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I am trying to show that the groups of rotations of a regular tetrahedron and a regular hexagonal prism are not isomorphic, not abelian and have order 12.

So the set of rotations of the regular tetrahedrons has 12 elements, so that has order 12. I am unsure how to show it is non abelian however. As for the hexagonal prism, I am familiar with the rotations of a regular hexagon, but not a hexagonal prism and have no idea what the group of rotations would be.

P.S. A similiar small question I had is that is the group of rotations of a regular tetrahedron isomorphic to $\mathbb{Z}_{12}$? They are both of order 12. However, the tetrahedron group is apparently not abelian so that would mean they are not isomorphic, correct? Thanks!

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  • $\begingroup$ The prism has the rotations of the hexagon plus the various rotations that switch ends. $\endgroup$ – Joffan Oct 27 '17 at 14:19
  • $\begingroup$ @Joffan Does that mean there are 12 rotations? The identity, 5 rotations of the normal hexagon and 6 rotations that switch ends? $\endgroup$ – niso9411 Oct 27 '17 at 14:25
  • $\begingroup$ Yes, the rotations with axis parallel to the end faces, all self-inverse. $\endgroup$ – Joffan Oct 27 '17 at 14:55
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Hints:

  • The hexagonal prism has a rotation of order $6$. The tetrahedron not. Remember that an isomorphism maps element of a certain order to an element with the same order.
  • To show a group is non-abelian, it suffices to find $2$ elements that not commute. Think about this geometrically. When I apply one transformation, and then another, does it yield the same result when I reverse the order? Translate back to group theory.
  • $\mathbb{Z}_{12}$ is cyclic, and therefore abelian. So, if both of your groups are non abelian, they can't be isomorphic to this group.
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  • $\begingroup$ I see, thanks! What if both groups are order 12, non abelian. Can they be isomorphic? Such as the group of rotations of a regular tetrahedron and the dihedral group $D_6$ of symmetries of a regular hexagon? $\endgroup$ – niso9411 Oct 27 '17 at 15:02
  • $\begingroup$ If 2 groups are isomorphic, they must have the same order. So, to be isomorphic, it is a necessary condition that the 2 groups have the same size, but not a sufficient one. I.e., in general, if 2 groups have the same order, they are not isomorphic (for example, $Z_2 \times Z_2$ is not isomorphic with $Z_4$, but they have the same size, this is an abelian example though). In your case, the 2 groups are not isomorphic. Show that one of the groups has an element of a certain order that the other one does not have. $\endgroup$ – user370967 Oct 27 '17 at 15:10
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The hexagonal prism has a rotation with period $6$. Does the tetrahedron?

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  • $\begingroup$ From what I understand, the tetrahedron has an order 3 rotation and order 2 rotation. $\endgroup$ – niso9411 Oct 27 '17 at 14:14

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