In Dummit and Foote's Abstract algebra a problem is given as-
Let $G$ be the group of rigid motions in $\mathbb R^3$ of a cube. Show that $|G|$=24.
For an n-gon, all the points on the n-gon can be determined by fixing the two vertex points on the n-gon.
For a 3-D solid, by fixing three adjacent vertex points on the solid all other points on the solid can be determined. This is easy to prove.
So if we label the vertices ("1","2",...., "8") of the solids and take into consideration the vertex "1". It has eight possibilities to take position, vertex "2" has 3 possibilities and the vertex "3" has 2 possibilities. So total symmetries are 8×3×2=48.
Out of 48 some are rotations and some are reflections.
How to conclude that 24 symmetries are rotations?
Like in planar n-gons, the rotation $r^k=\frac{2\pi k}{n}$, so there comes out to be n rotations. But in cube like how we define these rotations?
I have seen related post here, but some theorems are used to prove this, I have not studied these theorems yet as I am beginner to the abstract algebra.
Please clarify the doubt.
Hint: If it's a rigid motion in $\Bbb{R}^3$, then it's orientation-preserving, so you don't actually have 2 possibilities for where vertex 3 goes, once you've picked vertex 1 and 2.
Edit: "Orientation-preserving" means, informally, left-handed things don't get changed to right-handed. Rotations are orientation preserving.
In the picture, I cannot rotate the cube in such a way that 1 and 2 stay put, but 3 and 4 swap places. I would need to do a reflection, which would reverse orientation. The vertices around 2 have to go 1, 3, 4 in clockwise order no matter how I rotate the cube, so if I pick locations for 1 and 2, that automatically determines where 3 and 4 end up as well.