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| + | Yeah, would have been nice if D12 and S4 as groups had the same order...but they don't so you have to do it that way | ||
Revision as of 19:00, 12 February 2009
Prove that S4 is not isomorphic to D12.
D12 has elements of order 12 whereas S4 does not and therefore they cannot be isomporphic. --Aifrank 17:05, 9 February 2009 (UTC)
What do the elements of D12 and S4 look like? I found some pictures of what D12 and S4 can look like, but I am really stuck on what the elements are. Are they the rotations and reflections? --Josie
Yes they are the rotations and reflections. There are 12 rotations (30 degrees each - 360/12) as well as reflections. So you can see that D12 has elements of order 12 from those rotations and reflections. Then in S4 you either have a 4-cycle, or a 3-cycle, or 2 2-cycles, or 1 2-cycle. The orders of these are 4, 3, 2, 2. So none of them are order 12.
--Nswitzer 16:37, 10 February 2009 (UTC)
Thank you very much for the help! --Josie
Can you also prove this one by showing the size of S4 isn't equal to the size of D12? -Paul
If by size you mean the order, then $ \scriptstyle\mid S_4\mid $ is actually equal to $ \scriptstyle\mid D_{12}\mid $, their orders being 24.
- --Nick Rupley 02:57, 12 February 2009 (UTC)
Yeah, would have been nice if D12 and S4 as groups had the same order...but they don't so you have to do it that way
