Does Every Infinite Abelian Group Have an Element of Infinite Order?

The statement is false. An abelian group may be infinite because it has infinitely many distinct elements, even though every individual element has finite order.

A standard counterexample is the direct sum of countably many copies of the cyclic group of order $2$:

$$G=\bigoplus_{n=1}^{\infty} \mathbb{Z}/2\mathbb{Z}.$$

This group is infinite and abelian, but it has no element of infinite order.

What “infinite order” means

The order of an element $g$ in a group is the smallest positive integer $m$ such that

$$m g=0$$

when the group operation is written additively. If no such positive integer exists, $g$ has infinite order.

An infinite group can therefore have either:

• at least one element of infinite order; or
• only elements of finite order, in which case it is called a torsion group.

The false claim confuses the size of the group with the orders of its individual elements.

Counterexample: an infinite-dimensional vector space over $\mathbb{F}_2$

The group $G$ can be described concretely as the set of all sequences

$$(a_1,a_2,a_3,\ldots),$$

where each $a_i$ is either $0$ or $1$, and only finitely many entries are $1$. Addition is performed coordinatewise modulo $2$.

For example,

$$(1,0,1,0,0,\ldots)+(0,1,1,0,0,\ldots) =(1,1,0,0,0,\ldots).$$

The restriction to finitely many nonzero coordinates is what makes this a direct sum, rather than a direct product.

Why the group is infinite

For every positive integer $n$, let $e_n$ be the sequence with a $1$ in position $n$ and $0$ everywhere else. The elements

$$e_1,e_2,e_3,\ldots$$

are all distinct. Therefore $G$ contains infinitely many elements.

Coordinatewise addition modulo $2$ is commutative, so $G$ is abelian.

Why every element has finite order

Take any element $x=(a_1,a_2,a_3,\ldots)$ in $G$. Since every coordinate belongs to $\mathbb{F}_2$, we have $a_i+a_i=0$ for every $i$. Consequently,

$$x+x=0.$$

Thus every nonzero element has order exactly $2$, while the zero element has order $1$. No element has infinite order.

This proves that an infinite abelian group need not contain an element of infinite order.

A smaller description of the same example

One can also write the counterexample as

$$G=\{\text{finite subsets of }\mathbb{N}\},$$

with the group operation given by symmetric difference. For finite subsets $A$ and $B$,

$$A+B=A\mathbin{\triangle}B.$$

Because $A\mathbin{\triangle}A=\varnothing$, every element satisfies $2A=\varnothing$. The group is infinite because it contains the distinct singleton sets

$$\{1\},\{2\},\{3\},\ldots.$$

This is the same group as the direct sum of countably many copies of $\mathbb{Z}/2\mathbb{Z}$.

A true statement for finitely generated groups

The original claim becomes true if “infinite abelian group” is replaced by “infinite finitely generated abelian group.”

Suppose $G$ is generated by finitely many elements $g_1,\ldots,g_r$. If every generator has finite order, say $g_i$ has order $n_i$, then every element of $G$ can be written as

$$a_1g_1+\cdots+a_rg_r,$$

where each coefficient $a_i$ can be reduced modulo $n_i$. There are at most

$$n_1n_2\cdots n_r$$

possible reduced combinations. Hence $G$ would be finite.

Therefore, if a finitely generated abelian group is infinite, at least one of its generators must have infinite order. In fact, the structure theorem gives the more precise form

$$G\cong \mathbb{Z}^r\oplus T,$$

where $T$ is finite. The group is infinite exactly when $r\geq 1$, and then any nonzero element in the $\mathbb{Z}$-part has infinite order.

For example,

$$\mathbb{Z}\oplus \mathbb{Z}/6\mathbb{Z}$$

is infinite and contains elements of infinite order, such as $(1,0)$. By contrast, the direct sum of infinitely many copies of $\mathbb{Z}/2\mathbb{Z}$ is not finitely generated, which is why the finite-generation argument does not apply.

Another torsion example

The group $\mathbb{Q}/\mathbb{Z}$ is also infinite, abelian, and torsion. Every element has finite order: if $q\in\mathbb{Q}$, write $q=a/b$ with integers $a$ and $b>0$. Then

$$b(q+\mathbb{Z})=a+\mathbb{Z}=\mathbb{Z},$$

so the coset $q+\mathbb{Z}$ has finite order dividing $b$.

This example shows that the failure is not limited to groups in which every nonzero element has the same order. In the direct-sum example, every nonzero element has order $2$; in $\mathbb{Q}/\mathbb{Z}$, finite orders of arbitrarily large size occur. Neither group contains an element of infinite order.

Common mistake

Do not argue that an infinite group must contain an element of infinite order merely because the sequence

$$g,2g,3g,4g,\ldots$$

has infinitely many terms. If $g$ has finite order $n$, this sequence repeats because $ng=0$ and $(n+k)g=kg$. An infinite group can instead have infinitely many different elements, each lying in a finite cyclic subgroup, as the counterexample demonstrates.