I'm unsure how to show how the units of a Euclidean domain form a group.
I know:
Let D be a domain. We shall call the function $\|x\|$ defined on the non-zero elements x of D with values in the non-negative integers $(0,1,2,\dots)$ a Euclidean Evaluation if
For every $a,b ≠ 0$, $\|ab\| \ge \|a\|$ and $\|ab\|\ge \|b\|$
For every $a,b ≠ 0$, there exists a quotient $q$ and remainder $r$ in $D$, such that $a = qb + r$ and $\|r\| < \|b\|$ or $r= 0$.
A domain with a specified Euclidean valuation is called a Euclidean domain.
We say $b$ divides $a$ if $b = ac$ for some $c$, and denote this by $b \mid a$. In that case the remainder is $r$.
How do I show the units of a Euclidean domain form a group?
Given a commutative ring $R,$ we claim that $U = \{r \in R \,|\, rs = 1 \text{ for some } s \in R \}$ is a multiplicative group. Evidently, we have that $1 \in U,$ so $U$ is nonempty. Given any $r,u \in R,$ we have that $rs = 1$ and $uv = 1$ for some $s,v \in R.$ Observe that $(rv)(us) = (rs)(uv) = 1,$ hence we have that $ru^{-1} \in R.$ We conclude that $U \leq R.$ Of course, a Euclidean domain is a commutative ring, hence in particular, the units of a Euclidean domain $R$ form a multiplicative subgroup of $R.$