# Showing that AxB is isomorphic to BxA

• I
• Mr Davis 97
In summary, the isomorphism ##\phi## between ##A \times B## and ##B \times A## can be proven easily either by showing that ##\gamma## is its inverse or by showing that ##\phi## is injective and surjective.
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?

Mr Davis 97 said:
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.

Mr Davis 97

## 1. What does it mean for two sets to be isomorphic?

When two sets, A and B, are isomorphic, it means that there exists a one-to-one and onto mapping between the elements of A and B. In simpler terms, the elements of A can be paired with the elements of B in such a way that all elements in both sets have a unique pairing.

## 2. How do you show that AxB is isomorphic to BxA?

To show that AxB is isomorphic to BxA, you need to find a bijective function, also known as an isomorphism, that maps elements from AxB to BxA. This can be done by swapping the order of the elements in the Cartesian product, as the order of elements does not affect the bijectivity of the function.

## 3. Can AxB and BxA be isomorphic if A and B are not the same size?

No, AxB and BxA cannot be isomorphic if A and B are not the same size. In order for two sets to be isomorphic, they must have the same number of elements, otherwise the bijective mapping between them cannot exist.

## 4. What is the significance of showing that AxB is isomorphic to BxA?

Showing that AxB is isomorphic to BxA is important because it helps us understand the relationship between the two sets. It also allows us to easily compare the elements and operations of the two sets, as they are essentially the same, just in a different order.

## 5. Are all Cartesian products isomorphic to each other?

No, not all Cartesian products are isomorphic to each other. Isomorphism is a specific type of mapping between sets and not all sets can be mapped in a bijective way. Additionally, as mentioned before, the sets must have the same number of elements in order for them to be isomorphic.

• Linear and Abstract Algebra
Replies
3
Views
3K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
12
Views
1K
• Linear and Abstract Algebra
Replies
4
Views
1K
• Linear and Abstract Algebra
Replies
3
Views
1K
• Linear and Abstract Algebra
Replies
4
Views
2K
• Linear and Abstract Algebra
Replies
1
Views
1K
• Linear and Abstract Algebra
Replies
2
Views
2K
• Linear and Abstract Algebra
Replies
2
Views
3K