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?
Best Answer
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.