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Sep 29, 2016 at 13:22 vote accept stupid_question_bot
Sep 29, 2016 at 13:21 history edited stupid_question_bot CC BY-SA 3.0
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Sep 28, 2016 at 21:28 comment added HJRW @ArturoMagidin, of course, thanks for the reminder.
Sep 27, 2016 at 22:00 comment added Arturo Magidin @HJRW: There are no varieties in which all groups are finite, but there certainly are varieties in which the finitely generated relatively free groups are finite (for example, the Burnside varieties of exponents $3$, $4$, and $6$ are locally finite, hence the finitely generated relatively free groups are finite); likewise, the relatively free groups of finite rank in $\mathfrak{N}_2\cap\mathfrak{B}_n$ are all finite (as are, trivially, the finitely generated relatively free groups in the varieties of abelian groups of exponent $n>1$).
Sep 27, 2016 at 21:57 answer added Luc Guyot timeline score: 7
Sep 27, 2016 at 20:30 comment added Derek Holt I think that $C_n \times C_n$ works only for small $n$ (probably $n=2,3,4,6$) because all automorphisms induced by elements of ${\rm Aut}(F)$ have determinant $\pm 1$.
Sep 27, 2016 at 20:04 comment added HJRW @ArturoMagidin, iirc, there are no varieties of finite groups (and in particular no varieties in which the relatively free groups are finite).
Sep 27, 2016 at 19:26 answer added Derek Holt timeline score: 6
Sep 27, 2016 at 18:16 comment added Arturo Magidin Would $K$ verbal work? The quotient would be the relatively free group of rank $2$ in the relevant variety of groups (your example being the case of $K$ the verbal subgroup generated by $xyx^{-1}y^{-1}$ and $x^n$). (You'd want a variety where the relatively free group is finite, of course...)
Sep 27, 2016 at 17:32 history undeleted stupid_question_bot
Sep 27, 2016 at 17:30 history deleted stupid_question_bot via Vote
Sep 27, 2016 at 17:30 history asked stupid_question_bot CC BY-SA 3.0