Let $A = \{ m2^n \mid m,n \in \mathbb{Z} \} \le (\mathbb{Q},+)$ and let $T=\langle t\rangle$ be infinite cyclic that acts on $A$ by the rule $at=2a$. Then let $G = T \ltimes A$. What I would like to show that it has the maximality condition on normal subgroups, or equivalently to show that it does not have an infinite proper chain of normal subgroups $K_1 < K_2 <$ ...
My suspicion is that the only proper normal subgroup of $G$ is $A$, and was trying to show it. Could you please give me some suggestion on how to proceed?
I was not sure what the rule $at=2a$ really meant so I tought it could have meant that the operation induced is $$(k_1t,m_12^{n_1})(k_2t,m_22^{n_2}) = ((k_1+k_2)t, m_12^{n_1 + k_2} + m_22^{n_2})$$ if I am wrong could you please clarify?
My idea was to show that the only possible subgroups of $G$ are subgroups of $A$ and then show that subgroups of $A$ are those that have "bounded denominators", for example $2^n\mathbb{Z}$ for all $n\in \mathbb{Z}$ (so includining negatives). And then show that coniugating these "finite denominators" groups, we "go out of bounds", showing that the only normal subgroup of $G$ is $A$.