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I am in need of:

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

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It should be pointed out that all of the examples below are "infinitely related", which means that there is no generating set which has a finite set of "rules" which describes how the generators interact. A presentation consists of a set of generators and a list of rules. It is an open problem if there exist infinite, finitely related presented groups all of whose elements have finite order. – user1729 Jul 7 '13 at 14:27
@user1729 You can thus ask a new question! – awllower Jul 8 '13 at 8:46
@awllower I perhaps should have said "a famous open problem"! See this MathOverflow question. – user1729 Jul 8 '13 at 9:05

11 Answers 11

up vote 26 down vote accepted

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.

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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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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 Jun 9 '12 at 16:59
Yes, I know... I just thought of polynomials first. – fretty Jun 9 '12 at 17:02

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.

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the set of all roots of unity forms a group but order is infinite but order of every element is finite

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This has already been given as an example. – Pedro Tamaroff Feb 6 '15 at 3:26

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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very interesting – Leon Nov 22 '13 at 1:51

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.

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How come? Because they're hideous! Any noone can read the proof... – user1729 Aug 10 '12 at 10:24

Take any infinite set $T$. For instance, $\Bbb{Z}$.

Then the set of all permutations of $T$ (denoted $S_T$) forms a group under function composition. The subset of $S_T$ consisting of those permutations that move a finite number of elements in $T$ then makes up a subgroup of $S_T$.

Why is that?

Let's first explain what "moving an element" means. A permutation $\sigma$ $\in$ $S_T$ moves the element $t$ $\in$ $T$ iff $\sigma$$(t)$ $\neq$ $t$. Thus, permutations that moves a finite number of elements sends "most" elements of $T$ to themselves.

The subset of permutations that move finitely many elements in $T$ is a subgroup of $S_T$, because: the identity permutation moves zero elements, and if say, $\sigma$ $\in$ $S_T$ moves exactly the set $T_1$ $=$ $\{t_1, t_2, ..., t_m\}$ and $\tau$ $\in$ $S_T$ moves exactly the set $T_2$ $=$ $\{t'_1, t'_2, ..., t'_n\}$ where $T_1$, $T_2$ are subsets of $T$, then $\sigma$$\tau$$(t)$ $\neq$ $t$ for $t$ $\in$ $T$ can occur only if $\tau$$(t)$ $\in$ $T_1$ or $\ t$ $\in$ $T_2$. This means that $\sigma$$\tau$ moves $\lvert T_1\cup T_2 \lvert$ or less elements, implying that $\sigma$$\tau$ moves finitely many elements, and therefore implying that our set is closed under composition. Further, the inverse of $\sigma$ moves exactly as many elements as $\sigma$, which I'm sure you can see.

We now establish that in this group all elements are of finite order, yet the group is infinite.

For the first part, assume that $\sigma$ is an element that moves finitely many elements in $T$. Let's say... it moves $m$ elements, and let's call the set consisting of those $m$ elements $T'$. We realize that $\sigma$ can be viewed as a permutation in $S_{T'}$ when its domain is restricted to $T'$, and $T'$ being finite gives us $S_{T'}$ being a finite group of $\lvert T' \lvert$$!$ elements. It is intuitive that we can find a subgroup of $S_T$ that is isomorphic to $S_{T'}$; just choose the subgroup that fixes all elements not in $T'$ and permutates $T'$ (verifying that this is a subgroup isomorphic to $S_{T'}$ should be no problem). This subgroup that is isomorphic to $S_T'$ is finite; its elements are thus finite. It is a subgroup of our group of permutations that moves finitely many elements of $T$, and therefore $\sigma$ is of finite order, being in this group.

Finally, note that for any natural number $n$ and any subset of $S_T$ consisting of $n$ elements, say $n$ $=$ $90$ billions, or $n$ $=$ $43$, we can find a permutation that moves exactly those $n$ elements of that particular subset. Furthermore, we can find a permutation cycle that moves those $n$ elements, and such a permutation cycle makes up a cyclic subgroup of order $n$. The point is, for each $n$ $\in$ $\Bbb N$, there is an element in our subgroup of $S_T$ that is of order $n$. The infinity of $\Bbb N$ gives the infinity of our constructed group.

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The Burnside groups as discussed on the Wikipedia page.

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Would this work?

Consider a group of 2x2 diagonal matrices with diagonal elements of the form $e^{i\pi\frac pq} $. The identity element will be the identity matrix. And the order will always be finite due to the property of $e^{i\pi} $ function and the multiplicative property of diagonal matrices. Please point out if there's a mistake

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protected by Pedro Tamaroff Feb 6 '15 at 3:24

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