So I need to show that for groups $A,B$ that $$A\times B\cong B\times A$$ So i gather that I have to show that there exists a bijective homomorphsim between the two groups, and so the natural morphism I thought was $$\phi((a,b))=(b,a)$$
But I have to show first that this IS a homomorphism. So I consider two elements of $A\times B$. So let $(a,b), (c,d)\in A\times B$ where $a,c\in A, b,d\in B$. Then $$\phi((a,b)*(c,d))=\phi(a\square c,b\circ d)=(b\circ d,a\square c)=\phi((b,a)*(d,c))$$
I feel this is getting me nowhere.