The problem is a group theory proof that involves showing two fundamental properties of groups:

1. The identity element in a group is unique.
2. Every element in the group has a unique inverse.

Proof:

1. Uniqueness of the Identity Element

Let $G$ be a group with binary operation $*$, and let $e_1$ and $e_2$ be two identity elements in $G$.
By definition of the identity element:

• $e_1 * a = a$ for all $a \in G$,
• $e_2 * a = a$ for all $a \in G$.

Now, consider $e_1 * e_2$:

• Using $e_1$ as an identity element: $e_1 * e_2 = e_2$,
• Using $e_2$ as an identity element: $e_1 * e_2 = e_1$.

Thus, $e_1 = e_2$, proving that the identity element is unique.

2. Uniqueness of Inverses

Let $a \in G$, and suppose $b$ and $c$ are both inverses of $a$.
By definition of the inverse:

• $a * b = e$ and $b * a = e$,
• $a * c = e$ and $c * a = e$.

Now, consider $b$:
b = b * e \quad \text{(since e is the identity element)}. Substitute $e = a * c$ (from the definition of the inverse): $b = b * (a * c).$ Using the associativity property of the group operation: $b = (b * a) * c.$ Since $b * a = e$ (from the definition of inverse): $b = e * c.$ Finally, $b = c$, proving that the inverse of $a$ is unique.

Conclusion:

1. The identity element in a group $G$ is unique.
2. Every element in $G$ has a unique inverse.

Do you have questions or need further clarification?
Here are 5 related questions to deepen understanding:

1. What are the axioms that define a group?
2. Can a group have more than one identity element if the operation changes?
3. How is the proof of inverse uniqueness affected by associativity?
4. What would happen in a set with an operation that is not a group?
5. Why does this proof fail if the set is not closed under the operation?

Tip: Always ensure the operation satisfies all group axioms (closure, associativity, identity, and inverses) when working with groups.