So, I found this problem in Scott: Group Theory, basically all I could think of was this:
I can see that if $G$ has an element of infinite order, then it has infinite elements of infinite order, but not necessarily an $|G|=\mathrm{card}(G)$ elements of infinite order. However, using Lagrange´s theorem for infinite groups, and being $\mathrm{cl}(x)=\{\text{conjugates of }x\}$ and $c(x)=\text{centralizer of }x$ , it is: $$|G|=[G:c(x)]|c(x)|$$ but, because of group actions, it is also true that $|cl(x)|=[G:c(x)]$. This must mean it is $|G|=|cl(x)|\cdot|c(x)|$. Now, accepting AC, the last two cardinals must be comparable, so that the last product must be equal to either $|cl(x)|$ or $|c(x)|$. Now, because conjugates have the same order, in the first case it is $|G|=|cl(x)|$, and so the problem is over. But in the other case, it is $|G|=|c(x)|$, and I have no idea how to proceed. The only result I didn't use in the chapter this problem is would be the Higman-Neumann-Neumann theorem, but I do not see how that theorem would be of any use.
Any help would be appreciated.
If $y\in c(x)$ has finite order, then $xy$ has infinite order, because $(xy)^k = x^ky^k$, and for $k\neq 0$, $x^k$ has infinite order and $y^k$ has finite order, so $y^k\neq x^{-k}$.
Let $A=\{z\in c(x)\mid |z|\lt \infty\}$ and $B=c(x)\setminus A$.
If $|B|=|G|$, you are done. If $|B|\lt |G|$, then since $|G|=|c(x)| = |A|+|B| = \max(|A|,|B|)$, then...