It is known that every finitely generated nilpotent group is residually finite. Why finitely generated hypothesis is essential?
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2 Answers
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It is essential in the proof, that the group is finitely generated. A finitely generated nilpotent group is supersolvable, and every supersolvable group is residually finite( a result of K.A. Hirsch). For the first part, the proof uses that the $k$-th group $G_k$ in the lower central series of the group $G$ is finitely generated for each $k$, so that each $G_k/G_{k+1}$ is finitely generated Abelian, i.e., a direct product of finitely many cyclic groups. This is not true for nilpotent groups which are not finitely-generated.
Edit: A possible counterexample is the additive group of $\mathbb{Q}$. It is abelian, hence nilpotent, but not residually finite.
answered Oct 4, 2013 at 18:23
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1$\begingroup$ The question is not about the demonstration. It needed a counter-example? $\endgroup$ Oct 4, 2013 at 18:30
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1$\begingroup$ I added a counterexample. The proof, however, gives more insight (and shows why finitely-generated is essential). $\endgroup$ Oct 4, 2013 at 19:58
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Consider the additive group of rational numbers. It is abelian, but has no nontrivial finite quotients since it is divisible and every quotient of a divisible group is again divisible. Hence, this group is not residually finite.
answered Oct 4, 2013 at 18:47