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Theorem
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.)

Edit
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 $G$.

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    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...2012-12-31
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    @DonAntonio: He uses what you noted first. See http://math.stackexchange.com/a/209526/85812012-12-31
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    $G'\subseteq Z(G)\Longleftrightarrow L_3(G)=[G',G]=1$2012-12-31
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    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...2012-12-31
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    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. :)2012-12-31
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    Why $G/G'$ is a torsion group? It can be a direct sum of (among the other) copies of the additional group of rational, or not? (I'm glad I'm not the only one who has to think about it xD)2012-12-31
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    Rational? I don't think so because it is of finite rank. Isn't it? I am saying this according to what Alexander proved in above link.2012-12-31
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    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 $G$. Sorry for the misunderstanding (if it was), the term rank, today, it's something to refers to with extremely attention. In this case $\mathbb Q$ is locally cyclic and so of finite rank.2012-12-31
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    Anyway I think that "of finite rank" can be ometted without altering the result.2012-12-31
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    Can I ask why you might expect such a result to be true? It doesn't seem very likely!2012-12-31
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    It's just an idea I had to solve another problem... I can't find a countexample anyway...2012-12-31

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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 $G''=1$.

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    Is $C_q$ the cyclic group of order $q$? If so, $C (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...2012-12-31
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    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 $0$.2012-12-31
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    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 $q$?2012-12-31
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    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!2013-01-01