= Solution
Take the cube with vertices $(\pm1,\pm1,\pm1)$, and choose the edge $E=\{(1,1,z):-1\leq z\leq1\}$. Here the <symmetry group of a cube> consists of all rigid symmetries, including reflections. A symmetry fixes the centre and permutes the three pairs of opposite faces, so its matrix is a <signed permutation matrix>.
A symmetry in the <stabilizer subgroup> of $E$ must preserve its midpoint $(1,1,0)$ and send its direction $(0,0,1)$ to itself or its negative. In this <cube symmetry action on edges>, its only choices are to interchange the first two coordinates and to reverse the third coordinate. Thus the <stabilizer subgroup> consists of
$$
(x,y,z)\mapsto(x,y,z),\quad(y,x,z),\quad(x,y,-z),\quad(y,x,-z).
$$
The two generating reflections commute and have order two, so this <subgroup> is a <Klein four-group>, isomorphic to $C_2\times C_2$ and has order four. Permuting coordinates and changing signs carries $E$ to any edge. Its <orbit of a group action> is therefore the set of all twelve edges. The <orbit-stabilizer theorem> gives
$$
\boxed{|G|=12\cdot4=48.}
$$
If one restricts to orientation-preserving symmetries, only two of the four stabilizer elements remain, giving the <rotational symmetry group of a cube> of order $24$; the full symmetry <group> asked for here has order $48$.
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