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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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    @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.2012-10-17

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