= Solution
Use the same triangle and let $G$ consist of <translations> $x\mapsto x+m$ for $m\in\mathbb Z^2$. All its vertices have <integer> coordinates, and one vertex is the origin, so
$$
S=\bigcup_{m\in\mathbb Z^2}(m+T)=\mathbb Z^2.
$$
The <rotation> through a quarter-turn $(x,y)\mapsto(-y,x)$ preserves this <integer lattice>, and hence belongs to $H$. It fixes the origin and is not the identity, whereas the only <translation> fixing the origin is the identity. It is therefore outside $G$, proving \b[$G\subsetneq H$]. This is <symmetry enlargement under group translates> even though $T$ itself has trivial <stabilizer subgroup>.
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