# How is a symmetric group the subgroup of the group of isometries of three-dimensional space?

So I have this question to solve. I've already shown that the group of rotations of a cube is isomorphic to $S_4$. I need to prove that these two groups are not conjugate when considered as subgroups of the group of isometries of 3D space. First off, I'm having a hard time figuring out how $S_4$ can be a subgroup of the group of isometries of 3D space. All the isometries in 3D can be represented as a product of reflections, so, this group would be a group containing the reflections in 3D. Obviously, I can picture this as $S_4$ representing the symmetries of a cube, but I don't see how that would help me solve the question. Could anybody give me some tips as to how to approach this question?

• So you have an isomorphism from $S_4$ to "rotations of the cube". How are you supposed to consider $S_4$ other than as rotations of the cube? This question seems a little vague! – hayd May 23 '15 at 0:21
• A cube has 4 diagonals... – achille hui May 23 '15 at 0:21
• @achillehui So if we consider $S_4$ as the symmetry group of the four diagonals, how would we show that $S_4$ and the group of rotations of a cube are not conjugate? – dodo628 May 23 '15 at 0:27
• I'm not sure. The only thing which comes to my mind is conjugation in $O(3)$ preserve the orientation, it cannot send a rotation in $SO(3)$ to a reflection. An exchange of two diagonals (i.e a transposition in $S_4$) can be implemented as reflection. – achille hui May 23 '15 at 0:42
• I recommend watching this video (the appropriate start time is embedded within the link) youtu.be/VSB8jisn9xI?t=44m26s – eloiprime May 23 '15 at 1:10

Presumably the question is referring to the other faithful representation of $S_4$ in three dimensions, namely the group of all symmetries of a regular tetrahedron.