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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 a irreducible complex representation in which no non-identity element has th 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 direction 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 $H$).

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 $H$).

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 non-solvable 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 a irreducible complex representation in which no non-identity element has th 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 direction 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 $H$).

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 a irreducible complex representation in which no non-identity element has th 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 direction 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 $H$).

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 non-solvable 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 a irreducible complex representation in which no non-identity element has th 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 you students 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 direction 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 action $H$ 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 $H$).

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 non-solvable 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 a irreducible complex representation in which no non-identity element has th 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 direction 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 $H$).