Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $G$ be a group such that $G/G'$ is a divisible group of finite "general" rank. Suppose also that $G''={1}$. Then $G'\leq Z(G)$.

Is it possible? How can we show that? (I really have no ideas.)

Mal'cev (Mal'cev, "On groups of finite rank" Math. Sb. 22, 351-352 (1948)) defines the "general rank" of a group $G$ to be either $\infty$ or the least positive integer $R$ such that every finitely generated subgroup is contained in a $R$-generated subgroup of $G$.

share|cite|improve this question
This is a little confusing: did you actually mean Prufer rank? Because "rank of abelian group" doesn't go well with divisible groups, which they all are direct sums of the rationals and/or prufer groups... – DonAntonio Dec 31 '12 at 12:33
@DonAntonio: He uses what you noted first. See – Babak S. Dec 31 '12 at 12:36
$G'\subseteq Z(G)\Longleftrightarrow L_3(G)=[G',G]=1$ – Babak S. Dec 31 '12 at 12:52
What are you suggesting me @Babak? Sorry, I don't get it. I know that relation but: how can we use it? In particular, I miss how can we use that $G/G'$ is a divisible group... – W4cc0 Dec 31 '12 at 13:29
Honestly, you put me in a challenging problem so I have been thinking of it. What I could get: $G$ is solvable, $G/G'$ is as a diret sums of $Z(p^{\infty})$ for some $p\in P$ and is torsion and has no maximal subgroup. I added that relation, maybe someone is inspired to solve the problem. :) – Babak S. Dec 31 '12 at 13:39
up vote 3 down vote accepted

Let $W$ be the restricted wreath product $C_q \wr H$, where $H = Z(p^\infty)$ and $p,q$ are distinct primes. So $W$ is a semidirect product $B \rtimes H$, where $B$ is the base group of the wreath product. Now $B$ is a (restricted) direct product of countably infinitely many copies of $C_q$, and it has a subgroup $C$ of index $q$, normal in $W$, consisting of those elements for which the sum of the coefficients is $0$ modulo $q$.

I think $G := C \rtimes H$ is a counterexample to your question. We have $G'=C$, $G/G' \cong H$ and $G''=1$.

share|cite|improve this answer
Is $C_q$ the cyclic group of order $q$? If so, $C<B$ (with index $q$) is the direct product of all the cited copies of $C_q$ except one. So itself is a $C_q$. Then $C$ isn't necessarily normal in $W$, I think... – W4cc0 Dec 31 '12 at 19:44
Yes $C_q$ is the cyclic group of order $q$. The subgroup $C$ that I have defined is certainly normal in $W$. If, for example, $q=5$, the it would contain elements like $(\ldots,0,0,2,0,1,2,0,0,\ldots)$, $(\ldots,0,0,1,3,0,2,4,0,0,\ldots)$, where all other entries are $0$. – Derek Holt Dec 31 '12 at 22:28
I see... Why $C$ has index $q$ in B? And $C$ is characteristic in $W$ I suppose... Take (...,0,0,2,0,3,...), $h\in H$, $2^h=2*((1')^h)$ and $3^h=3*((1'')^h)$; how can we say that $((1''))^h$ and $(1')^h$ is such that the sum of coefficients is still $0$ modulo $q$? – W4cc0 Dec 31 '12 at 22:52
The map sending an element of $B$ to the sum of its coefficients is a surjective homomorphism from $B$ to $C_q$ with kernel $C$, so $C$ has index $q$ in $B$. I don't understand what you mean by $2^h = 2*(1')^h)$. By definition of wreath product, $H$ acts on $B$ by permuting the coefficients, so it is clear that $C$ is normalized by $H$. 2013 has just arrived here, so Happy New Year! – Derek Holt Jan 1 '13 at 0:12

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.