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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 Robinson Nov 6 '18 at 13:46

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  • 2
    $\begingroup$ 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. $\endgroup$ – Derek Holt Nov 6 '18 at 14:49
  • $\begingroup$ @ Derek Holt. You are right. I revised the question as you mentioned. So sorry for bad writing. $\endgroup$ – 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 $0 < i < n$.


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