# The group of additive sequences of integers

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.

• I think this is saying that $\prod_{i=1}^{\infty} \mathbb{Z}$ is not a free $\mathbb{Z}$-module. And there might be MSE posts about that. Any proof that I know of is very complicated, like pages long. Commented Jan 19, 2021 at 14:37
• @DuduBob I know that result, but I don't know, I don't think that we need to use that one or that this is equivalent to that. Commented Jan 19, 2021 at 14:41
• After a few hours of thinking about this and talking to friends of mine (under- and grad students), I am really looking forward to an answer that does not utilize the fact that $\mathbb{Z}^{\mathbb{N}}$ is not free. Commented Jan 20, 2021 at 12:21
• @S.Dolan $\Sigma a_n$ is an infinite sum, so doesn't make sense (for a sequence of integers) unless all but finitely many $a_n$ are zero. Where would your function send the sequence $1,1,1,1,\dots$? Commented Jan 21, 2021 at 16:22
• This follows from Prop 94.2 of Fuchs's Infinite Abelian Groups II. If I have time later I can write out how it works. It hinges on the fact that the intersection of $nG$ is 0, where $G=Z$. Commented Jan 21, 2021 at 19:41

This is a proof which is completely elementary (although not too simple). The idea is to prove that, for $$n$$ sufficiently big, $$f(e_{n})=0$$.

Consider a sequence of the kind $$x=(2^{n_{1}},2^{n_{2}},...)$$, where $$(n_{i})$$ is an increasing sequence of natural numbers (I'll explain later how I'm going to choose that sequence). For every $$k$$, we have

$$f(x)=f(2^{n_{1}},...,2^{n_{k}},2^{n_{k+1}},...)=\sum_{i=1}^{k}2^{n_{i}}f(e_{i})+f(0,...,0,2^{n_{k+1}},2^{n_{k+2}},...)=\sum_{i=1}^{k}2^{n_{i}}f(e_{i})+2^{n_{k+1}}f(0,...,0,1,2^{n_{k+2}-n_{k+1}},2^{n_{k+3}-n_{k+1}},...).$$

Call $$b_{k}=f(0,...,0,1,2^{n_{k+2}-n_{k+1}},2^{n_{k+3}-n_{k+1}},...)$$, so that

$$f(x)=\sum_{i=1}^{k}2^{n_{i}}f(e_{i})+2^{n_{k+1}}b_{k}.$$

Therefore,

$$b_{k}=\dfrac{1}{2^{n_{k+1}}}\left[f(x)-\sum_{i=1}^{k}2^{n_{i}}f(e_{i})\right].$$

Now, applying the Triangle Inequality,

$$|b_{k}|\leq \dfrac{1}{2^{n_{k+1}}}\left[|f(x)|+\sum_{i=1}^{k}2^{n_{i}}|f(e_{i})|\right].$$

By choosing $$(n_{i})$$ appropriately, we can make the right hand side of the above inequality tend to $$1/2$$. Simply choose by recursion the sequence so that for every $$k$$,

$$\sum_{i=1}^{k}2^{n_{i}}|f(e_{i})|<\dfrac{1}{2}2^{n_{k+1}},$$

and we have convergence to $$1/2$$. Since $$b_{k}$$ is an integer for every $$k$$, we have $$b_{k}=0$$ for $$k\geq N$$ ($$N$$ being a sufficiently large natural number).

Now, simply notice that $$b_{k}=f(e_{k+1})+2^{n_{k+2}-n_{k+1}}b_{k+1}$$ (this follows directly from the definition of $$b_{k}$$). Using this, we deduce immediately that $$f(e_{k})=0$$ for large $$k$$.

Hope this helps!

• For those of you who find the order of this proof confusing, logically speaking the first thing we do is choose the increasing sequence of natural numbers $(n_i)$ to satisfy recursively the strict inequality given towards the end. We can start with any value; $n_1=1$ will do. Then the triangle inequality bit implies that $b_k$ tends to 0. (This is an excellent solution by the way, well done). Commented Jan 22, 2021 at 11:09
• Exactly. The reason I didn't start by choosing the sequence is because it would seem too artificial, even for this proof. Thank you very much! Commented Jan 22, 2021 at 11:13
• @ne3886, that's not correct, see my earlier comment. Logically, the sequence is chosen first (it's defined recursively, but that's fine), and then $x$ is defined by the sequence. If you scan through the proof carefully you'll see that the sequence doesn't depend on $x$, but rather the other way around. Commented Jan 22, 2021 at 12:13
• @DarthLubinus The proof is correct, I have deleted all comments Commented Jan 22, 2021 at 22:02

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 $$1=k_1 such that $$\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 $$h_i$$.)

Suppose $$I = \{i \geq 1 \mid f(e_i) \neq 0\}$$ is infinite. We can suppose $$I = \mathbb{N}$$.

Let $$H_k = \prod_{i=1}^k\{0\} \times \prod_{i \geq k+1}\mathbb{Z}$$.

$$f:G \to \mathbb{Z}$$ induce an monomorphism $$\bar{f} : G/\ker{f} \to \mathbb{Z}$$

Now we consider the projective system $$\varphi: G/M_{k+1} \to G/M_k, \text{ where } M_k = H_k + \ker{f}$$

Let $$f_k: G/M_k \to \mathbb{Z}/n_k\mathbb{Z}$$ where $$f(M_k) = n_k \mathbb{Z}$$

We have a continuous monomorphism of topoligical groups : $$\varprojlim G/M_k \to \varprojlim\mathbb{Z}/n_k\mathbb{Z}$$

from $$H_k \subset M_k$$ we deduce a conitnuous monomorphism $$\varprojlim G/H_k \to \varprojlim G/M_k$$ By composition we have a continous monomorphism $$\varprojlim G/H_k \to \varprojlim\mathbb{Z}/n_k\mathbb{Z}$$

But $$\varprojlim G/H_k$$ is isomorph to $$G$$, and $$\varprojlim \mathbb{Z}/n_k\mathbb{Z}$$ is somthing like $$\prod_{k=1}^N\mathbb{Z}_{p_k}$$ where $$\mathbb{Z}_{p_k}$$ is $$p_k$$-adic ring!