Understanding the Direct Product of Groups: Applying Group Theory Axioms

[Monkeyfry180]

How do we know that the cartesian product of any two groups is also a group using the axioms of group theory?

 

[quasar987]

Exactly like that! You verify that all the axioms defining a group are satisfied. But this is if you know what the group operation is on the product.

But perhaps your question is "how do we define a binary operation on the cartesian product of two groups that makes it into a group"?

 

[Monkeyfry180]

Well let's say we have the two groups G1 and G2 with operations *1 and *2, respectively, and we do the cartesian product to get
G1 x G2 = { (a,b) : a is an element of G1, and b is an element of G2} = G

with the binary operation, * let's say, defined by
(a,b) * (c,d) = (a *1 c, b *2 d)

Prove that (G, *) is a group.

I tried setting up a table of what this would look like, but I'm having some real issues with this

 

[Monkeyfry180]

It's easy to talk myself through associativity, but the other three are giving me trouble

 

[Gigasoft]

- Closure: This is obvious. a *1 c is a member of G1, and b *2 d is a member of G2. Thus, (a *1 c,b *2 d) is a member of G1 x G2 by definition.
- Identity element: If e1 is the identity in G1, and e2 is the identity in G2, the identity in G1 x G2 is (e1,e2).
- Inverse element: With a *1 c = e1 and b *2 d = e2, (a *1 c,b *2 d) = (e1,e2), and (c,d) is therefore the inverse of (a,b).

 

[Monkeyfry180]

Alright, I just did the proofs and got the same answer, thank you so much.

Also, using those same values, if f1: G1 --> G is defined by f1(g) = (g, e2), how can we prove that f1 is an homomorphism, one to one, and onto?

 

[Landau]

Perhaps you can show what you have tried, instead of letting us do your homework?