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The problem asks to prove that \( H = \left\{ \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} : n \in \mathbb{Z} \right\} \subseteq GL(2, \mathbb{R}) \) is a cyclic subgroup of \( GL(2, \mathbb{R}) \) and to show that the matrices \( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \) and \( \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \) are generators of \( H \).

### Key Steps:

1. **Matrix Powers:**
   Each matrix in \( H \) is of the form \( \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \). For such a matrix, the product of two such matrices is:
   \[
   \begin{pmatrix} 1 & n_1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & n_2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & n_1 + n_2 \\ 0 & 1 \end{pmatrix}
   \]
   This implies that the group operation in \( H \) is essentially addition on \( n \in \mathbb{Z} \).

2. **Cyclic Nature:**
   Since the group operation corresponds to addition, \( H \) is isomorphic to \( \mathbb{Z} \), which is a cyclic group. Therefore, \( H \) is also a cyclic group.

3. **Generators:**
   - Consider the matrix \( A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \). For any positive integer \( n \), we have:
     \[
     A^n = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}
     \]
   - Similarly, consider \( B = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \). For any negative integer \( n \), we have:
     \[
     B^n = \begin{pmatrix} 1 & -n \\ 0 & 1 \end{pmatrix}
     \]
   Hence, the matrices \( A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \) generate the entire group \( H \), since every matrix in \( H \) can be expressed as a power of \( A \) or \( B \).

Thus, \( H \) is a cyclic group, and both \( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \) and \( \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \) are generators.

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Would you like more details on the matrix group properties, or do you have any questions?

Here are some related questions:
1. What does it mean for a group to be cyclic?
2. How do we prove that a matrix group is isomorphic to \( \mathbb{Z} \)?
3. Can a cyclic group have more than one generator?
4. What is the significance of the general linear group \( GL(2, \mathbb{R}) \)?
5. How do matrix powers work in other types of matrix groups?

**Tip:** When working with matrix groups, keep in mind how matrix multiplication relates to group operations, often revealing underlying structures like isomorphisms to familiar groups.

Let H = { ( 1 n \0 1 ) | n ∈ Z } ≤ GL(2, R). H is an infinite cyclic group of GL(2, R). Show that (1 1 \0 1) and (1 -1 \0 1) are two generators of H.

Learn how to prove that H, a subset of GL(2, R), is a cyclic group generated by specific matrices. This problem explores key concepts in group theory, cyclic groups, and matrix powers, suitable for advanced students studying linear algebra or abstract algebra.

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