Groups with no bounds on the size of abelian subgroups without infinite ones Is there a group $G$ in which every abelian subgroup is finite and there is no upper bound on the sizes of its abelian subgroups?
Let me say that a counterpart of the question above has posed by Paul Erd\"os in 1975 as follows:
Is there a group $G$ in which every subset consisting of pairwise non-commuting elements is finite and there is no upper bound on the sizes of such subsets?
The asnwer was given by B. H. Neumann in 1976 in negative. It was proved that groups $G$ satisfying the first assumption of the latter question must be center-by-finite and so the index of the center is a upper bound for sizes of subsets consisting of pairwise non-commuting elements.
 A: One can repeat the construction of the free Burnside group from Olshanskii "Geometry of defining relations in groups" imposing relations $w^{n_w}=1$ where $n_w\gg 1$ odd, depending on $w$ instead of $w^n=1$. All the auxiliary properties of the free Burnside groups except finite exponent will hold. In particular all Abelian subgroups of it will be cyclic, but the orders of them will be unbounded. The proof carries verbatim. I am pretty sure somebody (a student of Olshanskii) has done it explicitly already: one can construct torsion groups satisfying many "strange" properties. A more general way is to use lacunary hyperbolic groups
(see our paper with Olshanskii and Osin in arXiv). There are many torsion lacunary hyperbolic groups, all Abelian subgroups there are cyclic but their orders unbounded.  
A: Yes, such groups exist.
Consider the disjoint union of cyclic groups of odd order 
$\mathcal{C}:=\{ \mathbb{Z}/(2n+1)\mathbb{Z} \mid n \in \mathbb{N} \}$. By a theorem  of A. Ol'shanskii (see Thm 35.1 from his book "Geometry of defining relations in groups"), there exists an odd integer $n_0$ for which this collection of groups can be embedded into a countable $2$-generated simple group $G$ such that every proper subgroup of $G$ either is cyclic of order dividing $n_0$ or is conjugate to a subgroup of a group from $\mathcal C$.
Clearly $G$ is non-abelian, but every proper subgroup of $G$ is cyclic of finite odd order. Since each group from $\mathcal C$ is a subgroup of $G$, one sees that there is no bound on the orders of the cyclic subgroups.
