Let $(G, *)$ be an abelian group, meaning that the operation $*$ is commutative, i.e., $a * b = b * a$ for all $a, b \in G$.

To show: $(a * b)^2 = a^2 * b^2$

1. Start with $(a * b)^2$: $(a * b)^2 = (a * b) * (a * b)$ By the definition of squaring an element in a group, this is just the product of $(a * b)$ with itself.

2. Now apply the associativity of the group operation: $(a * b) * (a * b) = a * (b * a) * b$

3. Since $G$ is abelian (commutative), $b * a = a * b$. So we can rewrite the expression as: $a * (a * b * b) = a * a * b * b$

4. Now simplify the expression: $a * a * b * b = a^2 * b^2$

Thus, we have shown that: $(a * b)^2 = a^2 * b^2$

Conclusion:

In an abelian group $(G, *)$, the equation $(a * b)^2 = a^2 * b^2$ holds due to the commutative property of the group operation.

Would you like further details or clarifications?

Here are 5 related questions to deepen your understanding:

1. How would the proof change if $(G, *)$ were not abelian?
2. What is the general form of $(a * b)^n$ for any positive integer $n$ in an abelian group?
3. Can you prove that the identity element in an abelian group is unique?
4. What are examples of non-abelian groups where $(a * b)^2 \neq a^2 * b^2$?
5. How does associativity affect the structure of groups in general?

Tip: In group theory, commutativity simplifies many expressions and proofs. Always check if a group is abelian before applying commutative properties!