Infinitely generated subgroup of a finitely one (weird) I'm solving this question in Hungerford's book:

8​. Let $G$ be the multiplicative group generated by the real matrices $a=\begin{pmatrix}2&0\\0&1\end{pmatrix}$ and $b=\begin{pmatrix}1&1\\0&1\end{pmatrix}$. If $H$ is the set of all matrices in $G$ whose (main) diagonal entries are $1$, then $H$ is a subgroup that is not finitely generated.

I've already proved that $H$ is a subgroup of $G$ and every element of $H$ is of the form: 
$$
        \begin{pmatrix}
         1 & x \\
        0 & 1 \\
        \end{pmatrix}
$$
I thought this question really weird, $H$ is a subgroup of a group generated by $a$ and $b$ but $H$ itself can't be generated by $a$ or $b$, how can be possible?
My question is how to show $H$ is infinitely generated? I didn't find any good candidates to generate $H$.
Thanks in advance
 A: Notice that $H$ is an Abelian group, and all its non-identity elements have infinite order. If it were finitely generated, it would be a direct sum of a finite number of copies of $\mathbb{Z},$ say generated by $\{ \left(\begin{array}{clcr} 1 & x_{i}\\0 & 1 \end{array}\right) : 1 \leq i \leq n \}.$ then every element of $H$ could be expressed (uniquely, though that is not essential here) in the form  $\left(\begin{array}{clcr} 1 & u \\0 & 1 \end{array}\right)$, where $u = \sum_{i=1}^{n}
z_{i}x_{i},$ with each $z_{i} \in \mathbb{Z}.$ Now the $(1,2)$-entry of any $h \in H$ is certainly a rational number ( in fact with denominator a power of $2$). But the denominator of every $\mathbb{Z}$-combination of the $x_{i}$ is at worst the lcm of the denominators of the $x_{i},$ so is bounded. However, $\left(\begin{array}{clcr} 1 & 2^{-i}\\0 & 1 \end{array}\right)$ lies in $H$ for any $i \in \mathbb{N},$ a contradiction.
A: Given $s, t \in \mathbb{Z}$, define
$$ h(s,t) := a^{s} b^ta^{-s} = \begin{pmatrix} 1 & 2^s t \\ 0 & 1\end{pmatrix} \in H.$$
Observe that $h(0,0) = 1$, and that for all $s,t,u,v \in \mathbb{Z}$, we have 
$$ (a^sb^t)(a^ub^v) = h(s',t')a^{s+u}, \quad \text{where} \quad s' = \min\{s,s+u\}, \ t ' = 2^{s-s'}t + 2^{s+u-s'}v.$$
Since each element of $G$ is a product of matrices of the form $a^mb^n$ with $m,n \in \mathbb{Z}$, the above relations imply that each matrix $g \in G$ can be represented in the form 
$$g = h(s,t)a^r \quad \text{with} \quad s,t,r \in \mathbb{Z}.$$
Note that for $g \notin \langle a\rangle$, this factorisation can be made unique if we require that $t$ be an odd integer. A matrix $g \in G$ belongs to $H$ if and only if $\det(g) = 2^r = 1$, or in other words, if $g = h(s,t)$ for some $s, t\in\mathbb{Z}$. 
For $g = h(s,t) \in H$ such that $g \neq 1$, define $\operatorname{ord}(g)$ to be the unique $s' \in \mathbb{Z}$  such that $g = h(s',t')$ with an odd integer $t'$; for $g = 1$ put $\operatorname{ord}(g) = \infty$. Thus, we have $\operatorname{ord} h(s,t) \geq s$ for all $s,t\in\mathbb{Z}$. Now, suppose that we are given $k \geq 1$ pairs of integers $(s_1,t_1),\ldots,(s_k,t_k)$ with $t_j \equiv 1 \!\pmod 2$ for $1 \leq j \leq k$, and let $s = \min\{s_1,\ldots,s_k\}$. For any $k$-tuple $(m_1,\ldots,m_k) \in \mathbb{Z}^k$, we have
$$ h(s_1,t_1)^{m_1}\cdots h(s_k,t_k)^{m_k} = h(s, \,2^{s_1-s}m_1t_1 + \cdots + 2^{s_k-s}m_kt_k).$$
Therefore, $\operatorname{ord}\left(h(s_1,t_1)^{m_1}\cdots h(s_k,t_k)^{m_k}\right) \geq s$ for all $(m_1,\ldots,m_k) \in \mathbb{Z}^k$. On the other hand, $\operatorname{ord} h(s-1,1) = s-1$, and therefore, $h(s-1,1) \in H$ does not belong to the group generated by $h(s_1,t_1), \ldots, h(s_k,t_k)$. Hence, $H$ is not finitely generated.
