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Dec 6, 2013 at 17:18 vote accept Xindaris
Sep 23, 2013 at 15:53 comment added Joy-Joy Dear Xindaris, I find that you are right after doing the calculation again. I am sorry to make the mistake.
Sep 23, 2013 at 15:51 history edited Joy-Joy CC BY-SA 3.0
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Sep 23, 2013 at 14:39 comment added Xindaris When I did the steps for smith normal form I got the diagonal matrix with entries 1, 1, 2. If those are the elementary divisors, wouldn't that make it isomorphic to Z/2Z? Or did I misunderstand/do something wrong?
Sep 23, 2013 at 0:51 history answered Joy-Joy CC BY-SA 3.0