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Nov 10, 2017 at 0:17 vote accept David Roberts
Dec 2, 2011 at 16:42 comment added Goldstern You posted the same question to FOM on Nov 20. I think it makes sense to link to such crosspostings, to avoid duplication of work. See cs.nyu.edu/pipermail/fom/2011-November/015964.html
Nov 21, 2011 at 17:09 answer added Goldstern timeline score: 2
Nov 21, 2011 at 4:40 history edited David Roberts CC BY-SA 3.0
added 451 characters in body; edited title; added 48 characters in body; Post Made Community Wiki
Nov 21, 2011 at 0:55 comment added Benjamin Steinberg Ok, maybe somebody will answer my wacky version.
Nov 21, 2011 at 0:53 comment added David Roberts @Benjamin, I'm thinking of a pair of groups $L \lt G$ of cardinality $\geq \kappa$ such that $L$ contains no normal $N \lt G$ of cardinality $\kappa$. But if I can just take $G$ simple, then I am done.
Nov 20, 2011 at 21:07 history edited David Roberts CC BY-SA 3.0
Fixed typo
Nov 20, 2011 at 20:53 comment added David Roberts Whoops, I did mean |N| < lambda. You've answered my question Goldstern, thanks.
Nov 20, 2011 at 16:31 comment added S. Carnahan Benjamin Steinberg's version of this question sounds interesting.
Nov 20, 2011 at 14:19 comment added Benjamin Steinberg Maybe he means $\lambda$-simple and not $\lambda'$-simple for $\lambda'<\lambda$. Some motivation would make this clearer.
Nov 20, 2011 at 13:41 history edited Goldstern
edited tags
Nov 20, 2011 at 12:47 comment added Goldstern Do you mean $\le \lambda$ or $<\lambda$? Anyway, it seems to me that every simple group is $\lambda$-simple, or have I misunderstood the question? There are simple groups of any cardinality $\kappa$, e.g. the set of "even" permutations on a set of size $\kappa$. (The smallest normal subgroup containing all 3-cycles.)
Nov 20, 2011 at 10:32 history asked David Roberts CC BY-SA 3.0