Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

All groups here are abelian and $p$ is a prime number; I'll say $P$ is a $p$-group if every element of $P$ has finite order which is a power of $p$.

Suppose $\mathrm{Hom}(G,P) = 0$ for every $p$-group $P$. Does it follow that $\mathrm{map}_*(K(G,n), K(P,m))\sim *$ for all $p$-groups $P$ and all $m,n\geq 1$?

Certainly it is true if $G$ is a finitely generated group or an arbitrary torsion group (all elements of finite order); and it is true for any $G$ in the special case $m = n$. But I worry about the possibility of some oddball group with elements of infinite order such that $[ K(G,n), K(P,n+k)] \neq *$ for some $k > 0$ and some $p$-group $P$.

share|improve this question
add comment

1 Answer

up vote 7 down vote accepted

Yes. Your hypothesis on $G$ means that $G$ is torsion and prime to $p$: every element of $G$ has finite order prime to $p$. This in turn implies that the integral homology group $H_mK(G,n)$ is torsion and prime to $p$, for every $m,n>0$. That in turn implies the conclusion ($H^m(K(G,n);P)=0$ for $m,n>0$ and $P$ a $p$-group) by universal coefficients, since both Hom and Ext vanish for two torsion abelian groups of which one is prime to $p$ and the other is a $p$-group.

I guess the main point is that if $G$ has an element of infinite order then it has a nontrivial homomorphism to a $p$-group, because (1) it then has a nontrivial homomorphism to $\mathbb Q$, and (2) every nontrivial subgroup of $\mathbb Q$ has a nontrivial homomorphism to a $p$-group.

share|improve this answer
    
The second paragraph is perfect for what I have in mind. Thanks! –  Jeff Strom Feb 6 '11 at 0:27
add comment

Your Answer

 
discard

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.