Tag Archives: infinite-groups

A group G isomorph to the product group G x G

Let’s provide an example of a nontrivial group \(G\) such that \(G \cong G \times G\). For a finite group \(G\) of order \(\vert G \vert =n > 1\), the order of \(G \times G\) is equal to \(n^2\). Hence we have to look at infinite groups in order to get the example we’re seeking for.

We take for \(G\) the infinite direct product \[
G = \prod_{n \in \mathbb N} \mathbb Z_2 = \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 \dots,\] where \(\mathbb Z_2\) is endowed with the addition. Now let’s consider the map \[
\begin{array}{l|rcl}
\phi : & G & \longrightarrow & G \times G \\
& (g_1,g_2,g_3, \dots) & \longmapsto & ((g_1,g_3, \dots ),(g_2, g_4, \dots)) \end{array}\]

From the definition of the addition in \(G\) it follows that \(\phi\) is a group homomorphism. \(\phi\) is onto as for any element \(\overline{g}=((g_1, g_2, g_3, \dots),(g_1^\prime, g_2^\prime, g_3^\prime, \dots))\) in \(G \times G\), \(g = (g_1, g_1^\prime, g_2, g_2^\prime, \dots)\) is an inverse image of \(\overline{g}\) under \(\phi\). Also the identity element \(e=(\overline{0},\overline{0}, \dots)\) of \(G\) is the only element of the kernel of \(G\). Hence \(\phi\) is also one-to-one. Finally \(\phi\) is a group isomorphism between \(G\) and \(G \times G\).

An infinite group whose proper subgroups are all finite

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 \(p\). Continue reading An infinite group whose proper subgroups are all finite

A (not finitely generated) group isomorphic to a proper quotient group

The basic question that we raise here is the following one: given a group \(G\) and a proper subgroup \(H\) (i.e. \(H \notin \{\{1\},G\}\), can \(G/H\) be isomorphic to \(G\)? A group \(G\) is said to be hopfian (after Heinz Hopf) if it is not isomorphic with a proper quotient group.

All finite groups are hopfian as \(|G/H| = |G| \div |H|\). Also, all simple groups are hopfian as a simple group doesn’t have proper subgroups.

So we need to turn ourselves to infinite groups to uncover non hopfian groups. Continue reading A (not finitely generated) group isomorphic to a proper quotient group