To prove that if $f: G \to G$, defined by $f(g) = g^2$, is a homomorphism, then $G$ must be abelian:

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Step 1: Homomorphism property

If $f$ is a homomorphism, it satisfies the condition: $f(g_1 g_2) = f(g_1) f(g_2) \quad \text{for all } g_1, g_2 \in G.$

Using the definition of $f(g) = g^2$, this becomes: $(g_1 g_2)^2 = g_1^2 g_2^2.$

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Step 2: Expanding $(g_1 g_2)^2$

Using the group operation, $(g_1 g_2)^2$ means $(g_1 g_2)(g_1 g_2)$. By associativity of the group operation, we can write: $(g_1 g_2)^2 = g_1 g_2 g_1 g_2.$

On the other hand, $g_1^2 g_2^2$ means: $g_1^2 g_2^2 = g_1 g_1 g_2 g_2.$

Thus, the homomorphism condition implies: $g_1 g_2 g_1 g_2 = g_1 g_1 g_2 g_2.$

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Step 3: Simplifying the equality

To simplify, cancel $g_1$ from the left of both sides: $g_2 g_1 g_2 = g_1 g_2 g_2.$

Cancel $g_2$ from the right of both sides: $g_2 g_1 = g_1 g_2.$

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Step 4: Conclusion

The equality $g_2 g_1 = g_1 g_2$ for arbitrary $g_1, g_2 \in G$ shows that the group $G$ is abelian, because the group operation is commutative.

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Final Statement

If $f(g) = g^2$ is a homomorphism, then $G$ must be abelian.

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Do you want further clarifications, or have any related questions? Here are some follow-up questions you might consider:

1. What are other examples of homomorphisms that imply specific group properties?
2. What happens if $f(g) = g^n$ for a general $n$?
3. Can $f(g) = g^2$ be a homomorphism in non-abelian groups?
4. How does this result generalize to finite groups?
5. What is the significance of $f(g) = g^2$ in cyclic groups?

Tip: Always verify the group properties when dealing with homomorphisms; it often reveals important structural insights.