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Let $G$ be a finite non-Abelian group. Is it true that `` there exists a proper solvable subgroup $A$ and a proper subgroup $B$ of $G$ such that $G=AB$?"

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  • $\begingroup$ Nope: Consider $\mathbb Z/p\mathbb Z$, where $p$ is prime. This group only has one proper subgroup, i.e., $\{1\}$, which obviously doesn't satisfy your criteria. $\endgroup$ – Don Thousand Nov 1 '18 at 15:57
  • $\begingroup$ Thanks. That is true i missed to write the assumption $G$ is a non-abelian group $\endgroup$ – Mohsen Nov 1 '18 at 15:58
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    $\begingroup$ Apparently the answer is yes. In fact, you can require both $A, B$ to be solvable. $\endgroup$ – Mees de Vries Nov 1 '18 at 16:52
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    $\begingroup$ @MeesdeVries Is it obvious that the answer here follows from the MO question? Here the product is considered, while there the subgroup generated by them, which might be a lot larger. $\endgroup$ – Tobias Kildetoft Nov 1 '18 at 16:56
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    $\begingroup$ Ah, my bad. I understand "$AB$" to mean "the subgroup generated by $A$ and $B$". $\endgroup$ – Mees de Vries Nov 1 '18 at 17:02

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