TITLE: Direct limit of $\mathbb{Z}$-homomorphisms
QUESTION [7 upvotes]: 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?

REPLY [10 votes]: 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.