Example of finitely generated subgroup from Hungerford's algebra? I found One (counter)example in Hungerford's Algebra.
Let $G$ be the multiplicative group generated by the real matrices 
$$a = \left(\begin{array}{l l} 
1 & 1\\
0 & 1 
\end{array}\right),\quad
b = \left(\begin{array}{l l} 
2 & 0\\
0 & 1 
\end{array}\right)
$$
Let $H$ be the subgroup of $G$ consisting of matrices that have $1$s on the main diagonal. Then $H$ is not finitely generated.
How to prove this?
 A: For all $n\in\mathbb{Z}$ we have $A^n=\left(\begin{matrix}1 & n\\0 & 1\end{matrix}\right)$ and $B^n=\left(\begin{matrix}2^n & 0\\ 0 & 1\end{matrix}\right)$. So $$B^{-m}A^nB^m=\left(\begin{matrix}\frac{1}{2^m} & 0\\ 0 & 1\end{matrix}\right)\left(\begin{matrix}1 & n\\0 & 1\end{matrix}\right)\left(\begin{matrix}2^m & 0\\ 0 & 1\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2^m} & \frac{n}{2^m}\\ 0 & 1\end{matrix}\right)\left(\begin{matrix}2^m & 0\\ 0 & 1\end{matrix}\right)=\left(\begin{matrix}1 & \frac{n}{2^m}\\0 & 1\end{matrix}\right)$$ 
So clearly $H$ contains some elements apart from powers of $A$. It is also clear that every element of $G$ has $0,1$ on the bottom row. So first question is whether we can get any other elements in $H$ apart from those given above. It is clear we can't because no element of $G$ has anything except a rational with denominator a power of 2 in the row 1, col 2 position.
The second question is whether we can generate these elements finitely. But again it is clear we cannot, because if you multiply two elements of $H$, one with $\frac{n_1}{2^{m_1}}$ at the top right, and the other with $\frac{n_2}{2^{m_2}}$ then the product will have $\frac{n_3}{2^{m_3}}$ at the top right with $m_3\le\max(m_1,m_2)$. So any finite set of generators can only give a finite set of denominators for the top right element, whereas $H$ has an infinite set.
