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 work through the problem again, ensuring clarity and correctness in each step.

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

To prove that \( E \) is a subgroup of \( G \), we need to demonstrate that \( E \) satisfies the three necessary conditions: containing the identity element, closure under the group operation, and containing inverses.

**Proof:**

1. **Identity Element:**
   - Let \( e_G \) be the identity element in \( G \). For \( g = e_G \), we have
     \[
     \phi(e_G) = e_H \quad \text{and} \quad \psi(e_G) = e_H,
     \]
     where \( e_H \) is the identity element in \( H \). Thus,
     \[
     \phi(e_G) = \psi(e_G),
     \]
     implying \( e_G \in E \). Therefore, \( E \) contains the identity element of \( G \).

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, \( g_1g_2 \in E \). Thus, \( E \) is closed under the group operation.

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, \( g^{-1} \in E \). Therefore, \( E \) contains inverses of its elements.

Since \( E \) contains the identity element, is closed under the group operation, and contains inverses, \( E \) is indeed 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 \( \theta(g) = \phi(g)\psi(g)^{-1} \). To show that \( \theta \) is a group homomorphism and that \( \ker(\theta) = E \), we proceed as follows:

**Proof:**

1. **Homomorphism Property:**
   - For \( g_1, g_2 \in G \), we need to verify that \( \theta(g_1g_2) = \theta(g_1)\theta(g_2) \). We have
     \[
     \theta(g_1g_2

Explore the mathematical concepts of subgroup E in groups G and H, defined by homomorphisms phi and psi. Understand the criteria for E being a subgroup of G and the properties of the homomorphism theta, with E identified as its kernel. Suitable for advanced undergraduate students studying group theory.

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