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Let $G\cong(\underbrace{\mathbb{Z}_{q}\times\mathbb{Z}_{q}\times\dots\times\mathbb{Z}_{q}}_{n\,\,times})\rtimes\mathbb{Z}_{p}$, where a subgroup of order $p$ acts irreducibly on the kernel( means $G$ has no proper subgroup of order $pq^{i}$, for $1\leqslant i\leqslant n-1$). Prove the following proposition or give a counterexample:

$p\nmid (q^{i}-1)$ for each $1\leqslant i\leqslant n-1$.

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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 $1\leqslant i\leqslant n-1$