Given a $p$-group $F$, we define the derived series as follows: $$F^{(1)} = F,\ F^{(n)} = [F^{(n-1)} , F^{(n-1)} ]. $$

I'm now given the lower central series: $F_n = [F,F_{n-1} ]$, $F_1 =F$.

If I know that $F^{(1)} \nsubseteq F_2 $, how can I prove by induction that also $ F^{(n-1)} \nsubseteq F_{n}$?

Does anyone have an idea?


  • 1
    $\begingroup$ Your inclusion appears to be backwards. $\endgroup$
    – user17794
    Sep 29, 2012 at 17:33
  • $\begingroup$ Did you see that it isn't an inclusion? $\endgroup$
    – joshua
    Sep 29, 2012 at 21:07
  • $\begingroup$ Erm, your "not inclusion" appears to be backwards. $\endgroup$
    – user17794
    Sep 30, 2012 at 0:58
  • $\begingroup$ Definitely not true for n=4. $\endgroup$
    – user641
    Sep 30, 2012 at 4:44
  • 2
    $\begingroup$ Note that $F^{(1)} = F \not\subseteq F_2$ is true in any nontrivial finite $p$-group, and $F^{(2)} = F_2$ is true by definition in any group. I am unable to guess what the intended question is. But note that the notation for the derived series is nonstandard - it is more usual to have $F^{(0)} = F$. $\endgroup$
    – Derek Holt
    Sep 30, 2012 at 11:14

1 Answer 1


There seems to be some confusion with the question, so let me write some thoughts to help you along.

The big idea is that iterating brackets completely to one side is the "slowest" way to descend through a group via commutators - for example, $[F,[F,[F,F]]]$ is generally going to be bigger than $[[F,F],[F,F]]$. This is an intuitive way of seeing that nilpotent groups (for which the slower $[F,[F,[F,F]]]$ type bracketing eventually converges to $1$) are a much smaller class than solvable groups (for which the much faster $[[F,F],[F,F]]$ type bracketing eventually converges to $1$). Note that your claim cannot be true for any group which is solvable but not nilpotent, since eventually the derived series will go to $1$ while the lower central series stabilizes nontrivially.

It should help if I prove that this is the slowest way. First, you can show inductively that $[F_i,F_j]\subseteq F_{i+j}$ - let us call this fact $\star$. (You should try this!) Now, if we have $n$ copies of $G$ bracketed together in any form, let us call that a commutator of weight $n$. (For example, $[[F,F],[F,F]]$ and $[F,[F,[F,F]]]$ would both have weight $4$.) What I want to show is that any weight $n$ commutator is contained in $F_n$. Proceed by induction on $n$. Any weight $n$ commutator has the form $[X,Y]$ for a weight $i$ commutator $X$ and a weight $j$ commutator $Y$, so $i+j=n$. By the inductive hypothesis we have $X \leqslant F_i$ and $Y\leqslant F_j$, so by $\star$, $[X,Y]\leqslant [F_i,F_j]\leqslant F_{i+j}=F_n$.

The first example I gave is $F_4$ in your notation, and the second is $F^{(3)}$. You say that you want to show $F^{(3)}\not\subseteq F_4$, but you can see that these are both weight $4$ commutators, so by the above this cannot be true. Furthermore, for the other cases, observe that $F^{(k-1)}$ is a weight $2^{k-2}$ commutator for any $k$. Thus $F^{(k-1)}\leqslant F_{2^{k-2}}$. Since $F_{i+1}\leqslant F_{i}$ for all $i$, it follows that $$F^{(k-1)}\leqslant F_{2^{k-2}}\leqslant F_{2^{k-2}-1} \leqslant \cdots \leqslant F_{k}.$$

  • $\begingroup$ Thanks a lot @Alexander Gruber! $\endgroup$
    – joshua
    Oct 3, 2012 at 9:18

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