Choose the correct statement for a group G:
2025
Choose the correct statement for a group G:
- A.
If for all x, y ∈ G, (xy)² = x²y² then G is Commutative.
- B.
If for all x ∈ G, x³ = 1, then G is Commutative. 1 is the identity element of G.
- C.
If for all x ∈ G, x⁵ = 1, then G is Commutative. 1 is the identity element of G.
- D.
If G is Commutative, the sub-group of G need not be Commutative."
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Correct answer: A
Answer: The statement "If for all x, y ∈ G, (xy)^2 = x^2 y^2 then G is Commutative." is correct; the other statements are false.
Proof that (xy)^2 = x^2 y^2 implies G is abelian:
Start from (xy)^2 = x^2 y^2, i.e. xyxy = x x y y.
Left-multiply by x^{-1}: x^{-1}(xyxy) = x^{-1}(x x y y), which simplifies to y x y = x y^2.
Right-multiply by y^{-1}: (y x y) y^{-1} = (x y^2) y^{-1}, giving y x = x y.
Since y and x were arbitrary, every pair of elements commutes, so G is abelian.
Why the other statements are false:
The claim that x^3 = 1 for all x implies commutativity is false. Counterexample: the Heisenberg group of 3x3 upper-triangular matrices with 1s on the diagonal over Z_3 (matrices of the form [[1,a,c],[0,1,b],[0,0,1]] with a,b,c in Z_3]) is non-abelian while every element has order dividing 3, so x^3 = 1 for all x.
The claim that x^5 = 1 for all x implies commutativity is false for the same reason: the Heisenberg group over Z_5 is a non-abelian group whose elements have orders dividing 5, so exponent 5 does not force abelian.
The claim that a subgroup of an abelian group need not be abelian is false. If G is abelian and H ≤ G, then for any a,b in H we have ab = ba because the same equality holds in G; hence every subgroup of an abelian group is abelian.