Is this group of matrices cyclic? Is the group $H$ consisting of matrices of the form 
$ 
  \left( {\begin{array}{cc}
   1 & n \\       0 & 1 \\      \end{array} } \right)
$ cyclic, where $n \in \mathbb{Z}$? If not, how would you show this?
 A: if we define:
$$
B=\left( {\begin{array}{cc}
   0 & 1 \\       0 & 0 \\      \end{array} } \right)
$$
then
$$
A = I+B \\
B^2 = 0
$$
so by the binomial theorem, for $n \ge 0$:
$$
A^n =(I+B)^n = I+nB
$$
note also that:
$$
(I+B)(I-B) = I-B^2=I
$$
so
$$
A^{-1}=I-B \\
A^{-n}=(I-B)^n =I-nB
$$
A: Hint:
Compute the product $\begin{pmatrix}1&n\\0&1\end{pmatrix}\begin{pmatrix}1&p\\0&1\end{pmatrix}$.
A: Observe that $\left( {\begin{array}{cc}1&1\\0&1\end{array}}\right)^m=\left( {\begin{array}{cc}1&m\\0&1\end{array}}\right)$ for each $m\in\Bbb Z$.
A: Here's to get you started. I propose that $H$ is cyclic with generator $A=\left( {\begin{array}{cc}
   1 & 1 \\       0 & 1 \\      \end{array} } \right)$.
Why is this so? Try calculating $A^2$, and in general, you can prove by induction that $A^n = \left( {\begin{array}{cc}
   1 & n \\       0 & 1 \\      \end{array} } \right)$.
Don't forget the matrices for negative $n$. What is $A^{-1}$? How do you get $\left( {\begin{array}{cc}
   1 & -n \\       0 & 1 \\      \end{array} } \right)$ from $A$?
A: The group is indeed cyclic and is generated by the matrix
$$
  J=\left( {\begin{array}{cc}
   1 & 1 \\       0 & 1 \\      \end{array} } \right)
$$
As you have for $n \in \mathbb Z$
$$
  J^n =\left( {\begin{array}{cc}
   1 & n \\       0 & 1 \\      \end{array} } \right)
$$
