Any hints are appreciated. Thanks in advance.
My Attempt:
Either $\alpha = \beta$ or $\alpha \neq \beta$. The fist case is logically equivalent to $A=B$. If I can prove that the second case is equivalent to $A \bigcap B = \emptyset$, the proof is complete. I didn’t have any luck deriving a contradiction by assuming $A \bigcap B \neq \emptyset$. Also I noticed that $A \bigcap B = \emptyset$ implies there is $\theta$ s.t. $\theta = u_1 + \alpha = u_2 + \beta$ for some $u_1, u_2 \in M$ but not sure how that helps.
Take an element $v\in A\cap B$ then $v=\alpha+u=\beta+w$ for some $u,w\in M$. Then $\alpha-\beta=w-u\in M$. This means that $\alpha\in \beta+M$. So $A=\alpha+M=(\beta+M)+M=\beta+M=B$. This is implies the result.