# A question on Frobenius groups [closed]

Please change the title if needed.

Let $$p$$ and $$q$$ be distinct primes and $$G\cong(\underbrace{\mathbb{Z}_{q}\times\mathbb{Z}_{q}\times\dots\times\mathbb{Z}_{q}}_{n\,\,times})\rtimes\mathbb{Z}_{p}$$, where a subgroup of order $$p$$ acts irreducibly on the kernel( means $$G$$ has no proper subgroup of order $$pq^{i}$$, for $$1\leqslant i\leqslant n-1$$). How can we show that $$p\nmid (q^{i}-1)$$ for each $$1\leqslant i\leqslant n-1$$

## closed as off-topic by Derek Holt, abx, HJRW, Chris Godsil, Geoff RobinsonNov 6 '18 at 13:46

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• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Derek Holt, abx, HJRW, Chris Godsil, Geoff Robinson
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• Just to clarify, my vote to close was partly based on the wording of this question "Prove or give a counterexample" , which sounds more like an exercise than a problem encountered while undertaking research. Also, there is no indication given as to whether $p$ and $q$ are supposed to be prime numbers. – Derek Holt Nov 6 '18 at 14:49
• @ Derek Holt. You are right. I revised the question as you mentioned. So sorry for bad writing. – H.Shahsavari Nov 9 '18 at 3:02

Assume that $$p \neq q$$, and let $$r$$ be the order of $$q$$ modulo $$p$$. The set of irreducible representations of $$G = \mathbb{F}_p$$ over $$\mathbb{F}_q$$ is in bijection with Frobenius-orbits of irreducible representations of $$G$$ over $$\overline{\mathbb{F}_q}$$. Here, the non-trivial irreps over $$\overline{\mathbb{F}_q}$$ are $$1$$-dimensional (corresponding to $$p$$-roots of unity in $$\overline{\mathbb{F}_q}$$), and their Frobenius-orbit has length $$r$$, so that the irrep over $$\mathbb{F}_q$$ corresponding to such an orbit has dimension $$r$$.
Thus there are exactly $$1 + \frac{p-1}{r}$$ irreps of $$G$$ over $$\mathbb{F}_q$$: the trivial one, and $$\frac{p-1}{r}$$ of dimension $$r$$. So in your question, $$n >1$$ implies $$n =r$$, and thus $$p$$ does not divide $$q^i -1$$ for $$0 < i < n$$.