= Solution
{c}
For a <subgroup> $H$ of $G$, the <left coset> represented by $g$ is
$$
\boxed{gH=\{gh:h\in H\}.}
$$
Suppose two <left cosets> intersect. There are $h_1,h_2\in H$ such that $g_1h_1=g_2h_2$, so
$$
g_1=g_2h_2h_1^{-1}.
$$
Because $H$ is a <subgroup>, $h_2h_1^{-1}\in H$. Thus, for every $h\in H$, $g_1h=g_2(h_2h_1^{-1}h)\in g_2H$, proving $g_1H\subseteq g_2H$. Interchanging the two representatives gives the reverse inclusion. Therefore \b[two <left cosets> are equal or disjoint].
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