# Show groups of symmetries of a cube and a tetrahedron are not conjugate in isometry group.

I've shown that the symmetry group of a cube and a tetrahedron are both isomorphic to S4, but I am now trying to show that they are not conjugate when considered as subgroups of isometries of 3D space.

I cannot think of any kind of criteria that gives when two groups are not conjugate?

EDIT: I meant to say the rotations of a cube! So I've shown the full symmetry group of the tetrahedron is isomporphic to the rotations of a cube, however I need to show that they are not conjugate.

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First, if you're talking about proper isometries, the group of the tetrahedron is $A_4$. If you're including reflections, then the group for the cube is $S_4\times\mathbb Z_2$. Why? – Ted Shifrin May 19 '13 at 20:51
I also count reflections, do these not count as isometries? I made an error in asking the question, it should make sense now! Thanks – Wooster May 19 '13 at 20:54

Something's wrong with your first statement. If you consider orientation preserving 3D movements, the symmetry group of the cube is twice as big as that of the tetrahedron. And the same holds if you disregard orientation for both solids. Note that taking $4$ out of $8$ vertices of a cube, you get a tetrahedron. Any symmetry of that tetrahedron induces a symmetry of the cube. However, the cube allows additional symmetries that map the tetrahedron to the "other" tetrahedron (the one with the other four vertices). In other words, the fact that half the vertices of a cube form a tetrahedron shows that the tetrahedron group is a subgroup of index 2 in the cube group.

However, you do get $S_4$ if you count

• arbitrary isomometries of the tetrahedron (that is, just permutations of the four vertices) and
• orientation preserving isometries of the cube (that can be identified with permutations of the four spatial diagonals).

These two subgroups of the group of isometries cannot be conjugate to each other because the conjugate of an orientation preserving isometry is orientation preserving.

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I also count reflections in the symmetry group, so not just rotations. – Wooster May 19 '13 at 20:50
@Wooster If you also count reflections, then the cube still has twice as many symmetries by the same argument (i.e. both the cube and the tetrahedron symmetry group grow by a factor of $2$) – Hagen von Eitzen May 19 '13 at 20:51
I've made an edit now, thanks! – Wooster May 19 '13 at 20:55
that was the question I meant to ask. I can see intuitively why that should follow, but I'm not quite sure how to show it. I guess I need to show there is no element in the group of isometries that sends the orientation preserving element to a non-orientation preserving element? – Wooster May 19 '13 at 21:02
It won't let me answer my own question but: I think I've answered it now. If we just think of all the group elements as matrices. Then we want to show that these two groups are not conjugate. So we try and find an element to "conjugate with" from the isometry group. So we get A = C^-1 B C where A is an element of the rotations of a cube, and B an element from the rotations and reflections of the tetrahedron. It then follows that det(A) = det(B) and since for rotations det(A) = 1, and for reflections det(B) = -1, then we see that certain elements cannot possibly be conjugate. – Wooster May 19 '13 at 21:13

I think I've answered it now. If we just think of all the group elements as matrices. Then we want to show that these two groups are not conjugate. So we try and find an element to "conjugate with" from the isometry group. So we get $A = C^{-1}BC$ where $A$ is an element of the rotations of a cube, and B an element from the rotations and reflections of the tetrahedron.

It then follows that $det(A) = det(B)$ and since for rotations $det(A) = 1$, and for reflections $det(B) = -1$, then we see that certain elements cannot possibly be conjugate.

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