= Solution
{c}
Every <left coset> $kH$ with $k\in K$ is also a <left coset> of $H$ in $G$. Moreover equality of two such <cosets> means the same thing in $K$ and in $G$:
$$
kH=k'H\iff k'^{-1}k\in H.
$$
Thus the map from the set of <left cosets> of $H$ in $K$ to the set of <left cosets> of $H$ in $G$, taking $kH$ to itself, is injective. Consequently the <index of a subgroup> satisfies
$$
\boxed{|K/H|\le|G/H|\le n.}
$$
These are sets of <cosets>; no <normal subgroup> assumption is needed.
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