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Is the wreath product of two nilpotent groups always nilpotent?

I know the answer is no due to a condition "The regular wreath product A wr B of a group A by a group B is nilpotent if and only if A is a nilpotent p-group of finite exponent and B is a finite p-group for the same prime p ", but I can't easily construct a counter example to show it.

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Take any p-group $P$ and q-group $Q$, and then $P\wr Q$ is not nilpotent, as it has no normal Sylow q-subgroup. –  user641 May 31 '12 at 11:12
    
@SteveD: You should mention your implicit assumption $p\ne q$ ;-). To user31899: To be even more concrete than Steve, take $A$ and $B$ as simple as possible, but violating the condition after "if and only if", e.g., take something like $A=C_3$, $B=C_2$. –  j.p. May 31 '12 at 14:25
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1 Answer 1

Following Jug's suggestion: let $\,\,A:=C_3=\langle a\rangle\,,\,B:=C_2=\langle c\rangle\,$ , with the regular action of $\,B\,$ on itself, and form the (regular) wreath product $$A\wr B\cong \left(C_3\times C_3\right)\rtimes_R C_2$$ Take the elements $$\pi=((1,1),c))\,\,,\,\,\sigma=((a,a^2),1)$$It's now easy to check that$$\pi^2=\sigma^3=1\,\,,\,\,\pi\sigma\pi=\sigma^2$$ so we get that $\,\,\langle \pi\,,\,\sigma\rangle\cong S_3\,\,$ and thus $\,\,A\wr B\,\,$ can't be nilpotent, though both $\,\,A,B\,\,$ are (they're even abelian...)

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