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
How small can a $\mathcal{C}_1$ subgroup of $PSL_2(q)$ containing elements of certain prime orders be?
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group-theory
finite-groups
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0Your 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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0Thanks! That's exactly the proof I'm looking for. – 2012-10-17