# Supplement in finite groups

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 $$G=AB$$?"

• 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. – Don Thousand Nov 1 '18 at 15:57
• Thanks. That is true i missed to write the assumption $G$ is a non-abelian group – Mohsen Nov 1 '18 at 15:58
• Apparently the answer is yes. In fact, you can require both $A, B$ to be solvable. – Mees de Vries Nov 1 '18 at 16:52
• @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. – Tobias Kildetoft Nov 1 '18 at 16:56
• Ah, my bad. I understand "$AB$" to mean "the subgroup generated by $A$ and $B$". – Mees de Vries Nov 1 '18 at 17:02