Showing that AxB is isomorphic to BxA

  • #1
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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?
 

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  • #2
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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.
 
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