Sure! Here’s a classic problem in abstract algebra:

**Problem:** Let \( G \) be a group, and let \( a, b \in G \). Show that if \( a^2 = e \) (where \( e \) is the identity element in \( G \)), then \( a^{-1} = a \).

**Solution:**

1. Since \( a^2 = e \), we can write:
   \[
   a \cdot a = e
   \]
2. To find \( a^{-1} \), we note that \( a \cdot a = e \) implies:
   \[
   a^{-1} = a
   \]
   because multiplying both sides of the equation \( a \cdot a = e \) by \( a^{-1} \) on the left gives:
   \[
   a^{-1} \cdot (a \cdot a) = a^{-1} \cdot e \implies (a^{-1} \cdot a) \cdot a = a^{-1} \implies e \cdot a = a^{-1} \implies a = a^{-1}.
   \]
Thus, \( a^{-1} = a \) holds true.

Would you like more details or have any questions? Here are five related questions for further exploration:

1. What are other examples of elements \( a \) in groups such that \( a^n = e \) for some integer \( n \)?
2. Can you prove similar properties for elements of finite order in groups?
3. How does the property of being an involution (where \( a^2 = e \)) relate to group homomorphisms?
4. What implications does this have for the structure of the group \( G \)?
5. Can you provide an example of a specific group where such elements exist, like \( S_3 \) or \( D_n \)? 

**Tip:** Familiarize yourself with group properties like order, identity, and inverses to better understand group structures.

Let G be a group, and let a, b ∈ G. Show that if a^2 = e (where e is the identity element in G), then a^{-1} = a.

Explore the proof that if a^2 = e in a group G, then a^{-1} = a. This page delves into the properties of group elements, including involutions and identity elements, suitable for high school and college students studying abstract algebra.

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