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Dec 4, 2015 at 13:45 answer added Yakov timeline score: 3
Aug 5, 2013 at 17:37 vote accept Yassine Guerboussa
Aug 5, 2013 at 17:34 vote accept Yassine Guerboussa
Aug 5, 2013 at 17:37
Aug 5, 2013 at 17:34 vote accept Yassine Guerboussa
Aug 5, 2013 at 17:34
Jul 25, 2013 at 6:19 history edited Alexander Chervov
tag p-groups
Jul 25, 2013 at 2:16 answer added Will Sawin timeline score: 5
Jul 24, 2013 at 23:32 answer added Marty Isaacs timeline score: 7
Jun 19, 2013 at 16:45 comment added Yassine Guerboussa @ Yves Cornulier: I mean the nilpotency class. Thank you for your last comment.
Jun 19, 2013 at 8:45 comment added YCor Anyway, OK I'm convinced that the dihedral group of order $2^n\ge 16$ satisfies $n(G)=2$. Note that it can be embedded in a group of upper triangular matrices over $\mathbf{Z}/2\mathbf{Z}$ with 1 on the diagonal, and the latter satisfies $n(G)=1$. In particular, $n$ can, unlike the nilpotency or solvability length, increase when passing to subgroups.
Jun 19, 2013 at 8:37 comment added YCor what do you call class? is it the nilpotency class, or something else?
Jun 18, 2013 at 14:10 comment added Yassine Guerboussa @Yves Cornulier : There exist p-groups of a maximal class with a maximal subgroup that is abelian (for instance the dihedral groups $D_{2^n}$), if G is such a group with A maximal abelian, then T(G)=A or G has class 2. So if the order of G is large enough, n(G)>1.
Jun 17, 2013 at 20:45 comment added YCor what are examples of $p$-groups $G$ with $n(G)>1$? (i.e, $T(G)\neq G$).
Jun 17, 2013 at 16:17 history edited user9072 CC BY-SA 3.0
lt instead of < to avoid display problems
Jun 17, 2013 at 14:37 history asked Yassine Guerboussa CC BY-SA 3.0