Showing that AxB is isomorphic to BxA

[Mr Davis 97]

Let $A$ and $B$ be groups. Prove that $A \times B \cong B \times A$. I want to show this by giving an explicit isomorphism. Let $\phi : A \times B \to B \times A$, where $\phi (a,b) = (b,a)$. First, it's clear that $\phi$ is a homomorphism because $\phi((a,b)(a',b')) = \phi (aa', bb') = (bb', aa') = (b,a)(b',a') = \phi(a,b) \phi (a',b')$.

The next step is what I have a question about. In this case would it be better to show that $\gamma (b,a) = (a,b)$ is the inverse of $\phi$, or show explicitly that $\phi$ is injective and surjective?

 

[fresh_42]

Let $A$ and $B$ be groups. Prove that $A \times B \cong B \times A$. I want to show this by giving an explicit isomorphism. Let $\phi : A \times B \to B \times A$, where $\phi (a,b) = (b,a)$. First, it's clear that $\phi$ is a homomorphism because $\phi((a,b)(a',b')) = \phi (aa', bb') = (bb', aa') = (b,a)(b',a') = \phi(a,b) \phi (a',b')$.

The next step is what I have a question about. In this case would it be better to show that $\gamma (b,a) = (a,b)$ is the inverse of $\phi$, or show explicitly that $\phi$ is injective and surjective?

Both would be o.k., but it's so obvious, that you don't really need a proof. For the version with $\gamma$ don't forget, that both have to be proven: $\gamma \phi = \operatorname{id}_{A \times B}$ and $\phi \gamma = \operatorname{id}_{B \times A}$. However, surjectivity is immediately clear and so is injectivity.