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Apr 13, 2017 at 12:19 history edited CommunityBot
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Jan 17, 2015 at 15:35 comment added YCor OK, I answered the question on StackExchange.
Jan 17, 2015 at 14:30 comment added Stefan Kohl @YCor: Why don't you write an answer, instead of putting your answer in a comment? -- This question certainly deserves it!
Jan 16, 2015 at 16:01 comment added YCor Yes. Take the Burnside group on 2 generators and exponent $2^k$ for large $k$, which is known to be infinite (it's hard!). By the restricted Burnside problem (it's hard too), it has a minimal finite index subgroup, say $H$; hence $H$ is infinite, finitely generated and has no nontrivial finite quotient. Hence $H$ admits a simple quotient, which is necessarily infinite.
Jan 16, 2015 at 15:58 review First posts
Jan 16, 2015 at 16:24
Jan 16, 2015 at 15:55 history asked W4cc0 CC BY-SA 3.0