So I have to prove the following that if G and H are Groups then
$G \times H \cong H \times G $
I was thinking about showing the homomorphism:
I would define $f: G \times H \rightarrow G$ by $f(a,b)=(a,b)$
$f(a,b)=(a,b)=...$
I was hoping to type more of my answer, but I am not sure on how to proceed any hints would be helpful.
On
Okay thanks so I gave it a try: Injective $f(a,b)=f(c,d) \implies (a,b)=(c,d) ?$. $f(a,b)=f(c,d) \implies (b,a)=(d,c) \implies (a,b)=(c,d)$ Injective
Onto: $(f(G \times H) = f((a,b)| a \in G \ and\ b \in H) = ((b,a)|(b,a) \in H \times G) = H \times G $ Hence onto
Homomorphism since $f((a,b),(d,c))=[(b,a),(d,c))=((b,d),(a,c))=((b,a),(d,c))=f(a,b)f(c,d)$
That doesn't make sense. What if $a\notin H$?
Hint: Try $f(a,b)=(b,a)$ instead.