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(1) Suppose $\pi$ is a set of primes and $G$ is a $\pi$-divisible nilpotent group, i.e., for any $g \in G$ and $p \in \pi$, there exists $x \in G$ such that $x^p = g$. Is it necessary that all the homology groups of $G$ for the trivial group action with coefficients in $\mathbb{Z}$ are also $\pi$-divisible groups? I am in particular interested in the second homology group $H_2(G;\mathbb{Z})$, which is the Schur multiplier of $G$. Note that the result is true for $H_1(G;\mathbb{Z})$, which is the abelianization of $G$.

I believe that the following example shows that the result is false for $H_2(G;\mathbb{Z})$, but I'm not sure whether I am applying results correctly. The example group $G$ that I have in mind is defined as follows. Let $UT(3,\mathbb{Q})$ denote the group of $3 \times 3$ upper triangular matrices over the rational numbers that have 1s on the diagonal. Let $Z$ be an infinite cyclic subgroup of $UT(3,\mathbb{Q})$ generated by any non-identity element in its center. Let $G = UT(3,\mathbb{Q})/Z$. Then, $G$ is divisible by all primes, on account of being a quotient of $UT(3,\mathbb{Q})$, which is divisible by all primes. $G$ admits $UT(3,\mathbb{Q})$ as a stem extension group and the base of the extension is $Z$, isomorphic to $\mathbb{Z}$, which is not divisible by any prime. The base of any stem extension is a quotient of the Schur multiplier, so the Schur multiplier (which I haven't computed precisely) should not be divisible by any prime.

(2) Suppose $\pi$ is a set of primes and $G$ is a $\pi$-powered nilpotent group, i.e., for any $g \in G$ and $p \in \pi$, there exists unique $x \in G$ such that $x^p = g$. Is it necessary that all the homology groups of $G$ for the trivial group action with coefficients in $\mathbb{Z}$ are also $\pi$-powered groups? I am in particular interested in the second homology group $H_2(G;\mathbb{Z})$, which is the Schur multiplier of $G$. Note that the result is true for $H_1(G;\mathbb{Z})$, which is the abelianization of $G$ (the proof of this uses that $G$ is nilpotent).

Can anything be said in general, in the non-nilpotent case?

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In part (2), do you really mean to say "$G$ is a $\pi$-powered nilpotent group", given that you final note is that "for $G$ nilpotent..."? That is, was (2) supposed to just say "$G$ is a $\pi$-powered group", without the assumption that it is nilpotent? –  Arturo Magidin Apr 11 '13 at 21:28
    
I'm interested in the nilpotent case, but would like more general results if they exist. I'll modify the potentially confusing language. –  Vipul Naik Apr 11 '13 at 23:26

2 Answers 2

up vote 5 down vote accepted

In the nilpotent case - YES: Let $\pi$ be any set of primes. A group $K$ is called $\pi$-local if for every $n$ with no prime divisors in $\pi$ the $n$th power function $k\mapsto k^n$ is a bijection $K\to K$. If $G$ is any nilpotent group then there exists a $\pi$-localization of $G$ - it is the initial homomorphism $$h:G\to G_\pi$$ into a $\pi$-local group. Initial means that if $k:G\to K$ is any homomorphism with $K$ being $\pi$-local then $k$ uniquely factors through $h$. Now every $\pi$ -localization of a nilpotent group induces a $\pi$-localization on the reduced homology $$h_\ast:H_\ast(G;\mathbb{Z})\to H_\ast(G_\pi;\mathbb{Z})$$ Your question follows. You can find the details in P. Hilton, G. Mislin and J. Roitberg. Localization of Nilpotent Groups and Spaces. North-Holland Math. Studies no. 15 (North-Holland, 1975).

I am not aware of any so elegant results along these lines for more general groups, you may read about virtually nilpotent groups in C. Casacuberta and M. Castellet. Localization methods in the study of the homology of virtually nilpotent groups. Math. Proc. Cambridge Philos. Soc. 112 (1992), no. 3, 551–564.

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Does your answer say, that $H_2(G,\mathbb{Z})$ is $\pi$-divisible if $G$ is nilpotent, $\pi$-divisible ? –  Demin Hu Apr 12 '13 at 13:58
    
@Demin - yes, if the "reduced homology" is unclear, I should have written "for homology in positive degrees". In the case of group homology these conditions coincide. Also the converse is true - if all the integral homologies in positive degrees are $\pi$-divisible then the nilpotent group $G$ is also $\pi$-divisible. –  Adam Przeździecki Apr 13 '13 at 5:10
    
Thanks for the explanation. –  Demin Hu Apr 13 '13 at 15:53
    
@Adam: Do you mean $\pi$-powered in the comment above, rather than $\pi$-divisible? –  Vipul Naik Apr 16 '13 at 2:52
    
$\mbox{Erratum to my comment above:}$ yes Vipul, I meant uniquely $\pi$-divisible (same as $\pi$-powered). Thank you for noticing. My response to Demin's comment is now - no, my answer does not say anything about nilpotent groups which are $\pi$-divisible, but not uniquely $\pi$-divisible. –  Adam Przeździecki Apr 16 '13 at 13:17

For $G=UT(3,\mathbb{Q})/Z$, I computed $H_2(G,\mathbb{Z})=\mathbb{Z}$, which confirms that $G$ is a counterexample to (1). (This doesn't contradict Adam's answer, since $G\supseteq \mathbb{Q}/\mathbb{Z}$ isn't uniquely divisible)

The essential steps in my computation are (coefficients are always $\mathbb{Z}$ which are suppressed):

  • The LHS spectral sequence of the central extension $\mathbb{Q}/\mathbb{Z} \hookrightarrow G \twoheadrightarrow \mathbb{Q}^2$ yields $H_3(G)=0$ and $H_2(G) = E^3_{2,0} \le H_2(\mathbb{Q}^2)=\mathbb{Q}$.

  • The LHS of the central extension $\mathbb{Q} \hookrightarrow UT(3,\mathbb{Q}) \twoheadrightarrow \mathbb{Q}^2$ yields $H_2(UT(3,\mathbb{Q}))=\mathbb{Q}^2$

  • The LHS of the central extension $\mathbb{Z} \hookrightarrow UT(3,\mathbb{Q}) \twoheadrightarrow G$ has $$E^2=\begin{array}{cccc} |0 & & & & \newline |\mathbb{Z} & \mathbb{Q}^2 & & \newline |\underline{\mathbb{Z}} & \underline{\mathbb{Q}^2} & \underline{H_2(G)} & \underline{0} \newline \end{array}$$ Since $UT(3,\mathbb{Q}) \twoheadrightarrow G$ induces an isomorphism on the abelianization, we have $E^3_{01}=0$. Hence $d^2: H_2(G)\to \mathbb{Z}$ is epi. For $K = \ker d^2$ there is a s.e. sequence $\mathbb{Q}^2 \hookrightarrow H_2(UT(3,\mathbb{Q})) \twoheadrightarrow K$. Tensoring yields $K\otimes \mathbb{Q}=0$. If $K \neq 0$ there is $\mathbb{Z} \hookrightarrow K$ (since $K\le H_2(G)\le \mathbb{Q})$ and tensoring gives the contradiction $\mathbb{Q} \hookrightarrow K\otimes \mathbb{Q}=0$. Hence $K=0$ and $H_2(G)\cong \mathbb{Z}$ as claimed.

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Thanks a lot! I haven't used spectral sequences much in the past, so I will need to read this later to understand it. I appreciate the work you put into confirming my example. –  Vipul Naik Apr 16 '13 at 2:53

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