Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $G/N$. – user27126 Nov 30 '12 at 23:00
There is a little typo : The serie starts at $G^0=G$ so it can't be increasing. – Thibaut Dumont Nov 30 '12 at 23:01
up vote 5 down vote accepted

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?

share|cite|improve this answer
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... – DonAntonio Dec 1 '12 at 0:22
@DonAntonio I am using the notation for the lower central series as given by OP. – Alexander Gruber Dec 1 '12 at 0:25
Oh, I see @Alexander. Thanks. – DonAntonio Dec 1 '12 at 0:26
  • 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 $G^{(j)}=\bigcap_{\ell\ge0} G^{(\ell)}\le N$.

share|cite|improve this answer
Thank you all a lot for the help!!! I'll now try to "connect the dots" :) – Sara Dec 1 '12 at 0:29
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. – Sara Dec 1 '12 at 17:00
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! – Sara Dec 1 '12 at 17:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.