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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 $j<f$. Let $G=Z_p^f:Z_\frac{q-1}{\gcd(2,q-1)}$ be the parabolic subgroup of $PSL_2(q)$, i.e., the subgroup stabilizing a $1$-dimensional subspace of $\mathbb{F}_q^2$. A Subgroup of $G$ is called a $\mathcal{C}_1$ subgroup of $PSL_2(q)$ (coming from Aschbacher's description of subgroups of classical groups). If a $\mathcal{C}_1$ subgroup $H$ of $PSL_2(q)$ contains elements of orders $p$ and $r$ (or equivalently $pr$ divides $|H|$), then how small can $H$ be? Does $H$ need to contain $Z_p^f:Z_r$?

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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, $H = Z_p^r:Z_r$. –  Derek Holt Oct 17 '12 at 12:38
    
Thanks! That's exactly the proof I'm looking for. –  Binzhou Xia Oct 17 '12 at 15:34

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