# Direct Product of two Groups

(A,*) and (B,$$\nabla$$)

So to prove associativity, since I know that both A and B are groups, their direct product will be a group. Could I do the following

ai , bi$$\in A,B$$

[(a1,b1)(a2,b2)](a3,b3)=(a1,b1)[(a2,b2)(a3,b3)]

Since A and B are groups, I know they have distributing everything via the proper binary operations (I got kind of lazy at this point). Can I just multiply both sides by an a-11b-11 and so on until i get something like b3=b3

I just want to make sure this is sort of hte process one uses to prove the direct product is a group. By the way this question comes from Dummit and Foote, it says that the proof of this is left as a straightforward excercise.

tiny-tim
Homework Helper
Hi chaotixmonjuish! [(a1,b1)(a2,b2)](a3,b3)=(a1,b1)[(a2,b2)(a3,b3)]

By the way this question comes from Dummit and Foote, it says that the proof of this is left as a straightforward excercise.

Yup, it is straightforward …

i] what is [(a1,b1)(a2,b2)]?

ii] so what is [(a1,b1)(a2,b2)](a3,b3)? Your hint kind of confused me.

After multiplying everything

I'm changing the triangle to *'

(a1*a2,b1*'b2)(a3,b3)=(a1,b1)(a2*a3,b2*'b3)

Then once more:

(a1*a2*a3,b1*'b2*'b3)=
(a1*a2*a3,b1*'b2*'b3)

Since we know (A,*) and (B,*') is a group, we know there exists identity elements ai^1 and bi-1. We can use those identities to get all the a's equal to each other or the b's equal. That's shows associativity.

tiny-tim
Homework Helper
hmm … a bit complicated …

try starting with [(a1,b1)(a2,b2)](a3,b3) = ([a1a2]a3,[b1b2]b3) Outside of that little thing, am I on the right track?

tiny-tim
… which I still don't see the need for. Just start with my line, use the associative law on each side of the comma separately, then separate out (a1,b1), and you're done 