Group Theory: Condition for Commutativity

This question asks us to identify the correct statement about a group G, which is a fundamental concept in abstract algebra. Let's analyze the given options.

Option 1: Analysis

The statement is: If for all $x, y \in G$, $(xy)^2 = x^2y^2$, then G is Commutative.

Let's verify this statement. Assume that for any two elements $x$ and $y$ in the group $G$, the property $(xy)^2 = x^2y^2$ holds.

1. Start with the given condition: $(xy)^2 = x^2y^2$
2. Expand the left side: $(xy)^2 = (xy)(xy)$. $xyxy = x^2y^2$
3. Multiply both sides by the inverse of $x$ (denoted $x^{-1}$) on the left. Since $x \in G$, $x^{-1}$ exists. $x^{-1}(xyxy) = x^{-1}(x^2y^2)$
4. Using the associative property, we can rearrange the left side: $(x^{-1}x)yxy = (x^{-1}x)xy^2$ $e \cdot yxy = e \cdot xy^2$ (where $e$ is the identity element of the group) $yxy = xy^2$
5. Now, multiply both sides by the inverse of $y$ (denoted $y^{-1}$) on the right. Since $y \in G$, $y^{-1}$ exists. $(yxy)y^{-1} = (xy^2)y^{-1}$
6. Using the associative property again: $y x (y y^{-1}) = x (y^2 y^{-1})$ $y x e = x (y y y y^{-1})$ $yx = x y$

We have shown that if the condition $(xy)^2 = x^2y^2$ holds for all $x, y \in G$, then $yx = xy$. The condition $yx = xy$ is the definition of a commutative group (also called an Abelian group). Therefore, the statement is correct.

Option 2: Analysis

The statement is: If for all $x \in G$, $x^3 = 1$, then G is Commutative. 1 is the identity element of G.

This statement is also mathematically correct. A theorem in group theory states that if the exponent of a group $G$ divides an odd number $n$ (meaning $x^n=e$ for all $x \in G$ and $n$ is odd), then $G$ is commutative. Here, $n=3$, which is odd. The part "1 is the identity element of G" is standard notation ($1$ or $e$ represents the identity). However, Option 1 is presented as the correct answer.

Option 3: Analysis

The statement is: If for all $x \in G$, $x^5 = 1$, then G is Commutative. 1 is the identity element of G.

Similar to Option 2, this statement is also mathematically correct because $n=5$ is an odd number. If $x^5 = e$ for all $x \in G$, then $G$ is commutative.

Option 4: Analysis

The statement is: If G is Commutative, the sub-group of G need not be Commutative.

This statement is incorrect. A fundamental property of groups is that any subgroup of a commutative group is also commutative. If $G$ is commutative, then $xy = yx$ for all $x, y \in G$. If $H$ is a subgroup of $G$, then any elements $x, y \in H$ are also elements of $G$. Therefore, $xy = yx$ must hold for all elements in $H$, meaning $H$ is also commutative.

Conclusion

Based on the analysis, Option 1 provides a correct condition that guarantees a group is commutative. The derivation shows that $(xy)^2 = x^2y^2$ directly leads to $xy=yx$. Options 2 and 3 are also correct statements about group theory, but typically, multiple-choice questions aim for a specific answer. Option 4 is definitively false.