proof question?

Question

let A and B be groups. Prove that AXB ~= BXA

NOTE: ~= means a horizontal bar under ~.

Answer 1

If a group is defined to be commutative under multiplication (and, since I can't remember the definition, I am not sure whether this is so), then it is true by inspection, since commutation is the only thing going on here.

Answer 2

consider the following map
F: A X B ---> B X A
F(a,b)=(b,a)

it is a homomorphism,
its kernel is (e_A,e_B)
and it is surjective
therefore it is an isomorphism

Answer 3

question not clear