Tarski monster group

Remarks

Let $g \ne 1$ in $G$. If $\langle g \rangle = G$, then $G \cong \Z$, which has the subgroup $2\Z$ of infinite order; this is impossible. So $\langle g \rangle$ is a proper subgroup other than $\set{1}$, and $g$ has order $p$. Two elements $x, y$ that do not lie in one subgroup of order $p$ generate $G$, since $\langle x, y \rangle$ is neither trivial nor cyclic of order $p$. So $G$ is generated by $2$ elements and has exponent $p$, and it answers the bounded Burnside problem in the negative for this $p$.

$G$ is simple. Let $N \ne \set{1}$ be a normal subgroup with $N \ne G$; then $N$ has order $p$. Take $x \notin N$; then $\langle x \rangle N$ is a subgroup of order at most $p^2$, so it is proper since $G$ is infinite. It contains $N$ and $x$, so it is not cyclic of order $p$, a contradiction.