TITLE: The group of additive sequences of integers
QUESTION [15 upvotes]: Let $G$ be the additive group of the sequences $(a_n)_{n\ge 1} \subset \mathbb{Z}$ and let $f:G \to \mathbb{Z}$ be a group homomorphism. We denote by $e_i$ the sequence $(0,0,...,0,1,0,0...)$ (we just have a $1$ on the $i$-th position, everything else is zero). Show that the set $\{i\ge 1 | f(e_i) \ne 0\}$ is finite.

I thought that I should consider the set of sequences of finite support i.e. the subgroup generated by the $e_i$s. Let's denote it by $H$. We have that $f(H)$ is a subgroup of $\mathbb{Z}$, so $f(H)=m\mathbb{Z}$ for some integer $m$. We also have that $f(G)=n\mathbb{Z}$ for some positive integer $n$. But I don't know how to proceed any further. I think that we should aim for a contradiction. I can only think about something having to do with divisibility, but I don't know.

REPLY [3 votes]: So this all appears in Section 94 of Fuchs's Infinite Abelian Groups II, and this is due to Sasiada. The general result is that this property holds whenever the image $G$ is countable and reduced (i.e., has no non-trivial divisible subgroup, where a divisible group is one where, given any $x\in G$ and $n\in \mathbb N$, there exists $y\in G$ such that $x=n\cdot y$). I will write the proof in the case of $G=\mathbb Z$, but it is a simple matter to generalize this.
Let $P$ be the product of countably many copies of $\mathbb{Z}$ and $G=\mathbb{Z}$. Let $\eta:P\to G$ be a homomorphism such that $\eta(e_n)\neq 0$ for infinitely many $n$. Of course, by removing those for which $\eta(e_n)=0$, we may assume that $\eta(e_n)$ is non-zero for all $n$. Since $\bigcap_m mG=0$, there exists a sequence of integers
$$ 1=k_1<k_2<\cdots$$
such that $\eta(2k_n!e_n)\not\in k_{n+1}G$ for each $n=1,2,\dots$. The set of elements $(g_1,g_2,\dots)$ in $P$ with each $g_i$ equal to either $0$ or $k_i!$ has cardinality $2^{\aleph_0}$, and $G$ has strictly smaller cardinality. Thus there there are two different elements of this set that have the same image under $\eta$. Their difference is an element $x=(h_1,h_2,\dots)$ with each $h_i$ either $0$, $\pm k_i!$ or $\pm 2k_i!$.
But now we obtain a contradiction. If $j$ is the first index for which $h_j\neq 0$ (and there must be such an index as $x\neq 0$) then $\eta(h_j)\not\in k_{j+1}G$, and
$$ \eta(h_j)=\eta(x)-\eta(0,\dots,0,h_{j+1},h_{j+2},\dots)=-\eta(0,\dots,0,h_{j+1},\dots)\in k_{j+1}G.$$
(There appears to be a typo in the proof of 94.2 in Fuchs. He only requires that $\eta(k_i!e_i)\not\in k_{i+1}G$ for all $i$, whereas it seems to me that you need the factor of $2$ because it looks like he gets the possibilities for the entries of $x$ wrong, and omits the potential $2k_i!$, and only has $0$ or $\pm k_i!$ as options for $h_i$.)