Solution
= Solution
{c}
Let the <finite set> $G/H$ contain $r$ distinct left <cosets>. By <6/a/i>, they form a disjoint <partition of a set> $G$, and by <6/a/ii> each has $|H|$ elements. If $H$ is also finite, counting this finite disjoint <set union> gives
$$
\boxed{|G|=\sum_{W\in G/H}|W|=r|H|=|G/H|\,|H|.}
$$
Thus $G$ is a <finite group>. This is the counting argument behind <Lagrange's theorem>.