# Proving uniqueness of inverse by identity (Groups)

1. Sep 13, 2013

### YABSSOR

1. Which of the following is a group?
To find the identity element, which in these problems is an ordered pair (e1, e2) of real numbers, solve the equation (a,b)*(e1, e2)=(a,b) for e1 and e2.

2. (a,b)*(c,d)=(ac-bd,ad+bc), on the set ℝxℝ with the origin deleted.

3. The question also asks for you to find the inverse and I think implicitly for associativity and then for commutativity. I've got the other three down, but the identity axiom is giving me trouble.

(a,b)*(a,b)-1=(aa-1-bb-1,ab-1+ba-1)
Unless I'm really screwing up here, I think this implies that e1-e2 is the first value, which would be equivalent to e1. However, I don't know how to resolve the ab-1+ba-1) part.

I think I've been dropped from this class for a lack of prerequisites, but I think I'm still going to try and finish the class nonetheless.

2. Sep 13, 2013

### pasmith

That certainly is a complex multiplication rule.

That won't work; it's not the case here that $(a,b)^{-1} = (a^{-1},b^{-1})$ (what inverses did you assign to $(1,0)$ and $(0, 1)$, both of which are members of this alleged group?).

It is a consequence of the definitions that you cannot assign an inverse to an element until you have found the identity.

As suggested, you need to set $(a,b)*(e_1,e_2) = (ae_1 - be_2, ae_2 + be_1) = (a,b)$ and solve for $e_1$ and $e_2$.

Then, to find $(a,b)^{-1} = (c,d)$, set $(a,b)*(c,d) = (e_1,e_2)$ and solve for $c$ and $d$.