faithful irreducible $FH$-module $V$ for $F$ a field of prime order. Then the semidirect product $VH$ is a $2$-Frobenius group according to your definition,because the irreducibility of $V$ ensures that $C_{V}(a) = 1$ for all non-identity $a \in A.$

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Yes, sorry that is a typo, that is what I meant. –  Alexander Gruber Oct 20 '12 at 14:44
    
@GeoffRobinson So, you've given an example of the family of groups I'm talking about, which I am actually already familiar with... is there anything we know about the characters of those groups? (For example, I asked in the MSE question whether the formula for Frobenius groups' characters could be applied here.) Am I being dumb and the answer is staring me in the face, or is it simply that the characters of $2$-Frobenius groups are complicated/not well understood? –  Alexander Gruber Oct 21 '12 at 22:07
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I did think a little about this. I think you just have to look at inertial subgroups and use Clifford theory, but I couldn't say anything too coherent, though I had started to try. For example, if you look at an irreducible character $\mu$ of$F_{2}$ which does not contain £F_{1} in its kernel, then its inertial subgroup in $G$ is totally controlled by what goes on with the restriction to $F_{1}.$ But then ..? –  Geoff Robinson Oct 21 '12 at 22:48
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