Question 27

Evaluating each statement for a group $G$:

• (A) TRUE: Given $(xy)^2 = x^2 y^2$ for all $x, y \in G$. Expanding both sides gives $xyxy = xxyy$. Left-multiplying by $x^{-1}$ yields $yxy = xyy$. Right-multiplying by $y^{-1}$ yields $yx = xy$. Since this holds for all $x, y \in G$, $G$ is commutative (abelian).
• (B) TRUE: If $x^2 = 1$ for all $x \in G$, then every element is its own inverse ($x = x^{-1}$). For any $x, y \in G$, $(xy) \in G$, so $(xy)^{-1} = xy$. By the reversal law of inverses, $(xy)^{-1} = y^{-1} x^{-1} = yx$. Thus $xy = yx$, making $G$ commutative.
• (C) TRUE: A group of prime order 2 contains only the identity and one non-identity element, $G = \{1, a\}$. Since $1a = a1 = a$ and $a^2 = 1$, the operation is trivially commutative.
• (D) FALSE: If $G$ is commutative, any subgroup $H \subseteq G$ inherits the binary operation of $G$, and since $xy = yx$ holds for all $x, y \in G$, it must hold for all $x, y \in H$. Thus every subgroup of an abelian group is necessarily abelian.