This is a follow up to this question
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 $1\leqslant i\leqslant n-1$
js21 answered the question:
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 $0 < i < n$.
I have a few question about the answer and I would be very thankful if someone kindly help me about them. Could someone introduce me a reference for the first statement: "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 $\overline{\mathbb{F}_q}$." . Also I don't understand what he mean of "Frobenius-orbit" and also what's the meaning of $\overline{\mathbb{F}_q}$. I apologize if my questions are elementary. Thank you in advance.
Let $X$ be a generator of $\mathbb{Z}_p$, acting as $X\in Aut(\mathbb{Z}_q^n)=GL(n,q)$. Then saying that $X$ acts irreducibly is the same as saying the characteristic polynomial $g_X(t)=\det(tI-X)$ is an irreducible factor of $t^p-1\in\mathbb{F}_q[t]$.
Let $L$ be the splitting field of $g_X(t)$ over $\mathbb{F}_q$. Thus $L=\mathbb{F}_q(\zeta)$ where $\zeta$ is a primitive root, and $\zeta^p=1$. Then the Galois group of $L$ over $\mathbb{F}_q$ is cyclic, generated by the Frobenius automorphism $\sigma(a)=a^q$.
If $p$ divides $q^r-1$, then $\sigma^r(\zeta)=\zeta$, and $$ h(t)=\prod_{i=1}^r\left(t-\sigma^i(\zeta)\right) $$ is a member of $\mathbb{F}_q[t]$ because $\sigma$ permutes the terms, and is a factor of $$ g_X(t) = \prod_{i=1}^n\left(t-\sigma^i(\zeta)\right)$$
This contradicts the irreducibility of $g_X(t)$.