If (G, *) is an abelian group, show that(a*b)^2 =a^2*b^2.	

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!

If (G, *) is an abelian group, show that (a * b)^2 = a^2 * b^2.

Explore a detailed proof showing that in an abelian group (a * b)^2 equals a^2 * b^2, using commutative and associative properties of group operations. This is suitable for undergraduate students studying group theory.

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