D4 has order 8 and d12 has order 24. so D12 possibly has a subgroup of order 8. I am stuck at that point, any hints or suggestions?

Printable View

- May 4th 2011, 10:31 AMbarf6969showing D12 has a subgroup isomorphic to d4
D4 has order 8 and d12 has order 24. so D12 possibly has a subgroup of order 8. I am stuck at that point, any hints or suggestions?

- May 4th 2011, 10:42 AMDeveno
does the subgroup of rotations of D12 has a subgroup of order 4? can you pick a suitable subset of the reflections in D12 that might correspond to the reflections of D4?

perhaps you might consider what happens if you regard D12 as a subgroup of S12, and then regard 1,2,3 as "being the same number", then 4,5,6 and "being a 2nd number", etc.

equivalently, taking a 12-gon with vertices 1 through 12, considering only rigid motions that permute vertices 1,4,7 and 10, the other vertices just "going along for the ride". - May 4th 2011, 10:59 AMDeveno
the symmetries of a regular hexagon is a group of order 12. there is competing notation for dihedral groups. some authors use Dn, some authors use D2n, for the symmetries of a regular n-gon.

since you indicate D4 has order 8, it is the symmetries of a square. by the same token, D12 is thus the symmetries of a 12-gon.

there is no subgroup of the symmetries of a regular hexagon isomorphic to the symmetries of a square. the rotation group of the symmetries of a hexagon, has order 6, and thus cannot possibly contain a rotational subgroup of order 4. - May 4th 2011, 12:39 PMSwlabr
This is by far the neatest proof. Basically, if n=mk then you can think of n as an m-gon by twisting "k" times as opposed to twisting once. The can be made clear by drawing your m-gon inside your n-gon.

If you do not get any marks for this proof, then your lecturer is a criminal. But you can just blame Bourbaki. Everyone else does.