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I want to solve the following exercise:

Let $G$ be a group. $G^{0}\geq G^{1} \geq G^{2}\dots$ is a lower central series, s.t. $G^{0} = G$ and $G^{i+1} = [G,G^{i}]$. Let $N$ be a normal subgroup of $G$.

a) Then $G/N$ nilpotent $\Rightarrow$ $\bigcap\limits_{i\geq 0}G^{i} \subset N$.
b) With $G$ finite, $\bigcap\limits_{i\geq 0}G^{i} \subset N$ $\Rightarrow$ $G/N$ nilpotent.


So far I just have ideas to a):

I found a Corrolary that says: $N$ normal, $G/N$ nilpotent and $N\leq Z_{i}(G)$ for some $i$ (where it notates the upper central series), then $G$ is nilpotent.

I thought, if I could show that $N\leq Z_{i}(G)$, then with $G$ nilpotent, the lower central series terminates and the intersection would be trivial (? is that right?) and so contained in $N$.

Is that the right way? Or should I better approach that exercise differently?

Thanks and best, Sara

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    $\begingroup$ A lower central series of $G$ can be transferred to a lower central series for $G/N$, by considering $G^{i}N/N$. Try to relate this lower central series with nilpotence of $G/N$. $\endgroup$ – user27126 Nov 30 '12 at 23:00
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    $\begingroup$ There is a little typo : The serie starts at $G^0=G$ so it can't be increasing. $\endgroup$ – Thibaut Dumont Nov 30 '12 at 23:01
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No need for any of the book's lemmas! Instead, understand what's going on. Intuitively, what you're being asked to prove is that whenever $\bigcap_{i\geq 0} G^i$ is nontrivial (and thus $G$ is not nilpotent), you can make $G$ into a nilpotent group by quotienting out by $\bigcap_{i\geq 0} G^i$. In fact, $\bigcap_{i\geq 0} G^i$ is the smallest such subgroup so that this quotient is nilpotent.

What do you need to do to prove this?

  • What is a nilpotent group?

This one depends on what definition you're using, as there are a bunch of equivalent definitions, but I assume you're using this one: a group $K$ is nilpotent iff $K^n=1$ for some $n$, where $K^n$ (as in your notation) is the $n$th term in the lower central series.

  • What is $\bigcap_{i\geq 0} G^i$?

Surely $G^{i+1}\subseteq G^i$. (Prove this real quick.) Thus, if $\bigcap_{i\geq 0} G^i$ is not trivial, there must be some $m$ for which $G^m= G^k$ for every $k\geq m$, and so $G^m=\bigcap_{i\geq 0} G^i$.

  • Determine the relationship between $G^i$ and $(G/N)^i$.

What does it say about $G^i$ when $(G/N)^i$ is trivial? What does it say about $(G/N)^i$ when $G^i\leqslant N$? (If you're still stuck: can you rewrite $(G/N)^i$ in terms of $G^i$ and $N$)?

Can you connect the dots for me?

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  • $\begingroup$ I seem to not understand your definition of nilpotency: what is for you $\,K^n\,$ when $\,K\,$ is a group? For me it is the subgroup generated by the $\,n-$th powers of all elements of $\,K\,$, but this would make your definition false, since for example $\,S_3^6=1\,$ but, of course, $\,S_3\,$ isn't nilpotent... $\endgroup$ – DonAntonio Dec 1 '12 at 0:22
  • $\begingroup$ @DonAntonio I am using the notation for the lower central series as given by OP. $\endgroup$ – Alexander Gruber Dec 1 '12 at 0:25
  • $\begingroup$ Oh, I see @Alexander. Thanks. $\endgroup$ – DonAntonio Dec 1 '12 at 0:26
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  • Sub-exercise 1. Show for any subset $A$ and subgroup $B$ of $G$, $AB\le B\implies A\le B$.

  • Sub-exercise 2. Show that for any $L,\,M\le G$ and $N\trianglelefteq G$, we have $$\left[\frac{L}{N},\frac{M}{N}\right]=\frac{[L,M]N}{N}.$$

  • Sub-exercise 3. By induction and the above, show that $(G/N)^{(i)}=G^{(i)}N/N$.

As corollary, deduce part (a) by relating $(G/N)^{(i)}$, $G^{(i)}$ and $N$ together using $(G/N)^{(i)}=N/N$, which comes from the hypothesis that $G/N$ is nilpotent.

  • Sub-exercise 4. Argue that if $G$ is finite, then $G^{(j)}=G^{(j+1)}=\cdots$ for some $j$. Subsequently, prove that this final term $G^{(j)}$ is equal to the intersection $\bigcap_{\ell\ge0}G^{(\ell)}$.

Thus deduce part (b) as corollary by considering $(G/N)^{(j)}$ in view of $G^{(j)}=\bigcap_{\ell\ge0} G^{(\ell)}\le N$.

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  • $\begingroup$ Thank you all a lot for the help!!! I'll now try to "connect the dots" :) $\endgroup$ – Sara Dec 1 '12 at 0:29
  • $\begingroup$ I must admit I feel still a little confused with that, but here is my try: First: $(G/N)^{i} = G^{i}N/N$. I just do the induction step: $i \mapsto (i+1):$ $(G/N)^{(i+1)} = ((G/N)^{i})'$ (with $G' = [G,G]$.). $((G/N)^{i})'$ is generated by $[xN,yN] = [x,y]N$ with $x,y\in G^{i}$, $N$ normal in $G$. Since $\{[x,y]| x,y\in G^{i}\}$ is the system of generators of $G^{(i+1)}$ follows the assumption. $\endgroup$ – Sara Dec 1 '12 at 17:00
  • $\begingroup$ Now I assume $(G/N)$ is nilpotent. Then $(G/N)^{i}=e$ for some $m$, and for all $k\geq m:$ $(G/N)^{k}=(G/N)^{m}=e$. Then with the above property: $(G/N)^{m} = G^{m}N/N = e$. This is the case, if $G^{m}\subset N$. And with $G^{k}N/N=(G/N)^{k}=(G/N)^{m}=G^{m}N/N$ follows $G^{m}=G^{k}$ for some $m$ and all $k\geq m$. Hence $\bigcap\limits_{i\geq 0}G^{i} = G^{m}\subset N$. Is that ok like this? Thanks a lot! $\endgroup$ – Sara Dec 1 '12 at 17:05

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