Proving Associativity of Direct Product of Two Groups

[chaotixmonjuish]

Suppose you had the following:

(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

a_i , b_i$$\in A,B$$

[(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]

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)?

…

 

[chaotixmonjuish]

Your hint kind of confused me.

 

[chaotixmonjuish]

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 a_i^{^1} and b_i^-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]

hmm … a bit complicated …

try starting with [(a1,b1)(a2,b2)](a3,b3) = ([a1a2]a3,[b1b2]b3)

 

[chaotixmonjuish]

Outside of that little thing, am I on the right track?

 

[tiny-tim]

hmm … I got very confused by your …

Since we know (A,*) and (B,*') is a group, we know there exists identity elements a_i^{^1} and b_i^-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.

… which I still don't see the need for.

on looking back, your method seems to be the same as mine (with the first line left out)

This really is a "straightforward exercise" (as your book calls it), but you seem to have a knack for making it complicated.

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