Let 𝔽7 = ℤ/7ℤ be the finite field with seven elements, and let GL2
(𝔽7
) be the group of
invertible 2 × 2 matrices over 𝔽7 with matrix multiplication. Let 𝐻 = ⟨2⟩ = {2𝑛
| 𝑛 ∈ ℤ}
be the (multiplicative) subgroup of 𝔽∗
7
generated by 2, and let

𝐺 = {(
𝑎 𝑏
0 𝑎)
|
𝑎 ∈ 𝐻 , 𝑏 ∈ 𝔽7}.

2. What is the order of 𝐺?
3. Show that the map 𝜙 ∶ 𝔽7 → 𝐺 given by 𝜙(𝑏) = (
1 𝑏
0 1) is a group homomorphism

Let's address the problem step by step.

### Problem 2: Finding the order of \( G \)

Given:
- \( \mathbb{F}_7 = \mathbb{Z}/7\mathbb{Z} \) is the finite field with 7 elements.
- \( H = \langle 2 \rangle = \{ 2^n \mid n \in \mathbb{Z} \} \) is the multiplicative subgroup of \( \mathbb{F}_7^* \) generated by 2.
- \( G = \left\{ \left( \begin{matrix} a & b \\ 0 & a \end{matrix} \right) \mid a \in H, \, b \in \mathbb{F}_7 \right\} \).

#### Step 1: Determine the structure of \( H \)
- \( H = \langle 2 \rangle \) is the group generated by 2 under multiplication modulo 7.
- The elements of \( \mathbb{F}_7^* \) are \( \{1, 2, 3, 4, 5, 6\} \).
- \( 2^1 \equiv 2 \), \( 2^2 \equiv 4 \), \( 2^3 \equiv 8 \equiv 1 \mod 7 \).

Thus, \( H = \{1, 2, 4\} \), which has order 3.

#### Step 2: Count the number of elements in \( G \)
- For each \( a \in H \) (3 choices), \( b \) can be any element of \( \mathbb{F}_7 \) (7 choices).

Hence, the order of \( G \) is:
\[
|G| = |H| \times |\mathbb{F}_7| = 3 \times 7 = 21
\]

### Problem 3: Show that the map \( \phi: \mathbb{F}_7 \to G \) given by \( \phi(b) = \left( \begin{matrix} 1 & b \\ 0 & 1 \end{matrix} \right) \) is a group homomorphism

#### Step 1: Verify the group operation
- \( \mathbb{F}_7 \) under addition.
- \( G \) under matrix multiplication.

#### Step 2: Verify the homomorphism property
For \( b_1, b_2 \in \mathbb{F}_7 \), we need to show:
\[
\phi(b_1 + b_2) = \phi(b_1) \cdot \phi(b_2)
\]
where:
\[
\phi(b_1 + b_2) = \left( \begin{matrix} 1 & b_1 + b_2 \\ 0 & 1 \end{matrix} \right)
\]
and:
\[
\phi(b_1) \cdot \phi(b_2) = \left( \begin{matrix} 1 & b_1 \\ 0 & 1 \end{matrix} \right) \cdot \left( \begin{matrix} 1 & b_2 \\ 0 & 1 \end{matrix} \right) = \left( \begin{matrix} 1 \times 1 + b_1 \times 0 & 1 \times b_2 + b_1 \times 1 \\ 0 \times 1 + 1 \times 0 & 0 \times b_2 + 1 \times 1 \end{matrix} \right) = \left( \begin{matrix} 1 & b_1 + b_2 \\ 0 & 1 \end{matrix} \right)
\]

Since \( \phi(b_1 + b_2) = \phi(b_1) \cdot \phi(b_2) \), the map \( \phi \) is indeed a group homomorphism.

### Summary
1. The order of \( G \) is 21.
2. The map \( \phi: \mathbb{F}_7 \to G \) given by \( \phi(b) = \left( \begin{matrix} 1 & b \\ 0 & 1 \end{matrix} \right) \) is a group homomorphism.

Would you like more details on any of these steps, or do you have any questions?

### Relative Questions
1. How would the problem change if \( \mathbb{F}_7 \) were replaced by \( \mathbb{F}_5 \)?
2. Can you describe the kernel of the homomorphism \( \phi \)?
3. How would you describe the image of the homomorphism \( \phi \)?
4. What is the significance of the group \( G \) in the context of linear algebra?
5. How does the order of \( H \) influence the structure of \( G \)?
6. What are some examples of other matrices that belong to \( G \)?
7. How would the analysis change if \( H \) had more or fewer elements?
8. Can you identify other homomorphisms from \( \mathbb{F}_7 \) to \( G \)?

### Tip
When checking if a map is a homomorphism, always ensure to verify the group operation compatibility by explicitly performing the operations on both sides.

Explore the structure of GL2(F7), the group of invertible 2x2 matrices over the finite field F7, including subgroup analysis and homomorphism mapping. Detailed solutions provided for specific queries regarding order determination and homomorphism verification.

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