1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – hayd May 23 '15 at 0:21
  • $\begingroup$ A cube has 4 diagonals... $\endgroup$ – achille hui May 23 '15 at 0:21
  • $\begingroup$ @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? $\endgroup$ – dodo628 May 23 '15 at 0:27
  • $\begingroup$ 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. $\endgroup$ – achille hui May 23 '15 at 0:42
  • $\begingroup$ I recommend watching this video (the appropriate start time is embedded within the link) youtu.be/VSB8jisn9xI?t=44m26s $\endgroup$ – eloiprime May 23 '15 at 1:10
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Could you use that to prove that the group of rotations of a cube is not conjugate with the group of all symmetries of a tetrahedron? $\endgroup$ – dodo628 May 23 '15 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.