# Why are the cardinalities of direct and inverse limits of $\mathbb{Z}/p^k\mathbb{Z}$ different?

The inverse limit of the finite cyclic group $$\mathbb{Z}/p^k\mathbb{Z}$$ leads to the group of $$p$$-adic integers $$\mathbb{Z}_p$$, while the direct limit of $$\mathbb{Z}/p^k\mathbb{Z}$$ leads to the Pruefer $$p$$-group $$\mathbb{Z}(p^\infty)$$. For any given $$p$$, the $$p$$-adic integers are uncountable, having the cardinality of the real numbers, while the Pruefer $$p$$-group is countable, having the cardinality of the rationals (hence the natural numbers). Why is it that the cardinalities of the direct and inverse limits of $$\mathbb{Z}/p^k\mathbb{Z}$$ are not the same?

• It's because they are different. – Angina Seng Dec 15 '19 at 16:38
• I have never heard of an indirect limit. Perhaps you mean inverse limit? – Derek Holt Dec 15 '19 at 17:38
• @DerekHolt Yes, I meant inverse limit, my brain wasn't working when I wrote it up. – 1103_base_6 Dec 15 '19 at 17:54
• Is there a reason, except for the common word "limit", to think they should be the same? – Andreas Blass Dec 15 '19 at 20:07

The difference in the cardinalities of the two groups is due to the fact that the constructions of the inverse limit and the direct limit of $$\mathbb{Z}/p^k\mathbb{Z}$$ use different operations upon the constituent groups. The inverse limit of $$\mathbb{Z}/p^k\mathbb{Z}$$ involves the categorical product for abelian groups, the direct product of countably infinity many groups of $$\mathbb{Z}/p\mathbb{Z}$$, which is of cardinality $$p^\mathbb{N} \sim |\mathbb{R}|$$ or the cardinality of the reals. On the other hand, the direct limit of $$\mathbb{Z}/p^k\mathbb{Z}$$ involves the categorical coproduct for abelian groups, the direct sum of countably infinity many groups of $$\mathbb{Z}/p^k\mathbb{Z}$$, which is of cardinality $$\sum_{i=0}^{\infty} p^i \sim |\mathbb{N}|$$. This is because in the infinite case, the direct sum of abelian groups is only a subset of the direct product of abelian groups, namely those with finitely many non-zero terms.
It can be proven that the underlying set of the Pruefer $$p$$-group is isomorphic to the natural numbers in a base-$$p$$ positional notation system, while the underlying set of the $$p$$-adic integers is isomorphic to the circle group (the real numbers modulo one) in a base-$$p$$ positional notation system.