1. **Problem Statement:** Show that SO(2, \mathbb{Q}) is a subgroup of GL(2, \mathbb{Q}). SO(2, \mathbb{Q}) is defined as the set of all 2\times 2 orthogonal matrices with rational entries and determinant 1. GL(2, \mathbb{Q}) is the group of all invertible 2\times 2 matrices with rational entries. 2. **Step 1: Elements of SO(2, \mathbb{Q})** An element $A \in SO(2, \mathbb{Q})$ satisfies: $$ A^T A = I \quad \text{and} \quad \det(A) = 1, $$ where $A^T$ is the transpose of $A$ and $I$ is the identity matrix. Both conditions mean $A$ is orthogonal with determinant 1. 3. **Step 2: Closure under multiplication** If $A, B \in SO(2, \mathbb{Q})$, then: $$ (AB)^T (AB) = B^T A^T A B = B^T I B = B^T B = I $$ since both $A$ and $B$ are orthogonal. Also, $$ \det(AB) = \det(A) \det(B) = 1 \times 1 = 1. $$ Because the product $AB$ has rational entries (since $A$ and $B$ do), $AB \in SO(2, \mathbb{Q})$. 4. **Step 3: Identity matrix** The identity matrix $I \in SO(2, \mathbb{Q})$ since: $$ I^T I = I \quad \text{and} \quad \det(I) = 1, $$ and $I$ has rational entries. 5. **Step 4: Inverses** For any $A \in SO(2, \mathbb{Q})$, its inverse is $A^{-1} = A^T$ because $A$ is orthogonal. Since $A$ has rational entries, transpose $A^T$ also has rational entries. Also, $$ \det(A^{-1}) = \frac{1}{\det(A)} = 1, $$ so $A^{-1} \in SO(2, \mathbb{Q})$. 6. **Step 5: Conclusion** Since SO(2, \mathbb{Q}) contains the identity, is closed under multiplication, and every element has an inverse in SO(2, \mathbb{Q}), it is a subgroup of GL(2, \mathbb{Q}).