Let 𝐺 and 𝐻 be groups, and let 𝜙 ∶ 𝐺 → 𝐻 and 𝜓 ∶ 𝐺 → 𝐻 be two group homomorphisms.
Define
𝐸 = {𝑔 ∈ 𝐺 | 𝜙(𝑔) = 𝜓 (𝑔)}.
1. Show that 𝐸 is a subgroup of 𝐺.
Assume now that 𝐻 is abelian, and let 𝜃 ∶ 𝐺 → 𝐻 be given by 𝜃(𝑔) = 𝜙(𝑔)𝜓 (𝑔)−1
.
2. Show that 𝜃 is a group homomorphism, and that 𝐸 is its kernel.

Let's solve the problem in two parts as requested.

### Part 1: Showing that \( E \) is a Subgroup of \( G \)

To show that \( E \) is a subgroup of \( G \), we need to verify that \( E \) satisfies the subgroup criteria:

1. **Identity:** The identity element of \( G \) is in \( E \).
2. **Closure:** If \( g_1, g_2 \in E \), then \( g_1g_2 \in E \).
3. **Inverses:** If \( g \in E \), then \( g^{-1} \in E \).

**Proof:**

1. **Identity:**
   - Let \( e_G \) be the identity element in \( G \). We need to show that \( e_G \in E \). Since \( \phi \) and \( \psi \) are group homomorphisms,
     \[
     \phi(e_G) = e_H \quad \text{and} \quad \psi(e_G) = e_H,
     \]
     where \( e_H \) is the identity element in \( H \). Hence,
     \[
     \phi(e_G) = \psi(e_G),
     \]
     which implies \( e_G \in E \).

2. **Closure:**
   - Let \( g_1, g_2 \in E \). We need to show that \( g_1g_2 \in E \). Since \( g_1, g_2 \in E \), we have \( \phi(g_1) = \psi(g_1) \) and \( \phi(g_2) = \psi(g_2) \). Then,
     \[
     \phi(g_1g_2) = \phi(g_1)\phi(g_2) = \psi(g_1)\psi(g_2) = \psi(g_1g_2).
     \]
     Therefore, \( g_1g_2 \in E \).

3. **Inverses:**
   - Let \( g \in E \). We need to show that \( g^{-1} \in E \). Since \( g \in E \), we have \( \phi(g) = \psi(g) \). Then,
     \[
     \phi(g^{-1}) = \phi(g)^{-1} = \psi(g)^{-1} = \psi(g^{-1}).
     \]
     Thus, \( g^{-1} \in E \).

Since \( E \) contains the identity, is closed under the group operation, and contains inverses, \( E \) is a subgroup of \( G \).

### Part 2: Showing that \( \theta \) is a Group Homomorphism and \( E \) is its Kernel

We define \( \theta \colon G \to H \) by \( \theta(g) = \phi(g)\psi(g)^{-1} \). We need to show that \( \theta \) is a group homomorphism and that \( \ker(\theta) = E \).

**Proof:**

1. **Homomorphism Property:**
   - For \( g_1, g_2 \in G \), we have
     \[
     \theta(g_1g_2) = \phi(g_1g_2)\psi(g_1g_2)^{-1}.
     \]
     Since \( \phi \) and \( \psi \) are homomorphisms, we get
     \[
     \phi(g_1g_2) = \phi(g_1)\phi(g_2) \quad \text{and} \quad \psi(g_1g_2) = \psi(g_1)\psi(g_2).
     \]
     Therefore,
     \[
     \theta(g_1g_2) = \phi(g_1)\phi(g_2)\psi(g_2)^{-1}\psi(g_1)^{-1}.
     \]
     Since \( H \) is abelian, we can rearrange the terms:
     \[
     \theta(g_1g_2) = \phi(g_1)\psi(g_1)^{-1}\phi(g_2)\psi(g_2)^{-1} = \theta(g_1)\theta(g_2).
     \]
     Thus, \( \theta \) is a group homomorphism.

2. **Kernel of \( \theta \):**
   - We need to show that \( \ker(\theta) = E \). The kernel of \( \theta \) consists of all elements \( g \in G \) such that \( \theta(g) = e_H \), where \( e_H \) is the identity in \( H \). Therefore,
     \[
     \theta(g) = e_H \iff \phi(g)\psi(g)^{-1} = e_H \iff \phi(g) = \psi(g).
     \]
     Thus, \( \ker(\theta) = E \).

Hence, \( \theta \) is a group homomorphism and \( E \) is its kernel.

---

Do you want more details on any part of the solution? Here are some further questions to explore:

1. What are the implications of \( \ker(\theta) \) being a normal subgroup?
2. How does the abelian nature of \( H \) affect the properties of \( \theta \)?
3. Can you provide an example where \( \phi \) and \( \psi \) are not equal and describe \( E \)?
4. What is the quotient group \( G/E \) in this context?
5. How does the First Isomorphism Theorem apply to \( \theta \)?
6. Can \( \theta \) be surjective, and under what conditions?
7. How would the problem change if \( H \) were not abelian?
8. What can you infer about the structure of \( G \) if \( \theta \) is trivial?

**Tip:** When verifying homomorphisms, always check the operation preservation property and ensure the mapping respects the group structure.

Explore group theory concepts by proving that E, defined by the equality of two group homomorphisms, is a subgroup of G. Understand how the abelian nature of H affects the homomorphism theta and why E serves as its kernel. Suitable for advanced undergraduate students and researchers in mathematics.

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