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Another problem about nilpotent groups I cannot get a grip on:


Let $G$ be a nilpotent group. If $G/D(G)$ is finite (resp. countable), then so is $G$.


I've tried to use induction looking at the derived series, but nothing has come of it, and that's not a surprise, since it says "nilpotent" and not "solvable". Apart from that, I'm completely lost.

Any kind of help will be highly appreciated.

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    @Arturo: Thanks for the hint, it has helped me a great deal. There is one problem left, I will ask it as a comment to Derek's answer.2011-11-28

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Let me expand a little on Arturo's comment. Let $G = G_1 > G_2 > \cdots > G_{r+1}=1$ be the lower central series of $G$, where $G$ is nilpotent of class $r$. (So $G_2$ is the derived group.) By definition, $G_{k+1} = [G_1,G_k]$. Using the basic commutator identities, you can show that the commutator map $(g,h) \to [g,h]$ induces a bilinear map $G_1/G_2 \times G_{k-1}/G_k \to G_k/G_{k+1}$, and hence there is a homomorphism $G_1/G_2 \otimes G_{k-1}/G_k \to G_k/G_{k+1}$ So if $G_1/G_2$ and $G_{k-1}/G_k$ are both finite, then so is $G_k/G_{k+1}$. It follows by induction on $k$ that if $G_1/G_2$ is finite or countable then so is $G_k/G_{k+1}$ for all $k$ and hence so is $G$.

Similarly, if $G_1/G_2$ is a torsion group then so is $G$, and you can use this to show that the torsion elements of any nilpotent group form a subgroup.

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    @Stefan: A little bit, but not much; you just need to know that you have a group. $G_k$ is generated by elements of the form $[x,g]$ with $x\in G_1$ and $g\in G_{k-1}$; the set-theoretic map $G_1\times G_{k-1}\to G_k$ then factors through $G_{1}/G_2 \times G_{k-1}/G_k$ and has image that contains a generating set. Once you know that this map induces a homomorphism from the group $(G_1/G_2)\otimes (G_{k-1}/G_k)$ (by showing the map on the cartesian product is bilinear), then the image contains a generating set, hence is the whole thing.2011-11-29