We study some properties of the Prüfer \(p\)-group \(\mathbb{Z}_{p^\infty}\) for a prime number \(p\). The Prüfer \(p\)-group may be identified with the subgroup of the circle group, consisting of all \(p^n\)-th roots of unity as \(n\) ranges over all non-negative integers:
\[\mathbb{Z}_{p^\infty}=\bigcup_{k=0}^\infty \mathbb{Z}_{p^k} \text{ where } \mathbb{Z}_{p^k}= \{e^{\frac{2 i \pi m}{p^k}} \ | \ 0 \le m \le p^k-1\}\]

\(\mathbb{Z}_{p^\infty}\) is a group

First, let’s notice that for \(0 \le m \le n\) integers we have \(\mathbb{Z}_{p^m} \subseteq \mathbb{Z}_{p^n}\) as \(p^m | p^n\). Also for \(m \ge 0\) \(\mathbb{Z}_{p^m}\) is a subgroup of the circle group. We also notice that all elements of \(\mathbb{Z}_{p^\infty}\) have finite orders which are powers of \(p\).

Now for \(z_1,z_2\) elements of \(\mathbb{Z}_{p^\infty}\), there exist \(k_1,k_2 \ge 0\) with \(z_1 \in \mathbb{Z}_{p^{k_1}}\) and \(z_2 \in \mathbb{Z}_{p^{k_2}}\). We can suppose without loss of generality that \(k_1 \le k_2\). Hence \(z_1, z_2\ \in \mathbb{Z}_{p^{k_2}}\) and \(z_1 z_2^{-1} \in \mathbb{Z}_{p^{k_2}} \subseteq \mathbb{Z}_{p^\infty}\) which proves that \(\mathbb{Z}_{p^\infty}\) is a subgroup of the unit circle.

All subgroups of \(\mathbb{Z}_{p^\infty}\) are finite

Let \(H\) be a proper subgroup of \(\mathbb{Z}_{p^\infty}\). We prove that \(H\) is equal to one of the \(\mathbb{Z}_{p^n}\) for \(n \ge 0\). If the set of the orders of elements of \(H\) is infinite, then for all element \(z \in \mathbb{Z}_{p^\infty}\) of order \(p^k\), there would exist an element \(z^\prime \in H\) of order \(p^{k^\prime} > p^k\). Hence \(H\) would contain \(\mathbb{Z}_{p^\prime}\) and \(z \in H\). Finally \(\mathbb{Z}_{p^\infty}\) would be included in \(H\) in contradiction with the hypothesis that \(H\) is a proper subgroup.

Therefore the set of the orders of the elements of \(H\) is finite. If \(h \in H\) is an element with maximum order \(p^n\), we have \(H = \mathbb{Z}_{p^n}\).

In conclusion:

• \(\mathbb{Z}_{p^\infty}\) is an infinite group whose proper subgroups are all finite.
• It is a non-cyclic group whose all proper subgroups are cyclic.