# Every inﬁnite abelian group has at least one element of inﬁnite order?

Is the statement true?

Every inﬁnite abelian group has at least one element of inﬁnite order.

I am searching for an infinite abelian group with all elements having finite order. Please help me to find such groups.

• Don't "search" for one; make one yourself! – anon Dec 13 '14 at 4:51
• I was very confused about the title since it contains an obviously false statement, therefore replaced "." by "?" – Martin Brandenburg Dec 13 '14 at 9:05

Let $G=\{z\in \mathbb{C}:z^n=1$ for some positive integer $n\}$. Then $G$ with complex multiplication is an infinite abelian group but every element has finite order.

• |G|=$n$, i think... as it has n soln in $\mathbb C$? is n varying on $\mathbb Z$? – yio kk Dec 13 '14 at 3:55
• No, $|G|\not = n$, the $n$ in the definition of $G$ is not fixed. It says "for some $n$". Thus, $\alpha \in G$ and $\beta \in G$, even if $\alpha^3 = 1$ and $\beta^5 = 1$. $G$ consists of all possible such powers you can have. – Nicolas Bourbaki Dec 13 '14 at 3:58
• $G=\{e^{ir}\,|\,r\in\mathbb{Q}\}$ – hjhjhj57 Dec 13 '14 at 4:35
• Rather $G=\{e^{2\pi ir} :r\in \mathbb{Q}\}$. – Bingo Dec 13 '14 at 4:37

Consider $(\mathbb{Q},+)$ and the subgroup $(\mathbb{Z},+)$. Now take the quotient $\mathbb{Q}/\mathbb{Z}$.

• This is isomorphic to your other answer; perhaps it would be better to edit it in there. – Milo Brandt Dec 13 '14 at 3:56
• This is the only answer I gave, perhaps you are confusing me with another user. – Hubble Dec 13 '14 at 3:56
• Oops, my bad. (The answer I got confused with was the one by alphad, which is indeed isomorphic to this one) – Milo Brandt Dec 13 '14 at 3:59
• What is this group?.....I dont understand. – yio kk Dec 13 '14 at 4:04
• @yiokk, this is the group of all elements of the form $p/q + \mathbb{Z}$ under addition. Each element has order at most $q$. – Hubble Dec 13 '14 at 4:41

Take an infinite number of generators, and specify that any element squared is the identity.

• Or take the set of all finite subsets of $N$ with the operation of symmetric difference, where $N$ is an infinite set. – bof Dec 13 '14 at 3:56
• @bof Who said anything about $\mathbb N$? This generates groups of any cardinality with only elements of order $1$ and $2$! – Milo Brandt Dec 13 '14 at 3:59

Define the set $G$ to be the collection of all functions $f:\mathbb{N}\to \mathbb{Z}/2\mathbb{Z}$. We can then define $f+g$, for any two $f,g\in G$, in the natural way. Define $e$ to be the map such that $e(n) = 0$ for all $n\in \mathbb{N}$. It is not hard to see that $f+f = e$ for all $f\in G$.

There are many counter examples, some of the more natural ones are:

1. $( \mathbb{N}, \oplus )$ where $\oplus$ is bitwise Exclusive OR of the binary representations between two natural numbers. Every non-zero numbers has order $2$.

2. Consider any polynomial rings over any finite field. If we strip away its multiplication, we get an infinite abelian group with respect to the addition. Every non-zero element has order $p$, the characteristic of the underlying field.

• “1.” is identical to @Meelo’s answer and, moreover, is a case of “2.” with F₂ as the ground field. – Incnis Mrsi Dec 13 '14 at 11:07
• More precisely, to Meelo’s answer for countable number of generators. – Incnis Mrsi Dec 13 '14 at 11:27

As mentioned in various answers it is not true that every infinite abelian group contains an element of infinite order; let me try to structure the examples a bit along a standard distinction.

An abelian group in which every element has finite oder is called a torsion abelian group; more generally, the subsets of elements of finite order form a subgroup called the torsion subgroup.

Thus what you are looking for are infinite torsion abelian groups.

A further thing to observe is that even when the order of each element is finite, it can still happen that there is no common bound for the order of the elements.

This is the case in, e.g., $\mathbb{Q}/\mathbb{Z}$ and $\oplus_{n\in \mathbb{N}^{\ast}} \mathbb{Z}/n\mathbb{Z}$ (note the direct sum, the direct product would not be a torsion group).

A torsion abelian group where there is a common bound for the order of the elements is called a bounded abelian groups and the maximal order of an element, which is in fact the least common multiple of all orders, is called the exponent of the group.

There are also examples of infinite bounded abelian groups, such as $(\mathbb{Z}/n\mathbb{Z})^{\mathbb{N}}$ for some positive integer $n$.

• I do not understand your remark for several reasons. First, I did not say anywhere in my answer that the direct product of identical torsion groups is not a torsion group: where I mention the direct product the direct sum is not over identical torsion groups but over $Z/nZ$ and the $n$ is the index of the sum too. And, at the very end I even give an example of an (infinite) direct product of identical torsion groups and say it is torsion. Second, and more importantly, an infinite direct product of identical torsion groups is in fact not torsion in general (you'd need bounded). – quid Dec 13 '14 at 13:05
• Yes, because it is bounded, as I said. And I gave $Z/n Z$ raised to countable cardinality as an example of a torsion group at the end. Earlier however I did not consider the direct sum of identical torsion groups, but of $Z/nZ$ as $n$ varies over all positive integers; and I commented there that the direct product would not be torsion. – quid Dec 13 '14 at 13:45

The statement is false Consider the Power Set of Natural number with group operation of symmetric difference.Then the Group clearly is infinite,abelian,and no element is of infinite order more so each element has order $2$.

Consider $(\mathbb{F}_p[X],+)$, the polynomial ring over $\mathbb{F}_p$ under addition. This is an abelian infinite group with exponent $p$, which means that $a^p=0$ for every $a \in \mathbb{F}_p[X]$. Since we're looking at "$+$" $a^p$ is actually $p \cdot a$ in this case.

Notice that this is a stronger property, than "every element has finite order".

• Why to repeat same things for p th time where p is definitely greater than 1? – Incnis Mrsi Dec 13 '14 at 11:09

The most natural example to me is $\prod_{n=1}^\infty\mathbb Z_2$, where every element has order $2$.

• Notice that this coincides with the example posted by "I Love Mr. Paul". – Martin Brandenburg Dec 13 '14 at 9:06
• And also with achille hui's second example, if the field is $\mathbf{F}_2$. (Of course, every non-identity element has order $2$.) – fkraiem Dec 13 '14 at 9:52
• @fkraiem: the “∏” operation results in continuum cardinality, if I remember correctly. The example of Meelo–achille hui has countable cardinality. – Incnis Mrsi Dec 13 '14 at 11:16
• Oh, yes, silly me, a polynomial has only finitely many non-zero coefficients... – fkraiem Dec 13 '14 at 11:32