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The problem provided is from group theory and asks to prove that a function defined on a group is bijective, and to verify certain properties. Here's a breakdown of the problem:

**Given:**
- \( G \) is a group.
- \( a \in G \) is a fixed element of the group.
- The function \( h_a: G \to G \) is defined by \( h_a(x) = axa^{-1} \).

You are asked to:
1. Prove that the function \( h_a(x) = axa^{-1} \) is bijective.
2. Verify the following properties:
   - \( h_a(xy) = h_a(x) h_a(y) \) for any \( x, y \in G \).
   - \( h_a(e) = e \), where \( e \) is the identity element of \( G \).
   - \( h_a(x^{-1}) = (h_a(x))^{-1} \) for all \( x \in G \).

---

### **1. Proving that \( h_a(x) = axa^{-1} \) is bijective:**
- **Injectivity**: To prove that \( h_a \) is injective, assume that \( h_a(x_1) = h_a(x_2) \). This gives us:
  \[
  ax_1a^{-1} = ax_2a^{-1}
  \]
  Multiply both sides on the left by \( a^{-1} \) and on the right by \( a \) to get:
  \[
  x_1 = x_2
  \]
  Therefore, \( h_a \) is injective.

- **Surjectivity**: For surjectivity, for any \( y \in G \), we want to find an element \( x \in G \) such that \( h_a(x) = y \). That is, we want:
  \[
  axa^{-1} = y
  \]
  Multiply both sides on the left by \( a^{-1} \) and on the right by \( a \) to solve for \( x \):
  \[
  x = a^{-1}ya
  \]
  Therefore, for any \( y \in G \), there exists \( x = a^{-1}ya \in G \) such that \( h_a(x) = y \). Thus, \( h_a \) is surjective.

Since \( h_a \) is both injective and surjective, it is bijective.

---

### **2. Verifying the properties:**
- **Homomorphism property**: Verify that \( h_a(xy) = h_a(x) h_a(y) \) for any \( x, y \in G \):
  \[
  h_a(xy) = a(xy)a^{-1} = axa^{-1}aya^{-1} = h_a(x)h_a(y)
  \]
  Therefore, \( h_a \) preserves the group operation.

- **Identity element**: Verify that \( h_a(e) = e \):
  \[
  h_a(e) = aea^{-1} = aa^{-1} = e
  \]
  Thus, \( h_a(e) = e \), as required.

- **Inverses**: Verify that \( h_a(x^{-1}) = (h_a(x))^{-1} \) for all \( x \in G \):
  \[
  h_a(x^{-1}) = a(x^{-1})a^{-1}
  \]
  On the other hand:
  \[
  (h_a(x))^{-1} = (axa^{-1})^{-1} = a(x^{-1})a^{-1}
  \]
  Therefore, \( h_a(x^{-1}) = (h_a(x))^{-1} \).

---

### Conclusion:
All required properties of the function \( h_a(x) = axa^{-1} \) are verified, and we have shown that it is bijective.

---

Would you like any further clarification or details on any of these steps? Here are five related questions for practice:

1. How does the concept of conjugation relate to group automorphisms?
2. Can every group have a bijective conjugation map, or does this only apply to specific types of groups?
3. What are the implications of \( h_a \) being a group automorphism for the structure of \( G \)?
4. How would you prove that the set of all conjugations forms a subgroup of the automorphism group of \( G \)?
5. Can this result be extended to infinite groups, and if so, how?

**Tip:** When working with groups, always verify whether a map preserves the group operation, as this is key to identifying homomorphisms and automorphisms.

Sean (G, ·) un grupo, a ∈ G. Demuestre que la aplicación h_a: G → G, dada por h_a(x) = axa^{-1} es biyectiva. Verifique que h_a(xy) = h_a(x) h_a(y) cualesquiera que sean x, y ∈ G, que h_a(e) = e, y que h_a(x^{-1}) = (h_a(x))^{-1} para todo x ∈ G.

Learn how to prove the bijectivity of the conjugation map h_a(x) = axa^{-1} in a group G and verify key properties like h_a(xy) = h_a(x) h_a(y), h_a(e) = e, and h_a(x^{-1}) = (h_a(x))^{-1}. This problem is suitable for undergraduate-level group theory.

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