Thread: Group Theory - newcomer needs help

1. Group Theory - newcomer needs help

Given that A = {}

Define * on A by (p,q) * (r,s) = (pr+qs, ps+qr)

Does anyone know what an identity element for * on A might be? And how do I show that (A,*) is a group other than having shown that it is associative and is closed under *?

Thank you in advance,
Aalborg

2. Originally Posted by Alborg
Given that A = {}

Define * on A by (p,q) * (r,s) = (pr+qs, ps+qr)

Does anyone know what an identity element for * on A might be? And how do I show that (A,*) is a group other than having shown that it is associative and is closed under *?

Thank you in advance,
Aalborg
I think $\displaystyle (1, 0)$ is an identity element for * on A.

3. Hello,

For the identity element, I think that you have to solve for p and q in :

$\displaystyle (p,q)*(r,s)=(r,s)$

4. Hi

And how do I show that (A,*) is a group other than having shown that it is associative and is closed under *?
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 $\displaystyle (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q)$)

($\displaystyle (e_1,\,e_2)$ : identity element)

5. Originally Posted by flyingsquirrel
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 pair $\displaystyle (p,\,q)$. (you can, for example, first solve $\displaystyle (p',\,q')*(p,\,q)=(e_1,\,e_2)$ for $\displaystyle (p'\,q')$ and the equality $\displaystyle (p,\,q)*(p',\,q')=(p',\,q')*(p,\,q)$ will come after)

($\displaystyle (e_1,\,e_2)$ : 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 $\displaystyle (1, 0)$.