The influence of the finiteness of a set on the conjugation classes of a group Let $G$ be a torsion group. Suppose that $G = \langle a,B \rangle$ where $a \in G$ and $B$ is a abelian subgroup of $G$. Denote by $a^B$ the set of the conjugates of $a$ by elements of $B$, i.e., $a^B = \{a^b | b \in B\}$. 
My question: If $a^B$ is a finite set, then is $a^G=\{a^g | g \in G\}$ a finite set?
If not true, I would like a counterexample.
Note that $\langle a^B \rangle = \langle a^G \rangle$, but the set $a^B$ can be different from $a^G$.

If $B$ is finite then clearly $a^B$ is finite, so that is a good way to look for counterexamples. Tarski Monsters are certainly counterexamples, but there are also less exotic examples, such as the Grigorchuk group. That is usually described with four generators $a,b,c,d$, but in fact $bc=d$, and $\langle b,c \rangle$ is a Klein $4$-group, so you could take $B=\langle b,c \rangle$.
