Which of the following statements is/are TRUE for a group $G$ ?

Let us analyze each statement individually:

• Statement (A): If for all $x, y \in G$, $(xy)^2 = x^2 y^2$, then:

  $$xyxy = xxyy$$

  Multiplying on the left by $x^{-1}$ and on the right by $y^{-1}$ gives:

  $$x^{-1}(xyxy)y^{-1} = x^{-1}(xxyy)y^{-1} \implies yx = xy$$

  Since $yx = xy$ holds for all $x, y \in G$, $G$ is commutative (abelian). Thus, (A) is TRUE.

• Statement (B): If for all $x \in G$, $x^2 = 1$, then every element is its own inverse, i.e., $x = x^{-1}$. For any $x, y \in G$, $xy \in G$, so:

  $$xy = (xy)^{-1} = y^{-1}x^{-1} = yx$$

  Hence, $G$ is commutative. Thus, (B) is TRUE.

• Statement (C): Any group of prime order is cyclic, and all cyclic groups are commutative. Since $|G| = 2$ is a prime number, $G$ is cyclic and hence commutative. Thus, (C) is TRUE.

• Statement (D): If $G$ is commutative, then the commutative property holds for every pair of elements in $G$. Since any subgroup $H \le G$ is a subset of $G$, the operation is automatically commutative on $H$. Therefore, every subgroup of an abelian group must be abelian (commutative). Thus, (D) is FALSE.

Hence, the correct statements are A, B, and C.