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

Question

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?

Answer 1

I believe you mean inverse limit for indirect limit. There is no such thing as 'indirect limit' in category theory.

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.