The task asks to find the order of the element

$g = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$

in the group $SL_2(\mathbb{Z})$, the group of $2 \times 2$ integer matrices with determinant 1.

Steps:

1. Definition of Order: The order of an element $g$ in a group is the smallest positive integer $n$ such that $g^n = I$, where $I$ is the identity matrix: $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$

2. Matrix Multiplication: Compute powers of $g$ to find $n$ such that $g^n = I$. Since $g$ is of the form: $g = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix},$ the matrix powers $g^n$ can be computed as: $g^n = \begin{pmatrix} 1 & -n \\ 0 & 1 \end{pmatrix}.$

3. Equality Condition: For $g^n = I$, the condition is: $\begin{pmatrix} 1 & -n \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$

   Comparing elements, we find:

   • Top-right element: $-n = 0 \implies n = 0$.

4. Order of $g$: The matrix does not return to the identity for any finite $n > 0$, as the top-right element $-n$ cannot equal $0$ for positive $n$. Thus, the element $g$ does not have finite order, meaning its order is infinite.

Final Answer:

$o(g) = \infty$

Would you like me to break this down further or discuss related concepts?

Related Questions:

1. What is the significance of the group $SL_2(\mathbb{Z})$?
2. How do we define infinite order elements in a group?
3. Are there conditions for elements in $SL_2(\mathbb{Z})$ to have finite order?
4. Can similar elements in other groups have finite order?
5. How does this relate to modular arithmetic?

Tip:

Always verify matrix powers by explicit computation when checking orders to avoid assumptions.