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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... –  DonAntonio Dec 31 '12 at 12:33
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@DonAntonio: He uses what you noted first. See math.stackexchange.com/a/209526/8581 –  Babak S. Dec 31 '12 at 12:36
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$G'\subseteq Z(G)\Longleftrightarrow L_3(G)=[G',G]=1$ –  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
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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. :) –  Babak S. Dec 31 '12 at 13:39

1 Answer 1

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 $G''=1$.

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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
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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! –  Derek Holt Jan 1 '13 at 0:12

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