Question

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

Answer 1

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

$x= y+z= (a_0, pa_1, \dots, p^{n-1}a_{n-1}, 0,0, \dots ) + p^n (0,0,..,0, a_n, pa_{n+1},..)$

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 $\phi \equiv 0$.

Answer 2

One can at least show that there exists a non-trivial 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 non-trivial homomorphism from the ultraproduct to the rationals.

Answer 3

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.

Answer 4

I agree with the first answer.

The ultraproduct in question is pure-injective (by a theorem of Jan Mycielski, Some compactifications of general agebras, Coll. Math., 13 (1964), 1-9). If an abelian group G contains a non-zero homomorphic image of a pure-injective 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), 131-140)), 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 torsion-free abelian group of power less than continuum contains a non-trivial 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.