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Shaun
Shaun
14:46
Please would someone give me some insight into why the following was downvoted?
-1
A: Let $G$ be a group and $A,B\leq G$ abelian subgroups such that $AB=G$. Show $A\cap B\leq Z(G)$

ShaunI will use the one-step subgroup test. Since $A,B\le G$, they are themselves groups and their identity is the identity in $G$. Thus $e\in A\cap B$. Hence $A\cap B\neq\varnothing$. Since $A$ and $B$ are abelian, all elements $x$ of $A\cap B$ commute with all elements of $ A$ and of $B$. But $G=AB$...

@FabrizzioMuzz It looks like your question is deleted. I'm sorry I didn't get there in time to give feedback.

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