dihedral groups and rigid motion

prove that the order of the group of rigid motions in R3 of a cube is 24.

there's one solution to this by saying that any one face can be mapped onto any one of the other six faces, and since each face has four vertices there are four motions on each so there's a maximum of 24 motions total, then you just plug the group into GAP and write down the permutations.

my question is whether you can prove this using the definition of dihedral groups, considering D2n where n=4 (for the squares on each face). since D2n is defined as having an order of 2n, it follows that for each square there are 8 elements in the group, then if you consider the cube there are 3 unique faces (since parallel faces should produce equivalent motions) with 8 possible positions for the squares on said 3 faces hence 8*3=24. i passed this by the prof and he said it was wrong but i don't recall why.

in any case i'm still interested in if the dihedral argument is a possible solution.

I've got your rigid motion right here

Any rigid motion of the cube must permute the four body diagonals somehow.  It's easy to show that all 24 permutations are possible.

Your argument about the dihedral groups is totally wrong for several reasons.  One is that it implicitly assumes that the faces can move independent of one another, like a Rubik's cube.   Another reason is that the dihedral group includes motions that flip the square over, which on the cube would turn the face inside out.  But no rigid motion of the cube does this.  Another reason is that it totally disregards the fact that the faces don't only move in place, and there are motions of the cube that exchange some faces with other faces.