Examples of infinite groups such that all their respective elements are of finite order.

Solution 1:

Here is one. Let $(\mathbb{Q},+)$ denote the groups of rational numbers under addition, and consider it's subgroup $(\mathbb{Z},+)$ of integers. Then any element from the group $\mathbb{Q}/\mathbb{Z}$ has elements of the form $\frac{p}{q} + \mathbb{Z}$ which is of order at-most $q$. Hence it's of finite order.

• Group of all roots of unity in $\mathbb{C}^{\times}.$

Here is a link from MathOverflow which might prove helpful.

• https://mathoverflow.net/questions/57493/is-there-an-infinite-group-whose-elements-all-have-finite-order

Solution 2:

$G=(\Bbb Z/2\Bbb Z)^\omega$, or indeed $H^\omega$ for any finite group $H$.

Let $H$ be a finite group, and let $G=H^\omega$, the set of infinite sequences of element of $H$ with multiplication defined componentwise. If the order of $H$ is $n$, then clearly $g^n=1_G$ for each $g\in G$.

Added: For a more interesting example, let $G_n=\Bbb Z/n\Bbb Z$ for $n\in\Bbb Z^+$, and let $G$ be the direct sum of the $G_n$’s. In other words, $G$ is the set of sequences $$\langle m_k:k\in\Bbb Z^+\rangle\in\prod_{k\in\Bbb Z^+}G_k$$ such that only finitely many $m_k$ are non-zero. Then $G$ is infinite, all of its elements have finite order, and for each $n\in\Bbb Z^+$ $G$ has an element of order $n$.

Solution 3:

Let $\mathcal{P}(X)$ be the power set of a infinite set $X$. Consider the operation of symmetric difference, $\triangle$, on $\mathcal{P}(X)$.

Then, for all $A,B \in \mathcal{P}(X)$ we have that $A\triangle B=(A\setminus B)\cup(B\setminus A)$. It can be seen that $(\mathcal{P}(X),\triangle)$ is a commutative group with every element having order two. Hence, every element has finite order but the group is infinite.