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Let $q=p^f$, $r$ be a primitive prime divisor of $p^f-1$, i.e., $r\mid p^f-1$ but $r\nmid p^j-1$ for $j

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    Your assumption that $r$ is a primitive prime divisor of $p^f$ ensures that $Z_p^r$ is irreducible as a $Z_rF$-module (where $F$ is the field of order $p$). So, if $pr$ divides $|H|$, then the intersection of $H$ with $Z_p^r$ is nontrivial and hence, by irreducibility of the action, must be the whole of $Z_p^r$. So yes, $H = Z_p^r:Z_r$.2012-10-17
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    Thanks! That's exactly the proof I'm looking for.2012-10-17

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