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Let $n$, $m\in \mathbb{N}$. Let $p$, $q$ be primes with $q^{n}|p-1$. Let $H$ the semidirect product of a cyclic group $A=C_{p^{\large{m}}}$ by a cyclic group $B=C_{q^{\large{n+1}}}$ which induces an automorphism of order $q^{n}$ on $A$ (i.e. I mean that if $B=<b>$ and $\alpha \in Aut(A)$ with $|\alpha|=q^{n}$, then $b$ acts on $A$ as $\alpha$ )

Has $H$ the structure of a Frobenius complement?

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  • $\begingroup$ The smallest instance is $q=2$, $p=3$, $m=n=1$, with $|H|=12$. In that case, the answer is yes. There is a Frobenius group with kernel elementary abelian of order $25$ and complement $H$. $\endgroup$
    – Derek Holt
    Jun 17, 2014 at 19:21

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Yes, I believe it always does. There is a complex irreducible character of degree $q^{n}$ of $H$ (which is induced from a faithful linear character of $A\Omega_{1}(B)$) such that no non-identity element of $H$ has $1$ as an eigenvalue in the associated representation.Reduce that representation (mod $r$) for some prime $r \neq p,q$. Regarding the associated module as an elementary Abelian $r$-group $R$ ( possibly of rank greater than $q^{n}$), the group $RH$ is a Frobenius group with complement $H.$

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  • $\begingroup$ Where I can find a more detailed proof? $\endgroup$
    – user53089
    Jun 18, 2014 at 8:11
  • $\begingroup$ Which details do you need? Induce any faithful linear character of $A\Omega_{1}(B)$ up to all of $AB,$ and that character will work. $\endgroup$ Jun 18, 2014 at 12:16
  • $\begingroup$ It is a general fact that a finite group $H$ is a Frobenius complement if and only if $H$ has an irreducible complex representation in which only the identity element has $1$ as a eigenvalue. $\endgroup$ Jun 19, 2014 at 8:03

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