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Let $(\mathbb{N},\leq$) be a directed set where $m\leq n$ if an only if $m$ divides $n$.

We define a directed system of groups where $G_{n}=\mathbb{Z}$ for all $n\in \mathbb{N}$ and $f_{mn}\colon \mathbb{Z} \to \mathbb{Z}$ is multiplication by $\frac{n}{m}$.

I am trying to compute the directed limit of that system. The definition that I know is $$\dot{\bigcup}_{n\in \mathbb{N}} G_{n} / \sim,$$ where $z \sim y$ if $f_{mn}(z)=f_{kn}(y)$ where $n$ divides $m$ and $n$ divides $k$.

What I have tried is the following: in $G_{2}=\mathbb{Z}$, $2n=n$ where $n\in G_{1}$. In $G_{3}=\mathbb{Z}$, $3n=n$ where $n\in G_{1}$. In $G_{4}$, $4n=n=2n$ where $n\in G_{1}$ and $2n\in G_{2}$. In general, in $G_{k}$, $kn=n=k_{2}n=k_{3}n=\cdots=k_{m}n$ where $k_{2},\cdots,k_{m}$ are the divisors of $k$, $n\in G_{1}$ and $k_{i}n\in G_{i}$. Nevertheless, I do not know how to continue.

Can someone help me, please? Thank you in advance!

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  • $\begingroup$ Everything divides $0$. $\endgroup$ – Derek Elkins Mar 14 at 21:20
  • $\begingroup$ @DerekElkins I am assuming that $0$ is not a natural number $\endgroup$ – Karen Mar 14 at 21:20
  • $\begingroup$ $\le$ is the worst possible symbol to denote divisibility in $\mathbf{N}$... $\endgroup$ – YCor Mar 14 at 22:29
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$$\underrightarrow{\lim}G_n\simeq \mathbf{Q},$$ the additive group of rational numbers.

Indeed, denote by $|$ divisibility in $\mathbf{N}_{>0}$. Since $\{n!:n\ge 0\}$ is cofinal in $(\mathbf{N}_{>0},|)$, the direct limit is the same as the direct limit restricted to this cofinal sequence $H_n=G_{n!}=\mathbf{Z}$ with given map $H_{n-1}\to H_{n}$ given by multiplication by $n$.

It is equivalent to write $H'_n=\frac{1}{n!}$ and consider the direct limit using identity maps. This makes clear that the direct limit is the union of this increasing sequence of subgroups of $\mathbf{Q}$, namely $\mathbf{Q}$ itself.

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  • $\begingroup$ Why is the direct limit equivalent to just taking the direct limit of cofinal terms? You have the same ordering for the construction of $\hat{\mathbb{Z}}$ but certainly this argument shouldn’t work there. $\endgroup$ – Santana Afton Mar 15 at 0:55
  • $\begingroup$ @SantanaAfton Think twice, one indeed has $\widehat{\mathbf{Z}}=\underleftarrow{\lim}\mathbf{Z}/n!\mathbf{Z}$. $\endgroup$ – YCor Mar 15 at 16:58
  • $\begingroup$ Hm. Is the isomorphism $(n_i)\mapsto ( n_1n_2\dots n_i )$? $\endgroup$ – Santana Afton Mar 15 at 21:06

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