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 $1\leqslant i\leqslant n-1$ 



A question on Frobenius groups
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 $0 < i < n$.


