Prove that this structure is a group.

Question

I have been having trouble solving the following question:

Consider the set $G=\{z \in \mathbb{C} \mid z=a+i\sqrt5b \}$, $a,b∈\mathbb{Q}$ with standard multiplication operation $\cdot$

Prove that $(G,\cdot)$ is a group

I have shown that closure, asociativity hold true and that there exist the neutral element $e=1+0i\sqrt5 = 1$ for each element of $G$.

I have found an inverse $z^{-1}=\frac{1}{a+i\sqrt5 b}$ , $a,b \in \mathbb{Q}$ , but this element does not seem to be in G.

Does there exist another inverse that is in G or is the inverse $z^{-1}$ in fact in $G$?

Answer 1

Just use the fact that$$\frac1{a+i\sqrt5\,b}=\frac{a-i\sqrt5\,b}{\left(a+i\sqrt5\,b\right)\left(a-i\sqrt5\,b\right)}=\frac{a-i\sqrt5\,b}{a^2+5b^2}.$$

Answer 2

Hint: In view of the inverse, take $a+i\sqrt 5b\in G$ and determine $c+i\sqrt 5 d$ such that

$(a+i\sqrt 5 b)(c+i\sqrt 5 d) = 1 = 1+0i = 1 + 0i\sqrt 5$,

i.e., by multiplying out: $ac - 5bd=1$ and $i\sqrt 5(ad+bc)=0.$

Two linear equations with two unknowns, $c$ and $d$.