Solution
= Solution
{c}
Choose $b_0\in B$ and put $H=b_0^{-1}B$. Then $e\in H$. For any $g_1,g_2\in G$,
$$
g_1H=(g_1b_0^{-1})B,\qquad g_2H=(g_2b_0^{-1})B,
$$
so these translates are equal or disjoint by the assumed property of $B$. Applying <5/b/i> shows that $H$ is a <subgroup>. Thus
$$
\boxed{B=b_0H,\qquad H=b_0^{-1}B\leq G.}
$$
This proves that $B$ is a <left coset>. Together with <5/a/i>, it gives the <equal-or-disjoint translates characterize cosets> criterion.