Let $G$ be a finite simple group. Suppose that there exist a minimal subgroup of $G$ say $L$ isomorphic to $\mathbb{Z}_{p}$ and a maximal subgroup of $G$ say $M$ isomorphic to $\mathbb{Z}_{p}\rtimes\mathbb{Z}_{q}$ for some primes $p$ and $q$ such that $M$ is the only nontrivial proper subgroup of $G$ containing properly $L$. If $L'$ and $M'$ are some other subgroups of $G$ isomorphic to $\mathbb{Z}_{r}$ and $\mathbb{Z}_{r}\rtimes\mathbb{Z}_{s}$ for some primes $r$ and $s$ with the same conditions as $L$ and $M$, then show that $L\cong L'$ and $M\cong M'$.( i.e. show that $p=r$ and $s=q$).
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$\begingroup$ Do you have an example of such $G$ with $G>100$? $\endgroup$– user6976Aug 8, 2017 at 1:23

$\begingroup$ @ Mark. For example $\mathbb{Z}_{7}$ in $L_{2}(7)$ or $\mathbb{Z}_{11}$ in $L_{2}(11)$. But this examples could not prove the result in general. $\endgroup$– H.ShahsavariAug 8, 2017 at 1:52

1$\begingroup$ I suspect that there are only finitely many such groups $G$ or that at least all such $G$ are (well) known. then the proof will follow by inspection. $\endgroup$– user6976Aug 8, 2017 at 2:01

1$\begingroup$ There are probably infinitely many such groups $G$ (for instance it will be sufficient that there are infinitely many Fermat primes), but I think that the second part of Mark Sapir's comment is likely correct. $\endgroup$– John ShareshianAug 8, 2017 at 2:15

$\begingroup$ Are you assuming the minimal subgroup is unique? further more could you please more explain on the semi direct product of cyclic groups of order $p,q$? What group morphism $\phi: \mathbb{Z}_p \to Aut (\mathbb{Z}_q)$ are you considering for $ \mathbb{Z}_p \rtimes_{\phi} \mathbb{Z}_q$?(sorry if my question is elementary) $\endgroup$– Ali TaghaviAug 21, 2017 at 13:22
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