# Characterization of Frobenius complements

I have learned that Frobenius complements are characterized (among finite groups) by having a fixed point free complex representation. That is, a finite group $G$ is a Frobenius complement if and only if there is a finite dimensional complex representation $\rho:G\to GL(V)$ so that the linear mapping $\rho(g)$ has no nonzero fixed points for any $g\in G\setminus\{1\}$. Is there a citable reference for this fact?

I have found references that mention that this is well-known. For example:

It is well-known that a finite group $H$ has the structure of a Frobenius complement if and only if it can act fixed-point-freely on some finite group $G$ (this means that non-trivial elements of $H$ fix only the identitty element of $G$). Equivalently, we may require that $H$ acts fixed-point-freely on some elementary abelian group $E$, or on a linear space $V$ over a field of characteristic zero. (Aner Shalev: A new characterization of frobenius complements)

I'm aware of Passman's book Permutation groups, but it does not state this characterization. I would prefer a reference that states this characterization explicitly (and preferably includes a proof). My intended audience does not have a strong background in representation theory, so references of a more implicit kind ("this can be inferred from various results in this book" or "it is folklore/well-known/true that…") would be inconvenient.

Edit: The references and descriptions given in the comments and the answer are very useful but do not contain an explicit reference for the representation theoretical characterization of Frobenius complements I'm looking for. The direction that Frobenius complements have fixed-point-free representations is given as Theorem 18.1v in Passman's book, but the other direction is still missing.

• Since (finite) Frobenius kernels are nilpotent (a result of Thompson), a minimal normal subgroup of a Frobenius group is elementary abelian, and so it is not hard to see that a finite group is a Frobenius complement if and only if it has a fixed point free representation over a finite field in coprime characteristic. I would guess that to get the equivalence with fixed point free complex representations, you would need to know about the equivalence between complex representations and representations in finite coprime characteristic. I understand why you prefer a direct reference! – Derek Holt Dec 30 '14 at 14:10
• @Derek Holt: You don't actually need Thompson's Theorem. Frobenius's original theorem, together with basic facts about coprime action are enough. If $G = KH$ is a Frobenius group with kernel $K$ and complement $H,$ then ${\rm gcd}(|K|,|H|) = 1$ and for each prime divisor $q$ of $|K|$, $K$ has an H-invariant Sylow $q$-subgroup ( on which the action of $H$ is still faithful and "Frobenius"). Hence we can find a Frobenius subgroup of the form $VH$ where $V$ is an elementary Abeliaan $q$-group on which $H$ acts irreducibly. – Geoff Robinson Dec 30 '14 at 16:55
• @GeoffRobinson Yes, good point! – Derek Holt Dec 30 '14 at 17:27
• @LSpice, the thing I'm writing about is related to this earlier post: mathoverflow.net/q/184606/55893 See the second motivation paragraph. The intended audience is mainly the inverse problems community. (Sorry for taking a week to write a simple comment.) – Joonas Ilmavirta Jan 7 '15 at 16:39
• Passman's book contains as Theorem 18.1.v. the statement that a Frobenius complement $G$ has a complex representation such that each non-identity element has no fixed points on the representation space. (The proof is as in Geoff Robinson's answer.) I don't know a reference for the other direction (but the proof is similar, as Geoff mentions). I looked into Huppert's "Endliche Gruppen", but I wasn't able to find the statement there. – Frieder Ladisch Jan 8 '15 at 16:11

Since you are asking for a justification of the characterization of Frobenius complements by a representation-theoretic property I am not sure that you can completely avoid representation theory altogether. Also this characterization of Frobenius complements is genuinely "folklore"-something which anyone who has worked seriously with Frobenius groups would be likely to know. I am not sure who first proved it-one place to look is work of H. Zassenhaus, who proved, for example, that ${\rm SL}(2,5)$ is the only perfect Frobenius complement.

Anyway, the proof of the direction " A Frobenius complement $H$ has an irreducible complex representation with each non-identity element acting without the eigenvalue 1" is as given in the comments of Derek Holt and myself- I know of no other way to do it, other than to show that $H$ has such an irreducible representation in characteristic $q \neq 0$ coprime to $|H|$ in which each non-identity element has no non-trivial fixed-points. This may not be absolutely irreducible, but it is a sum of Galois conjugate absolutely irreducible representations, still over a finite field, and no non-identity element of $H$ has the eigenvalue $1$. The theory of Brauer characters shows that $H$ has a complex irreducible representation with the same property.

The opposite direction is similar. If $H$ has an irreducible complex representation in which no non-identity element has the eigenvalue $1$, we can reduce the representation (mod $q$) for any prime $q$ not dividing $|H|$, obtaining an absolutely irreducible representation over a finite field of characteristic $q$. Possibly increasing the dimension, we obtain an irreducible representation of $H$ over ${\rm GF}(q)$ such that each non-identity element of $H$ has no non-zero fixed point on the underlying module, say $V$. Then the semidirect product $HV$ is a Frobenius group with complement $H.$

If you want to get by without telling your readers all details, it would probably be reasonable to sketch the proof of the first direction (the necessary facts about coprime action can be found in many texts, eg Gorenstein's "Finite Groups"), and then to point out that the representation theory of $H$ in coprime characteristic is "essentially the same" as the complex representation theory, then to point out that the proof in the other direction follows for the same reason. The technicalities about fields of realizability, and worrying about realizability over the prime field, are probably best omitted in any case, at that level.

(By the way, $H$ acting in a Frobenius manner on a vector space is a much stronger requirement than $H$ acting fixed point freely. A Frobenius action of $H$ means that no non-identity element of $H$ has a non-zero fixed vector. To say "$H$ acts fixed-point freely" means (in common usage) that there is no non-zero vector fixed by all of $H$).

• Thanks! My fear that this is a genuine folklore thing has come true. I actually need the citation for a paper which is intended to be readable for people who do not work with finite groups or representation theory or finite fields at all (but hopefully remember what such things are). I feel uneasy about giving a hand-wavy "this is folklore in this other field" reference and giving a coarse description of a proof, but I guess I'll do that if there is no better option. – Joonas Ilmavirta Dec 30 '14 at 22:13

I don't have enough reputation to comment, but if you're still looking for a reference, this is basically theorem 25.5 in Feit's Characters of Finite Groups.

Here is the exact statement of the theorem:

Let $$C\neq\{1\}$$ be a group. There exists a Frobenius group $$G$$ with kernel $$K$$ such that $$G/K\cong C$$ if and only if there exists an irreducible character $$\chi$$ of $$C$$ such that for every subgroup $$1, $$\langle \chi_H,1_H\rangle=0$$

• It would be really helpful is you could add the exact theorem statement, for those of us without access to this book. I accidentally also require such a reference. – A.B. Feb 21 '19 at 6:03
• @A.B.: Ok, I am out of the country at the moment, but when I'm back in a few days I'll write the statement. – Hempelicious Feb 21 '19 at 11:09
• Sorry, I know very little about character theory - could you explain why the highlighted box in your answer is equivalent to the highlighted box in the question? – A.B. Mar 6 '19 at 10:38
• @A.B.: This might be better suited to math.SE. The idea is simple though: the product with $1_H$ measures the size of the trivially invariant subspace under the action by $H$. – Hempelicious Mar 7 '19 at 20:50

This is (part of) Theorem 6.13 in Serre's Finite Groups: An Introduction, which says the following. Say that an action of a group $$H$$ on another group $$N$$ is almost free if the action on $$N \setminus \{ e \}$$ is free. Then TFAE:

1. $$H$$ is a Frobenius complement; equivalently, $$H$$ acts almost freely on another finite group $$N$$.
2. $$H$$ acts almost freely on a finite-dimensional vector space over some field.
3. $$H$$ acts almost freely on a finite-dimensional vector space over every field of characteristic not dividing $$|H|$$.
4. $$H$$ acts almost freely and irreducibly on a finite-dimensional vector space over $$\mathbb{C}$$.
5. $$H$$ acts freely and orthogonally on some sphere.