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I am trying to understand why the order of the stabilizer group of a body diagonal of a cube is 6 rather than 3. It is clear to me that rotations about that diagonal stabilize the diagonal, by definition, and the order of this group is 3 (including the identity). So that means there must be 3 other, non-identity, rotations that stabilize this diagonal. I am terrible at geometry and I cannot think of what these others would be. I don't think that any of the rotations about $i$, $j$, $k$ can stabilize the diagonal, and I can't see 3 elements of order 2 doing this either. But then there are no elements left to consider! I am confused.

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There is a "swap the poles" symmetry (180 degree rotation that swaps the two ends of the body diagonal), and if you allow reflections, the "inversion" should also work. (A good answer should have a picture too :-) – Jack Schmidt Oct 17 '12 at 19:29
have you considered rotary reflections? – user39280 Oct 17 '12 at 19:32
@JackSchmidt and dado, does the "swap the poles" symmetry correspond to a "double-rotation" about the $i$, $j$, and $k$ axis (st order of these guys is 2)? Or am I misunderstanding you? – tacos_tacos_tacos Oct 17 '12 at 20:17
@jshin47: $(i,j,k) \mapsto (-j,-i,-k)$ or $a\mathbf{i}+b\mathbf{j} +c\mathbf{k} \mapsto -b\mathbf{i}-a\mathbf{j}-c\mathbf{k}$ is the rotation in terms of the coordinates. The cube itself is all $(a,b,c)=a\mathbf{i}+b\mathbf{j} +c\mathbf{k}$ such that $|a| \leq 1$, $|b|\leq 1$, $|c|\leq 1$. I still think a good answer deserves a picture, but if no-one will post one, I'll write up the linear algebra version. – Jack Schmidt Oct 17 '12 at 23:38

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