(New page: For two groups to be isomorphic, there has to be a one-to-one mapping. This is impossible for <math>S_4</math> and <math>D_{12}</math> because <math>D_{12}</math> has an order of 12, and <...) |
|||
| Line 1: | Line 1: | ||
For two groups to be isomorphic, there has to be a one-to-one mapping. This is impossible for <math>S_4</math> and <math>D_{12}</math> because <math>D_{12}</math> has an order of 12, and <math>S_4</math> does not,<math>S_4=(14)(13)(12)</math>, but I could be wrong. -Dan | For two groups to be isomorphic, there has to be a one-to-one mapping. This is impossible for <math>S_4</math> and <math>D_{12}</math> because <math>D_{12}</math> has an order of 12, and <math>S_4</math> does not,<math>S_4=(14)(13)(12)</math>, but I could be wrong. -Dan | ||
| + | |||
| + | Yeah that's pretty much what I said. I put that elements of <math>S_4</math> can only have orders of 1,2,3, or 4 and since <math>D_{12}</math> has an element of order 12, they are not isomorphic. -John | ||
Latest revision as of 09:32, 25 September 2008
For two groups to be isomorphic, there has to be a one-to-one mapping. This is impossible for $ S_4 $ and $ D_{12} $ because $ D_{12} $ has an order of 12, and $ S_4 $ does not,$ S_4=(14)(13)(12) $, but I could be wrong. -Dan
Yeah that's pretty much what I said. I put that elements of $ S_4 $ can only have orders of 1,2,3, or 4 and since $ D_{12} $ has an element of order 12, they are not isomorphic. -John
