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Nov 4, 2009 at 3:55 comment added Autumn Kent I guess I don't understand what "middle" means.
Nov 4, 2009 at 1:44 comment added Autumn Kent Oh, just take G = <g_1, g_2, . . .> to be the kernel of the map from <a,b> to the free abelian group on a and b. Then G has (P) and K' lies in G, and so K' has (P) as well.
Nov 4, 2009 at 1:20 history answered Autumn Kent CC BY-SA 2.5