Let $G=AB$ be a periodic group product of its locally cyclic subgroups $A$ and $B$. Let's consider its subgroups $H=A^*B^*$, with $A^* \leq A$ and $B^* \leq B$; $R$ abelian and normal in $G$; $S= R \cap \langle A, B^* \rangle$ and $T= R \cap \langle A^*,B \rangle$ both normal in $G$ and such that $H \cap (S \cap T)=1$.
Assuming $H=A^*B^*$ is abelian, prove that $A^*$ centralizes $S$ and $B^*$ centralizes $T$, and consequently $H$ centralizes $S \cap T$.
I'd like to prove that $[A^*,S]=1$ and similarly $[B^*,T]=1$, but I only got that $[A^*,S] \leq [G,S] \leq S$ since $S \trianglelefteq G$.
About $H$ centralizes $S \cap T$, I don't have normality of $A^*$ and $B^*$ in $G$ which would allow me to say $[A^*B^*, S \cap T]=[A^*, S\cap T][B^*, S \cap T] \leq [A^*,S] [B^*,T]=1$.
Any suggestions are appreciated, thank you
$A$ is locally cyclic and hence abelian, so $A^* \le C_G(A)$. Also $H^*=A^*B^*$ is abelian, so $A^* \le C_G(B^*)$. So $A^* \le C_G( \langle A,B^* \rangle)$, and hence $A^* \le C_G(S)$.
Similarly $B^* \le C_G(T)$, and hence $H=A^*B^* \le C_G(S \cap T)$.