# The derived subgroup of an infinite nilpotent group has infinite index

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.

• Is $D(G) = [G,G]$, the derived/commutator subgroup? If $G_n$ is the $n$th term of the lower central series, then there is a surjection from $(G/G')\otimes\cdots\otimes(G/G')$ ($n$ factors) onto $G_n)/G_{n+1}$. Then use induction on the class. – Arturo Magidin Nov 27 '11 at 0:36
• @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. – Stefan Nov 28 '11 at 20:36

## 1 Answer

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.

• Thank you! Could you (or @Arturo Magidin) elaborate on the proof that $g\equiv g' (\text{mod } G_2)$ implies $[g,h]\equiv[g',h] (\text{mod } G_k)$? – Stefan Nov 28 '11 at 20:38
• @Stefan: $[G_k,G_{\ell}]\subseteq G_{k+\ell}$ for any $G$. Therefore, if $x\in G_2$ and $h\in G_{k-1}$, then $[x,h]\in G_{k+1}$. Since $[xy,z] = [x,z]^y[y,z]$, then $[gx,h] = [g,h]^x[x,h] \equiv [g,h]\pmod{G_{k+1}}$, since $G_{k}$ is central in $G/G_{k+1}$, so $[g,h]^x=[g,h]$, and $[x,h]\in G_{k+1}$. – Arturo Magidin Nov 28 '11 at 21:00
• @Stefan: For the simple case of $k=2$, you can just note that $[G_2,G_{\ell}] = [G,G,G_{\ell}]$. Since $[x,y^{-1},z]^y[y,z^{-1},x]^z[z,x^{-1},y]^x=1$, $[G,G,G_{\ell}]$ is generated by products of conjugates of elements in $[[G,G_{\ell}],G]$ and $[[G_{\ell},G],G]$, but both of these are plainly contained in $G_{\ell+2}$, so $[G_2,G_{\ell}]\subseteq G_{\ell+2}$. – Arturo Magidin Nov 28 '11 at 21:10
• @Arturo: Thank you. I think I understand the proof now. Do I assume rightly that the notion of the tensor product is indispensable in showing that the mapping $G_1/G_2 \times G_{k-1}/G_k \to G_k/G_{k+1}$ is surjective? – Stefan Nov 29 '11 at 20:25
• @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. – Arturo Magidin Nov 29 '11 at 20:30