Problem:
If $G$ is the multiplicative subgroup of $GL_2\mathrm{R}$ generated by $a = \bigg(\begin{matrix}2& 0\\ 0 & 1\end{matrix}\bigg)$ and $b = \bigg(\begin{matrix}1& 1\\ 0 & 1\end{matrix}\bigg)$, and $H$ is the subset $H\subset G$ of matrices whose diagonal entries are $1$, show that $H$ is not finitely generated.
Attempt:
So, I was able to show that $H=\bigg\{\bigg(\begin{matrix}1& m/2^n\\ 0 & 1\end{matrix}\bigg)\bigg| m,n \in Z\bigg\}$. But I'm unsure on how to proceed from this to finite generated-ness. The way I used to show that H is like that was based around how general elements in $G$ can have a determinant of $1$ and exploring around with the form of $a^{-n}b^ma^n$...The elements in the RHS of the above equality are all of the form $a^{-n}b^ma^n$. Is it enough to point out that finitely many pairs of $(n,m)$ cannot generate all terms in the RHS, only infinitely many can?
You already basically observed that $H$ is isomorphic to the additive structure of the ring $\mathbb Z [1/2]$. If this means nothing to you, $$\mathbb Z [1/2] = \{m/2^n : n, m \in \mathbb Z\}.$$ Now if this were finitely generated by $S = \{m_1/2^{n_1}, \dots, m_k/2^{n_k}\}$, let $n$ be the minimum of the $n_i$. But then, $2^{-(n+1)}$ is not a linear combination of elements of $S$ over $\mathbb Z$, a contradiction.