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This is part of a theorem in Dummit & Foote, but the proof is omitted. I want to prove it by myself but I'm starting to give up:

Def. $Z^0(G)=1$, $Z^1(G)=Z(G)$, ..., $Z^{i+1}(G)/Z^i(G)=Z(G/Z^i(G))$. The series $Z^0(G)\le Z^1(G)\le\dots$ is the upper central series of $G$. $G$ is of nilpotence class $c$ iff. $Z^c(G)=G$ but $Z^{c-1}(G)<G$.

Def. $\gamma_0(G)=G$, $\gamma_1(G)=[G,G]$, ..., $\gamma_{i+1}(G)=[G,\gamma_i(G)]$. The series $\dots\le\gamma_1(G)\le\gamma_0(G)$ is the lower central series of $G$.

Thm. $G$ is of nilpotence class $c$ iff. $\gamma_c(G)=1$ but $\gamma_{c-1}(G)>1$. If $G$ is of nilpotence class $c$, then $$Z^0(G)\le\gamma_{c-1}(G)\le Z^1(G)\le\gamma_{c-2}(G)\le Z^2(G)\le\dots\le Z^{c-1}(G)\le \gamma_0(G)\le Z^c(G)$$

I tried to show that the upper central series is the fastest ascending central series of $G$. Suppose $1=H_0\le H_1\le\dots\le\ H_n=G$ is a central series of $G$. By def., central series is a normal series.

I want to show that $H_i\le Z^i(G)$ for all $i$. Obviously the result is true for $i=0,1$. Suppose $H_k\le Z^k(G)$.

I'll show that $H_{k+1}\le Z^{k+1}(G)$, i.e. that $H_{k+1}\le\pi^{-1}(Z(G/Z^k(G)))$, i.e. $$H_{k+1}Z^k(G)/Z^k(G)\le Z(G/Z^k(G))-(*)$$ where $\pi$ is the natural projection homomorphism $G\rightarrow G/Z^k(G)$.

Now I only know that $[G,H_{k+1}]\le H_k\le Z^k(G)$, how to continue from here? Or is it even possible to do so?

I also tried to show that the lower central series is the fastest descending central series of $G$. Suppose $1=K_n\le\dots\le K_1\le K_0=G$ is a central series of $G$. Then i show that $\gamma_i(G)\le K_i$ for all $i$. Obviously the result is true for $i=0,1$. Suppose $\gamma_k(G)\le K_k$.

I'll show that $\gamma_{k+1}(G)\le K_{k+1}$. This is way easier: $\gamma_{k+1}(G)=[G,\gamma_k(G)]\le [G,K_k]\le K_{k+1}$. The last inclusion is due to an equiv. def. of a central series which I know how to prove.

But then how to conclude that the two series are of the same length?

For the 'alternating-term' part of the theorem, I have no idea. I can only prove that the result is true for $c=0,1,2$. And I can prove that $\gamma_{c-i}\le Z^i(G)$. We do a backward induction: The reselt is true for $i=c$. Suppose $\gamma_{c-k}\le Z^k(G)$, then $\gamma_{c-(k-1)}(G)=\gamma_{c-k+1}(G)=[G,\gamma_{c-k}(G)]\le [G,Z^k(G)]\le Z^{k-1}(G)$. The last inclusion is direct from how we define the centers of $G$. How to prove that $Z^i(G)\le\gamma_{c-i-1}(G)$??


P.S. As Derek Holt pointed out in the comments, $Z^i(G)\le\gamma_{c-i-1}(G)$ is false in general: let $H$ be a nilpotent group with nilpotence class $3$. Let $G=H\times\Bbb Z_p$. Then $G$ is also nilpotent of class $3$. $Z^1(G)=Z(H)\times\Bbb Z_p$, but $\gamma_1(G)=[G,G]=[H,H]\times[\Bbb Z_p,\Bbb Z_p]=[H,H]\times1\neq Z^1(G)$

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  • $\begingroup$ It seems to me that you have done all the hard work and then for some reason lost your nerve just before the finishing line! $[G,H_{k+1}] \le Z^k(G)$ implies $[G,H_{k+1}Z^k(G)] \le Z^k(G)$, which implies $(*)$. The fact that the upper central series is the fastest ascending shows that no other central series can be longer. Similarly for the descending central series, so they must both have the same length. $\endgroup$ – Derek Holt Sep 16 '17 at 9:04
  • $\begingroup$ @DerekHolt I know how $[G,H_{k+1}Z^k(G)]\le Z^k(G)$ implies $(*)$ now. The '$[G,H_{k+1}]$ implies $[G,H_{k+1}Z^k(G)]\le Z^k(G)$' part was not apparent to me though until after you've said so. It uses the fact that $Z^k(G)\unlhd G$. By the way, why is $Z^i(G)\le\gamma_{c-i-1}(G)$? I have absolutely no clue. $\endgroup$ – user441558 Sep 16 '17 at 9:20
  • $\begingroup$ @DerekHolt Plus, why upper central series being the fastest ascending shows that no other central series can be longer? Isn't it 'shorter' not 'longer'? And for the descending central series, it is also the 'shortest' one, and in fact since both the upper central series and the lower central series are the 'shortest' central series, they must be of the same length, right? $\endgroup$ – user441558 Sep 16 '17 at 9:43
  • $\begingroup$ Yes sorry, I wrote "longer " instead of "shorter" (so i would not do so well in an exam these days), but I think you understand that argument now. $\endgroup$ – Derek Holt Sep 16 '17 at 10:20
  • $\begingroup$ $Z^i(G) \le \gamma_{c-i-1}(G)$ is not true in general. Where did you get that from? For example if $H$ is a nilpotent group of class $3$ and $G = H \times C_p$, then $Z_1(G)$ contains the direct factor $C_p$, but that is not contained in $[G,G]$. $\endgroup$ – Derek Holt Sep 16 '17 at 10:23

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