Solution

The correct answer is: Option 1

Problem recap: Choose the correct statement about a group G (group operation written multiplicatively).

Why Option 1 is true – “If for all x, y ∈ G, (xy)² = x²y² then G is commutative.”

• Expand both sides using the group operation:
  (xy)^2 = xyxy and x^2y^2 = xxyy.
• Given equality: xyxy = xxyy.
• Left–multiply by x^{-1} (possible in a group):
  x^{-1}(xyxy) = x^{-1}(xxyy) ⇒ yxy = xyy.
• Right–multiply by y^{-1}:
  (yxy)y^{-1} = (xyy)y^{-1} ⇒ yx = xy.
• This holds for all x,y ∈ G, hence G is abelian (commutative).

Why the other options are false

Option 2 – “If for all x ∈ G, x³ = 1, then G is commutative. 1 is the identity element of G.”

• There exist non-abelian groups in which every element has order dividing 3 (exponent 3). Example: the group of 3×3 upper-triangular matrices with 1s on the diagonal over the field ℤ/3ℤ (often written UT3(3)). In this group, each element satisfies A^3 = I but the group is not abelian (take two elementary matrices; they do not commute).
• So “G is commutative” is false. (Also, writing the identity as “1” is just a notation; the statement as a whole is not correct.)

Option 3 – “If for all x ∈ G, x⁵ = 1, then G is commutative. 1 is the identity element of G.”

• Similarly false: there are non-abelian groups of exponent 5, e.g., the 3×3 upper-unitriangular matrices over ℤ/5ℤ (UT3(5)). Every element satisfies A^5 = I, but the group is non-abelian.

Option 4 – “If G is commutative, the subgroup of G need not be commutative.”

• False. Any subgroup H ≤ G of an abelian group is abelian: for any a,b ∈ H, since ab = ba in G, the same equality holds in H.

Conclusion: Only Option 1 is correct; therefore the answer is Option 1.