= Solution
{c}
Each left <coset> $gH$ is nonempty, since it contains $g=ge$, and every $g\in G$ lies in its own left <coset>. Suppose $aH$ and $bH$ intersect, say $ah_1=bh_2$ with $h_1,h_2\in H$. Then
$$
a=bh_2h_1^{-1}.
$$
Since $H$ is a <subgroup>, $h_2h_1^{-1}\in H$, and $aH\subseteq bH$. The reversed relation $b=ah_1h_2^{-1}$ gives $bH\subseteq aH$. Thus any two left <cosets> are either equal or disjoint, and together they cover $G$. Therefore \b[$G/H$ is a <partition of a set> $G$]. No finiteness or <normal subgroup> assumption is needed.
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