Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 5 / b

For $a\in A$, the element $a=ae$ belongs to both $A$ and $aA$. The assumed equal-or-disjoint property therefore gives $aA=A$.

If $a,b\in A$, then $ab\in aA=A$, so $A$ is closed under multiplication. Also $e\in A=aA$, so some $c\in A$ satisfies $ac=e$. Therefore $c=a^{-1}\in A$. Along with the given identity element, closure and existence of inverse elements establish that $A$ is a subgroup of $G$. This is the subgroup test. The proof even works without assuming that $G$ is finite.

Choose $b_0\in B$ and put $H=b_0^{-1}B$. Then $e\in H$. For any $g_1,g_2\in G$,so these translates are equal or disjoint by the assumed property of $B$. Applying i shows that $H$ is a subgroup. ThusThis proves that $B$ is a left coset. Together with i, it gives the equal-or-disjoint translates characterize cosets criterion.