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 $s=q$).

  • $\begingroup$ Do you have an example of such $G$ with $|G|>100$? $\endgroup$
    – user6976
    Aug 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. $\endgroup$ Aug 8, 2017 at 1:52
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    $\begingroup$ 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. $\endgroup$
    – user6976
    Aug 8, 2017 at 2:01
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    $\begingroup$ 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. $\endgroup$ Aug 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) $\endgroup$ Aug 21, 2017 at 13:22


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