Hi
You also need show that there exists $\displaystyle (p',\,q')$ such that $\displaystyle (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q)=(e_1,\,e_2)$ for all pairs $\displaystyle (p,\,q)$. (you can, for example, solve $\displaystyle (p',\,q')*(p,\,q)=(e_1,\,e_2)$ for $\displaystyle (p',\,q')$ and then deduce the equality $\displaystyle (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q)$)And how do I show that (A,*) is a group other than having shown that it is associative and is closed under *?
($\displaystyle (e_1,\,e_2)$ : identity element)
Given $\displaystyle (p, q) \in A$ let $\displaystyle (r, s) \in A$ be defined as such:
If $\displaystyle |p| > |q|$, then take $\displaystyle r = kp, s = -kq$ such that $\displaystyle k(p^2 - q^2) = 1$.
If $\displaystyle |q| > |p|$, then take $\displaystyle r = -kp, s = kq$ such that $\displaystyle k(q^2 - p^2) = 1$.
This shows that every element $\displaystyle x \in (A, *)$ has an inverse under the identity $\displaystyle (1, 0)$.