= Solution
To obtain a <crystal>'s <point group>, discard the translation parts of its space-group operations. A screw becomes a <rotation> and a glide becomes a mirror. Thus the tetragonal group's $4/n$, $2_1/m$ and $2/m$ operations give
$$
\boxed{4/mmm\equiv D_{4h}.}
$$
For the orthorhombic structure the three perpendicular twofold/glide or screw/mirror directions give
$$
\boxed{mmm\equiv D_{2h}.}
$$
The primitive <Bravais lattice> symbol does not itself add a point operation; the fractional translations matter for the <space group> but disappear from this point-group reduction.
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