Let $G$ be a group and denote by $Z_i(G)$ the subgroups of its upper central series, i.e. $$ Z_1(G) = Z(G), \quad Z_{i+1}(G) / Z_i(G) = Z(G / Z_i(G)) $$ i.e. the preimage of the center in the quotient $G / Z_i(G)$. Define the subgroup $$ Z_{\infty}(G) = \bigcup_{i=1}^{\infty} Z_i(G). $$ We have $Z_1(G) \le Z_2(G) \le Z_3(G) \le \ldots$.

I want to know if the center of the quotient $G/ Z_{\infty}(G)$ has trivial center, i.e. do we have $$ Z(G / Z_{\infty}(G)) = 1. $$ I can prove this for the finite case, i.e. if $G$ is finite, then we have $Z_{\infty}(G) = Z_k(G)$ for some $k$, as the series can not grow infinitely, then for this $k$ we have $Z_{k+1}(G) = Z_k(G)$, which says that the center of $G/Z_k(G)$ must be trivial.

But does the same hold for infinite groups? I can neither prove or disprove it.

Remark: As pointed out by Derek Holt in his comment, this is not the definition of the hypercentre for infinite groups. But I am particulary interested in this construction, as it is the "hypercentre" as defined in B. Huppert, Endliche Gruppen, page 260. I know this book is specifically about finite groups, but in one exercise it is asked to proof a special property about it if $G$ is finite, but I want to see if it fails if $G$ is infinite without altering the definition of $Z_{\infty}(G)$.

  • $\begingroup$ That is not the standard definition of the hypercenter of a group. You have to take the limit over all ordinals, not just the finite ordinals. See en.wikipedia.org/wiki/Central_series $\endgroup$ – Derek Holt Dec 26 '15 at 22:50
  • $\begingroup$ Thanks for pointing out. I did not knew this, because this is called the hypercentre in my book (B. Huppert, Endliche Gruppen I). I changed the text and avoided to speak about the hypercentre. But I am still interested in this particular construction and the center of its quotient. $\endgroup$ – StefanH Dec 26 '15 at 23:00
  • $\begingroup$ Your definition is equivalent to the general one for finite groups and Huppert's book is about finite groups. Your question is answered in the wikipedia article: "For every ordinal $\lambda$ there is a group $G$ with $Z_\lambda(G) = G$, but $Z_\alpha(G) \ne G$ for $\alpha < \lambda$, (Gluškov 1952) and (McLain 1956)". $\endgroup$ – Derek Holt Dec 26 '15 at 23:08
  • $\begingroup$ Just to relate this to my question, because I am not that familiar with ordinals: the $\alpha$'s are the natural numbers, and $Z_{\lambda}(G) = Z_{\infty}(G)$? But then it would not be a counter-example to the claim that the center of $G/Z_{\infty}(G)$ is always trivial? $\endgroup$ – StefanH Dec 26 '15 at 23:14
  • $\begingroup$ Take $\lambda = \omega+1$ to get a counterexample - $\omega$ is the first infinite ordinal. $\endgroup$ – Derek Holt Dec 26 '15 at 23:18

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