Is this group of matrices cyclic?

Question

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?

Answer 1

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 $n \in \mathbb Z$

$$ J^n =\left( {\begin{array}{cc} 1 & n \\ 0 & 1 \\ \end{array} } \right) $$

Answer 2

Hint:

Compute the product $\begin{pmatrix}1&n\\0&1\end{pmatrix}\begin{pmatrix}1&p\\0&1\end{pmatrix}$.

Answer 3

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 $A$?

Answer 4

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 $m\in\Bbb Z$.

Answer 5

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 $$ A^{-1}=I-B \\ A^{-n}=(I-B)^n =I-nB $$