If such a subgroup exists, then the number of elements in $G \setminus H$ is $1$. Take that element $g$. Now analyze the product $g \cdot h$ for $h \in H$.
Solution: Let $G$ be a group of order $n > 2$ and let $H$ be a subgroup of $G$ such that $\vert H \vert = n - 1$. Then, the number of elements in $G$ which are not in $H$ is $1$. Let us denote the element which is not in $H$ by $g$. Let $h \neq e$ be an element of $H$. Then
\[
g\cdot h \neq g \implies g \cdot h \in H \implies g\cdot h \cdot h^{-1} = g \in H,
\]
a contradiction. Thus, there can not be a subgroup of order $n-1$.