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Nov 2, 2013 at 10:45 vote accept Alireza Abdollahi
Nov 2, 2013 at 10:39 answer added YCor timeline score: 7
Nov 2, 2013 at 9:44 history edited Alireza Abdollahi CC BY-SA 3.0
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Nov 1, 2013 at 20:29 comment added YCor PS: my first comment implies however that every infinite residually finite solvable group has an infinite abelian normal subgroup.
Nov 1, 2013 at 20:23 comment added YCor However, the classical example of a group with finite derived subgroup but center of infinite index (see Derek's post in MathSE: math.stackexchange.com/questions/272152/…) is probably a source of infinite f.g. solvable groups with no infinite abelian normal subgroups, by taking the semidirect product of Derek's example with $\mathbf{Z}$ shifting the $x_i,y_i$.
Nov 1, 2013 at 20:20 comment added YCor The naive way to try to prove that these are only finite solvable groups results in: every infinite solvable group has an infinite step-2 nilpotent normal subgroup whose derived subgroup is finite.
Nov 1, 2013 at 19:59 comment added Stefan Kohl I think a classification and a presentation are two different things -- maybe you can clarify what kind of classification you are looking for? -- Up to isomorphism (seems asking a lot here), or what else?
Nov 1, 2013 at 19:38 history asked Alireza Abdollahi CC BY-SA 3.0