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Sep 2, 2014 at 19:36 comment added Ramiro de la Vega That´s much simpler @Emil, thank you.
Sep 2, 2014 at 19:33 comment added Emil Jeřábek Instead of the compactness argument, you can e.g. take for $H_0$ the direct sum of $\omega$ copies of a group of size $\mu$, and for $H_n$ its subgroup omitting the first $n$ copies.
Sep 2, 2014 at 19:08 comment added Ramiro de la Vega @unser47958, I added some details about that.
Sep 2, 2014 at 19:07 history edited Ramiro de la Vega CC BY-SA 3.0
Added some details to part of the proof.
Sep 2, 2014 at 17:32 comment added Minimus Heximus I do not understand why $G$ exists.
Sep 2, 2014 at 16:45 history answered Ramiro de la Vega CC BY-SA 3.0