Equal-or-disjoint translates characterize cosets ($gB\cap hB\ne \varnothing ⟹gB=hB$)

A nonempty subset $B$ of any group $G$ is a left coset of a subgroup if and only if any two left translates $gB,hB$ are equal or disjoint. No finiteness assumption is needed.

For the forward direction, write $B=bH$ with $H$ a subgroup. Then $gB=gbH$, and distinct left cosets of $H$ are disjoint. For the converse, first assume $e\in B$. For every $b\in B$, $b\in B\cap bB$, so $bB=B$. This proves closure under multiplication. Since $e\in B=bB$, there is $c\in B$ with $bc=e$, showing $b^{-1}\in B$. Thus $B$ passes the subgroup test.

For a general nonempty $B$, choose $b\in B$ and set $H=b^{-1}B$. It contains $e$, and all its left translates have the assumed property because $gH=(g{b}^{-1})B$. The previous argument makes $H$ a subgroup, and $B=bH$ is a left coset. This gives a recognition criterion for cosets from their translate intersections alone.