# Direct limit of $\mathbb{Z}$-homomorphisms

What is the direct limit of the following sequence of $\mathbb{Z}$-homomorphisms (as groups)? $$\mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{3} \mathbb{Z}\xrightarrow{5} \mathbb{Z}\xrightarrow{7} \mathbb{Z}\xrightarrow{11} \mathbb{Z}\xrightarrow{13}\cdots$$ Here the label $p$ indicates multiplication by $p$, and the sequence is just the sequence of primes. Is there an easy description of this limit? Would we get a different limit if we take a subsequence of the primes, or the sequence of primes squared?

-
I think I have figured out the answer to the first question, it should be the subgroup of $\mathbb{Q}$ consisting of all $p/q$ where $q$ is a square-free integer. – Lukas Geyer Oct 16 '12 at 18:57

It is the group of rational numbers with squarefree denominators.

Your system is $$Z_0 \xrightarrow{p_0} Z_1 \xrightarrow{p_1} Z_2 \xrightarrow{p_2} Z_3 \ldots$$ where $p_k$ is the $k$th prime number and $Z_k = \mathbb Z$

For every $k$, define the map $\phi_k : Z_k \to \mathbb Q$ by $x \mapsto x(p_0 \ldots p_{k-1})^{-1}$. This collection of maps is compatible with the system, and the union of the ranges of those maps are those rationals with squarefree denominators. Since the maps are also injective, it is easy to show that if $\psi_k : Z_k \to G$ is another compatible family of group morphisms, you can define a unique map $\psi : \bigcup \phi_k(\mathbb Z) \to G$ making the triangle commute.

Switching the order of the primes wouldn't change anything, and putting them more than once would allow for rationals with more denominators.

-

Let $(a_k)_{k\in\Bbb N}$ be a sequence of natural numbers. Consider the direct limit of the system

$$\Bbb Z\xrightarrow{a_1}\Bbb Z\xrightarrow{a_2}\Bbb Z\xrightarrow{a_3}\Bbb Z\xrightarrow{a_4}\cdots.$$

We may rewrite the above as an isomorphic system comprised of inclusions of subgroups of $\Bbb Q$:

$$\Bbb Z\hookrightarrow \frac{1}{a_1}\Bbb Z\hookrightarrow\frac{1}{a_1a_2}\Bbb Z\hookrightarrow \frac{1}{a_1a_2a_3}\Bbb Z\hookrightarrow\cdots.$$

Note the commutative diagrams that show this system is isomorphic.

$\hskip 1.5in$

(Here $\alpha$ is division by $a_1\cdots a_{n-1}$ and $\beta$ is division by $a_1\cdots a_n$.)

At this point the direct limit is just the union of ascending subgroups, which will give the subgroup of rationals with denominator dividing into some product of $a_i$'s.

-