# Group Axioms G4 Associative

• Mar 23rd 2011, 12:35 AM
Arron
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.
• Mar 23rd 2011, 01:30 AM
Swlabr
Quote:

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 $\displaystyle (x \circ y) \circ z$ and $\displaystyle x \circ (y \circ z)$ and see if they are the same!

So...do this...
• Mar 23rd 2011, 06:03 AM
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
• Mar 23rd 2011, 06:10 AM
Swlabr
Quote:

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 - $\displaystyle x \circ (y\circ z) = x\circ \frac{yz}{7}$ - you put in the "/7" as you perform the multiplication, and similarly with the other part.