Let G be the (nonprincipal) ultraproduct of all finite cyclic groups of orders n!, n=1,2,3,... . Is there a homomorphism from G onto the infinite cyclic group?

$\begingroup$ Is this ultraproduct unique (i.e., it doesn't depend on the ultrafilter)? It seems to me as though it should be, but I can't prove it. Then again, I don't have much experience with these things. $\endgroup$ – Peter Shor Aug 31 '10 at 15:38

$\begingroup$ @Peter Shor: the ultrapower of the reals (i.e. the hyperreals) is independent of the nonprincipal ultrafilter you choose...so long as you believe GCH. Maybe something similar is needed here. $\endgroup$ – M T Sep 9 '11 at 22:01
I think the answer is no. The ultraproduct $U$ is naturally a quotient of ${\mathbb Z}^{\infty}$, the direct product of countably many copies of ${\mathbb Z}$. In the obvious quotient map, the image of the direct sum is zero. Now, it is enough to show that:
Claim: Any homomorphism $ \phi: {\mathbb Z}^{\infty} \to {\mathbb Z}$ that vanishes on the direct sum is identically zero.
Proof: (I learned this from a book by T.Y.Lam):
For a prime number $p$, let $A_p$ be the set of elements in ${\mathbb Z}^{\infty}$ of the form $(a_0, pa_1, p^2a_2,...)$, i.e. the elements whose $i$th coordinate is divisible by $p^i$. Any element $x \in A_p$ can be decomposed as
$x= y+z= (a_0, pa_1, \dots, p^{n1}a_{n1}, 0,0, \dots ) + p^n (0,0,..,0, a_n, pa_{n+1},..)$
Now, $y$ is in the direct sum, hence $\phi(y)=0$. Also $\phi(z) \in p^n {\mathbb Z}$, which implies that $\phi (x) \in \cap_{n=1}^{\infty} p^n {\mathbb Z} = \{ 0 \}$
Now, choose two distinct primes $p$ and $q$. Since $\gcd(p^n,q^n)=1$, it is easy to see that $A_p+A_q= {\mathbb Z}^{\infty}$. This implies that $\phi \equiv 0$.

$\begingroup$ Why can you assume that $x$ is of the form $(a_0,2a_1,4a_2,\dots)$? $\endgroup$ – Stefan Geschke Aug 31 '10 at 10:53

$\begingroup$ I have just edited the question for clarity. I had seen this argument before, but I had forgotten it. Note that it actually implies that the dual of the product is the sum. $\endgroup$ – Tom Goodwillie Aug 31 '10 at 11:36

$\begingroup$ Sorry, I was in a run, so the proof was not clear. I just rewrote it. $\endgroup$ – Keivan Karai Aug 31 '10 at 11:41

$\begingroup$ Sorry Tom, I think we started to edit almost at the same time! $\endgroup$ – Keivan Karai Aug 31 '10 at 11:42

$\begingroup$ I guess I was in a hurry, too, when I wrote "question" for "answer". $\endgroup$ – Tom Goodwillie Aug 31 '10 at 12:27
One can at least show that there exists a nontrivial homomorphism to the rationals.
Call a sequence $a=(a_n)_{n \in {\mathbb N}}$ with $a_n \in {\mathbb Z}/n! {\mathbb Z}$ bounded if there exists a constant $C$ such that $\tilde a_n \leq C$ for all $n$, where $\tilde a_n$ is the representative of $a_n$ with $n!/2 < \tilde a_n \leq n!/2$. Let $L$ be the set of bounded elements in the ultraproduct.
For $a \in L$, $\phi(a):= \lim_{n \to \omega} \tilde a_n$ defines a homomorphism $\phi \colon L \to {\mathbb Z}$. Considered as a homomorphism to ${\mathbb Q}$ it has an extension to the whole ultraproduct, since ${\mathbb Q}$ is an injective abelian group. Hence, there exists a nontrivial homomorphism from the ultraproduct to the rationals.

$\begingroup$ Do you mean the ultrafilter limit with respect to the ultrafilter used to form the ultraproduct when you write $\lim_{n\to\omega}\tilde a_n$? $\endgroup$ – Stefan Geschke Aug 31 '10 at 13:26

$\begingroup$ Yes. However, to get the conclusion, one can also take an arbitrary nontorsion element and build a similar extension. (Which looks even easier.) $\endgroup$ – Andreas Thom Aug 31 '10 at 14:10
I think in certain cases, say if the set of prime numbers belongs to the ultrafilter, $G$ becomes a vector space over the field of rational numbers.

3$\begingroup$ The orders of the cyclic groups in the question are $n!$ not $n$. Hence, the ultraproduct always contains a lot of torsion. $\endgroup$ – Andreas Thom Aug 31 '10 at 9:53
I agree with the first answer.
The ultraproduct in question is pureinjective (by a theorem of Jan Mycielski, Some compactifications of general agebras, Coll. Math., 13 (1964), 19). If an abelian group G contains a nonzero homomorphic image of a pureinjective group, then G is not slender. However, the infinite cyclic group is slender (by a theorem of Specker from 1950 (E. Specker, Additive Gruppen von folgen ganzer Zahlen, Portugaliae Math. 9 (1950), 131140)), which is proved above as the Claim.
One might add: by a theorem of Sasiada (in L. Fuchs, Infinite Abelian Groups, vol.2, 1973), no reduced torsionfree abelian group of power less than continuum contains a nontrivial homomorphic image of the ultraproduct in question.
The results quoted above are also to be found in P.C. Eklof, A.H. Mekler, Almost Free Modules, rev. ed., 2002.