# Question About the Derived Series and Commutators

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?

Thanks !

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Your inclusion appears to be backwards. –  Tim Duff Sep 29 '12 at 17:33
Did you see that it isn't an inclusion? –  joshua Sep 29 '12 at 21:07
Erm, your "not inclusion" appears to be backwards. –  Tim Duff Sep 30 '12 at 0:58
Definitely not true for n=4. –  user641 Sep 30 '12 at 4:44
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 $F^{(0)} = F$. –  Derek Holt Sep 30 '12 at 11:14

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 $$F^{(k-1)}\leqslant F_{2^{k-2}}\leqslant F_{2^{k-2}-1} \leqslant \cdots \leqslant F_{k}.$$