ultrapowers - Ultraproducts of finite cyclic groups

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 $phi equiv 0$.