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Would you please construct a locally nilpotent group which is not nilpotent? An example of a locally nilpotent group which is not nilpotent?

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Hint: Try taking a direct product of infinitely many finite nilpotent groups. –  Derek Holt Oct 27 '12 at 15:20
@Derek: you mean "restricted direct product" (aka direct sum). (or you have a nonstandard definition of "direct product") –  YCor Oct 31 '12 at 21:41

2 Answers 2

The Fitting subgroup $\mathrm{Fit}(G)$ of a group $G$ is the product of all nilpotent normal subgroups of $G$. Since the product of finitely many nilpotent normal subgroups is again a nilpotent normal subgroup, $\mathrm{Fit}(G)$ is locally nilpotent. To find your example it therefore suffices to find a nonnilpotent group with $G=\mathrm{Fit}(G)$.

In fact, if $p$ is a prime and $E$ is an infinite elementary abelian $p$-group (i.e. $E$ is abelian of exponent $p$), then the wreath product $\mathbb{Z}_p\wr E$ satisfies this property, but the proof (and even the construction of the wreath product, if you haven't seen it before) is somewhat involved.

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I'll give another example. First, two definitions.

  1. The Prüfer $p$-group or $p$-quasicyclic group, denoted $\mathbb{Z}_{p^{\infty}}$ is the group defined by all complex $p^n$th roots of unity for any $n\in \mathbb{N}$. (Note that this is the Pontryagin dual group to the $p$-adic integers.)

  2. The generalized dihedral group of an abelian group $H$, denoted $\text{Dih}(H)$, is the semidirect product of $H$ by $\mathbb{Z}_2$, where $\mathbb{Z}_2$ acts by inversion on elements of $H$. In other words, in $\text{Dih}(H)=H\rtimes \mathbb{Z}_2$, $(h_1,0)(h_2,a)=(h_1h_2,a)$ and $(h_1,1)(h_2,a)=(h_1h_2^{-1},a\oplus1)$.

The example: $\text{Dih}(Z_{2^\infty})$ is locally nilpotent but not nilpotent.

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$Dih(H)$ is well defined iff $H$ is abelian. Otherwise the inversion map of $H$ is not an automorphism. –  YCor Oct 31 '12 at 21:40
Thanks @YvesCornulier, I forgot to mention that. –  Alexander Gruber Nov 1 '12 at 0:41

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