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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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    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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    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