# In a group $G$, if every elements are of finite order then $G$ is at most countable.

Let us consider a group $$G$$.

If $$G$$ is a finite group then every element of $$G$$ has finite order. But the converse need not be true, i.e., every elements in a group has finite order doesn't imply the group is finite.

Examples: The factor group $$\mathbb{Q}/\mathbb{Z}$$, the Prüfer $$p$$-group etc. (all these groups are countable).

Conjecture $$1$$: In a group $$G$$, if every element has finite order then $$G$$ is at most countable.

Conjecture $$2$$: Every uncountable groups contain an infinite cyclic group.

Let $$S$$ is a countable subset of $$G$$ (possible because of Axiom of Choice). Since $$G$$ is not countably generated i.e $$G\neq \langle S\rangle$$, $$\exists g_1\in G$$ such that $$g_1\notin \langle S\rangle$$.

Again $$\langle S, g_1\rangle$$ is countable implies $$G\neq \langle S, g_1\rangle$$.Hence $$\exists g_2\in G$$ such that $$g_2\notin \langle S, g_1\rangle$$

$$\vdots$$

Hence $$\exists (g_n) \in G$$ such that $$g_{n+1}\notin \langle S, g_1,\ldots,g_n\rangle$$

Let $$H=\{g_n\}_{n\in\Bbb{N}}$$

Claim: $$\langle H\rangle\cong \mathbb{Z}$$

I have no way to proceed further. Is it possible to define an isomorphism between $$H$$ and $$\mathbb{Z}$$ ?

Conjecture $$1$$ and $$2$$ are equivalent. I want to prove the second one using the fact that "an uncountable group can't be countably generated".

One final thought : Does $$G=\Pi_{i\in \Bbb{N}} \mathbb{Z}_2$$ provide a counter example?

• One final thought: Can $G=\Pi_{i\in \Bbb{R}} \mathbb{Z}_2$ provide a counter example? Commented Jul 28, 2022 at 8:18
• Oops. Foiled again. @egreg Commented Jul 28, 2022 at 8:43
• My knowledge and abstract algebra is always an $\epsilon$ distance away for every $\epsilon>0$ . It's a bad question, since I have already found a counter example by myself. Just a one step verification needed. My bad. Commented Jul 28, 2022 at 8:55
• You could do $\prod_{i\in\Bbb N}\Bbb Z_2$. Commented Jul 28, 2022 at 9:04
• @Cpc Yes. As the product is uncountable. It is enough to choose index set infinite. Commented Jul 28, 2022 at 9:07

Both conjectures are false.

The coproduct group $$(\mathbb{Z}/2\mathbb{Z})^{(\alpha)}$$, where $$\alpha$$ is any cardinal number, that is, direct sum of $$\alpha$$ copies of $$\mathbb{Z}/2\mathbb{Z}$$, has cardinality $$\alpha$$ and its elements have finite order two (except for the neutral element).

This obviously contradicts also the second conjecture.

• So my example is correct. Can you provide one more example? Commented Jul 28, 2022 at 8:39
• @LostinSpace The direct product you mention is good, but not as general as mine, because there are restrictions on the cardinalities of direct products. What additional example do you need? Commented Jul 28, 2022 at 8:44
• @LostinSpace I'm sure you can find it yourself. Commented Jul 28, 2022 at 8:47

Another counterexample to both conjectures: the group of permutations of an infinite set which have finite support (i.e. fix all but finitely many elements). This is nonabelian and centerless, and if we restrict to the index two subgroup of even permutations and the set is countable the group is simple too.

As we noted, $$G=\prod_{i\in \Bbb N}\Bbb Z_2$$ would indeed suffice for a counterexample. If you're familiar with the binary representation of real numbers; or by Cantor, $$\mid G\mid=2^{\aleph_0}\gt\aleph_0$$ it's an uncountable group. (@egreg's answer covers this I believe)