# Thread: Group Axioms G4 Associative

1. ## Group Axioms G4 Associative

Hi!

Could anyone help!

I need to prove G4 associativity for the following group and I know it exists

(Q*, o) where x o y = xy/7.

This group is closed so G1 holds
This group is has an identity e = 7 so G2 holds
This group has an inverses which is 49/x.

2. Originally Posted by Arron
Hi!

Could anyone help!

I need to prove G4 associativity for the following group and I know it exists

(Q*, o) where x o y = xy/7.

This group is closed so G1 holds
This group is has an identity e = 7 so G2 holds
This group has an inverses which is 49/x.
Associativity is the `easy' one, because you don't really have to think! All you have to do is calculate $(x \circ y) \circ z$ and $x \circ (y \circ z)$ and see if they are the same!

So...do this...

3. Thanks. I assume you mean the following.

x o (y o z) = x/7 o (yz/7)
= xyz/49

(x o y) o z = 1/7(xy/7) o z
= xyz/49

4. Originally Posted by Arron
Thanks. I assume you mean the following.

x o (y o z) = x/7 o (yz/7)
= xyz/49

(x o y) o z = 1/7(xy/7) o z
= xyz/49
Not quite - $x \circ (y\circ z) = x\circ \frac{yz}{7}$ - you put in the "/7" as you perform the multiplication, and similarly with the other part.