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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.

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    $\begingroup$ 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. $\endgroup$
    – DuduBob
    Commented Jan 19, 2021 at 14:37
  • $\begingroup$ @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. $\endgroup$
    – TheZone
    Commented Jan 19, 2021 at 14:41
  • $\begingroup$ 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. $\endgroup$ Commented Jan 20, 2021 at 12:21
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    $\begingroup$ @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 $1,1,1,1,\dots$? $\endgroup$ Commented Jan 21, 2021 at 16:22
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    $\begingroup$ 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 $G=Z$. $\endgroup$ Commented Jan 21, 2021 at 19:41

3 Answers 3

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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!

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    $\begingroup$ 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). $\endgroup$
    – Tom Sharpe
    Commented Jan 22, 2021 at 11:09
  • $\begingroup$ 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! $\endgroup$ Commented Jan 22, 2021 at 11:13
  • $\begingroup$ @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. $\endgroup$
    – Tom Sharpe
    Commented Jan 22, 2021 at 12:13
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    $\begingroup$ @DarthLubinus The proof is correct, I have deleted all comments $\endgroup$
    – user388615
    Commented Jan 22, 2021 at 22:02
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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 $h_i$.)

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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!

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