Let H = {M ∈ GL2(Z/7Z) : det(M) ∈ {1, 2, 4}}. Prove that GL2(Z/7Z)/H ≃ Z/2Z.

To prove that \( GL_2(\mathbb{Z}/7\mathbb{Z})/H \cong \mathbb{Z}/2\mathbb{Z} \), we need to demonstrate that the quotient group \( GL_2(\mathbb{Z}/7\mathbb{Z})/H \) has order 2 and that it is isomorphic to the cyclic group \( \mathbb{Z}/2\mathbb{Z} \).

### Step 1: Understanding the Groups Involved

- \( GL_2(\mathbb{Z}/7\mathbb{Z}) \) is the group of all \( 2 \times 2 \) invertible matrices over the field \( \mathbb{Z}/7\mathbb{Z} \). The order of \( GL_2(\mathbb{Z}/7\mathbb{Z}) \) is given by:
  \[
  |GL_2(\mathbb{Z}/7\mathbb{Z})| = (7^2 - 1)(7^2 - 7) = 48 \times 42 = 2016.
  \]
- \( H \) is defined as the subgroup of \( GL_2(\mathbb{Z}/7\mathbb{Z}) \) consisting of matrices \( M \) such that \( \det(M) \in \{1, 2, 4\} \).

### Step 2: Size of the Subgroup \( H \)

To understand \( |H| \), consider the following:

- The determinant map \( \det: GL_2(\mathbb{Z}/7\mathbb{Z}) \to \mathbb{Z}/7\mathbb{Z}^* \) is a homomorphism.
- The set \( \{1, 2, 4\} \) is a subgroup of \( \mathbb{Z}/7\mathbb{Z}^* \) with 3 elements (since \( \{1, 2, 4\} \) forms a multiplicative group of order 3 in \( \mathbb{Z}/7\mathbb{Z} \)).

The kernel of the determinant map is the set of matrices with determinant 1, which has order equal to \( |SL_2(\mathbb{Z}/7\mathbb{Z})| \), where \( SL_2(\mathbb{Z}/7\mathbb{Z}) \) is the special linear group of matrices with determinant 1. 

The order of \( SL_2(\mathbb{Z}/7\mathbb{Z}) \) is:
\[
|SL_2(\mathbb{Z}/7\mathbb{Z})| = |GL_2(\mathbb{Z}/7\mathbb{Z})| / |\mathbb{Z}/7\mathbb{Z}^*| = 2016 / 6 = 336.
\]
Since \( H \) is defined as matrices with determinant in \( \{1, 2, 4\} \), and each value of the determinant corresponds to a unique coset in \( GL_2(\mathbb{Z}/7\mathbb{Z}) \), the order of \( H \) is:
\[
|H| = |SL_2(\mathbb{Z}/7\mathbb{Z})| \times 3 = 336 \times 3 = 1008.
\]

### Step 3: Order of the Quotient Group

The quotient group \( GL_2(\mathbb{Z}/7\mathbb{Z})/H \) has order:
\[
\frac{|GL_2(\mathbb{Z}/7\mathbb{Z})|}{|H|} = \frac{2016}{1008} = 2.
\]
This means that \( GL_2(\mathbb{Z}/7\mathbb{Z})/H \) is a group of order 2.

### Step 4: Isomorphism to \( \mathbb{Z}/2\mathbb{Z} \)

A group of order 2 is necessarily isomorphic to the cyclic group \( \mathbb{Z}/2\mathbb{Z} \). Therefore, we can conclude:
\[
GL_2(\mathbb{Z}/7\mathbb{Z})/H \cong \mathbb{Z}/2\mathbb{Z}.
\]

### Conclusion

The quotient group \( GL_2(\mathbb{Z}/7\mathbb{Z})/H \) is isomorphic to \( \mathbb{Z}/2\mathbb{Z} \).

Would you like more details on any part of this proof? Here are some related questions:

1. How do you compute the order of \( SL_2(\mathbb{Z}/7\mathbb{Z}) \)?
2. What is the significance of the determinant map in linear algebra?
3. How do the elements of \( \mathbb{Z}/7\mathbb{Z}^* \) form a group?
4. What other quotient groups are commonly studied in linear algebra?
5. How can we generalize this result to other finite fields?
6. What are some other examples of groups with order 2?
7. How does the structure of \( GL_2(\mathbb{Z}/7\mathbb{Z}) \) influence its subgroups?
8. How would the problem change if we used a different modulus than 7?

**Tip:** When analyzing quotient groups, understanding the order of the original group and its subgroups is key to determining the structure of the quotient.

Learn how to prove GL2(Z/7Z)/H ≃ Z/2Z using group theory and the properties of matrix groups over finite fields. This problem involves understanding quotient groups, determinant maps, and the order of groups. Suitable for advanced undergraduate students and researchers.

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