# Example of infinite groups, such that all its elements are of finite order

I am in need of:

• Example of infinite groups, such that all its elements are of finite order
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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 $\mathbb{C}^{\times}.$

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This is my favourite example, but you may want to consider to changing that the order is at most $q$ – Belgi Jun 8 '12 at 8:34
@Belgi: Thanks for the comment. – user9413 Jun 8 '12 at 9:36
@Chandrasekhar: Would the $\mathbb Z_{p^{\infty}}$ be an example here? – Babak S. Jun 8 '12 at 10:38
@Babak Yes. In fact, $\,\mathbb{Z}_p(\infty)\,$ is the $\,p-\,$ torsion subgroup of the circle group. – DonAntonio Jun 9 '12 at 3:52

$G=(\Bbb Z/2\Bbb Z)^\omega$, 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 $n$.

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$\mathbf Q/\mathbf Z$, and also the group $G^{\mathbb N}$ of infinite sequences of elements in any given finite group $G$ (all whose elements have order dividing $\#G$).

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If you want a particularly evil/fascinating example, the Grigorchuk group is finitely generated, and every element has finite order, but it is still infinite. (This is related to Sam Nead's answer).

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How about the group of polynomials with coefficients from the integers mod $2$, under addition. Every element has order $2$, except $0$, which is the identity so has order 1.

You can do the same for polynomials with coefficients in any finite group. The addition of these polynomials is almost the same, just now the coefficients are calculated using the group product of the original coefficients. Each polynomial in this construction has to have finite order by finiteness of the group.

In fact you can do this construction but taking the coefficients to be in any group that answers your question and you get another group which answers your question.

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 $\left(\left(\mathbb{Z}/n\mathbb{Z}\right)[x],+\right) \cong \left(\mathbb{Z}/n\mathbb{Z}\right)^{\omega}$. – Jackson Walters Jun 9 '12 at 16:59 Yes, I know... I just thought of polynomials first. – fretty Jun 9 '12 at 17:02

I wonder how come Tarski monster groups haven't yet been mentioned: these are infinite groups in which all non-trivial finitely generated subgroups are cyclic of order some fixed prime $\,p\,$. These are examples of finitely generated infinite simple groups.