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Aug 20, 2019 at 21:59 history became hot network question
Aug 20, 2019 at 19:24 vote accept Santana Afton
Aug 20, 2019 at 19:23 answer added Derek Holt timeline score: 15
Aug 20, 2019 at 16:25 comment added YCor @DerekHolt great. I'd be happy if you post an answer then as I haven't gone much into that paper.
Aug 20, 2019 at 15:40 comment added Derek Holt Furthermore, in the example in Section 9 of that paper, $G_1$ has order $2$ and, for $i>1$, $Z(G_i) < G_i'$, so it is easy to see that $G$ and $G'$ are not isomorphic. So we do indeed have an example of a group with cyclical derived seriesof period $2$.
Aug 20, 2019 at 15:32 history edited YCor CC BY-SA 4.0
added remarks from discussion on MathSE to avoid repetition here.
Aug 20, 2019 at 15:15 comment added Derek Holt YCor's Question (a) is considered in this paper by B.H. Neumann, and some examples are constructed in Section 9 and 10.
Aug 20, 2019 at 15:02 comment added Santana Afton @YCor The free group on countably many generators should be one such example.
Aug 20, 2019 at 15:02 comment added Derek Holt @YCor what about a free group of countably infinite rank?
Aug 20, 2019 at 14:50 comment added user44191 @YCor a couple ideas that might make (a) easier: instead of terminating, you could use a biinfinite sequence. Alternatively, you could terminate in any perfect group, not the trivial one, if I'm not mistaken.
Aug 20, 2019 at 14:42 comment added YCor Let $(G_n)$ be a sequence of groups with $G_0=1$ and $[G_n,G_n]$ isomorphic to $G_{n-1}$. Consider $G=\bigoplus_{n\ge 1}G_{2n}$. Then its derived subgroup $G'$ is isomorphic to $\bigoplus_{n\ge 1}G_{2n-1}$, and its second derived subgroup is isomorphic to $G$. Next we have (a) to exhibit such a sequence $(G_n)$, and ensure (b) that $G$ is not isomorphic to $G'$. Quite surprisingly (a) is not obvious to me (I don't expect (b) to be an issue but it makes no sense before having an example).
Aug 20, 2019 at 13:49 history asked Santana Afton CC BY-SA 4.0