# Math Help - Group Axiom G3

1. ## Group Axiom G3

The group

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

G1 closure hold

G2 Idenitiy - I have found the identity to be 7

I just need to prove G3 inversers and am not sure how to do this.

Can anyone help?

Many thanks.

Arron

2. We have $x\circ x^{-1}=e$ ( $e$ is the unit), so $x\cdot x^{-1}/7=7$. Thus, $x^{-1}=49/x$.

3. Originally Posted by emakarov
$x\cdot x^{-1}/7=7$. Thus, $x^{-1}=49/x$.
What?

$\displaystyle x \circ x^{-1} = e \implies \frac{xx^{-1}}{7} = 1$

doesn't it?

-Dan

4. The OP has already established that the unit is 7.

5. No. The OP said that e=7 in his original post.

6. (Ahem!) Missed that. Thanks.

-Dan