Proof that units mod $n$ are closed

Question

I am trying to prove that the invertible elements of $\mathbb{Z}_n$ are closed under multiplication. Taking the definition for granted (though I am not sure how to "prove" something like this; I assume it is something that is typically taken as a definition), $$\overline{a} \cdot \overline{b} = \overline{ab},$$ it appears true by definition that $\overline{ab}$ is an element of $\mathbb{Z}_n$: $a, b \in \mathbb{Z}$, so $ab \in \mathbb{Z}$, and the "vertical bar" (notation I am not particularly accustomed to) forces this to be an element of $\mathbb{Z}_n$, even if $ab \in \{0, 1, \ldots, n-1\}$, as it is surely equivalent to one even if not. It suffices to show that this element is invertible. Since $\overline{a}$ and $\overline{b}$ are invertible (call the inverses $\overline{c}$ and $\overline{d}$, respectively) I claim that $\overline{dc}$ is the inverse of $\overline{ab}$. Using only the properties of $\mathbb{Z}$, we get: $$\overline{ab} \cdot \overline{dc} = \overline{(ab)(dc)} = \overline{(ac)(bd)} = \overline{ac} \cdot \overline{bd} = \overline{1} \cdot \overline{1} = \overline{1}.$$ How does this look? Have I skipped any steps?

Answer 1

If you know some abstract algebra, this can be proved in a more general sense. Note that $\mathbb{Z}_n$ is an example of a ring with identity. Let $R$ be an arbitrary ring with identity, and let $R^\times$ be the set of units in $R$. We will show $R^\times$ is closed under multiplication. Let $a,b\in R^\times$. Then $a,b$ are units in $R$, so there exist elements $c,d\in R$ such that $ac=bd=1$. We need to show $ab\in R^\times$, that is, $ab$ is a unit in $R$. Observe $$ (ab)(dc)=a(bd)c=ac=1. $$ Therefore $ab$ has an inverse in $R$, namely $dc$. Hence $ab\in R^\times$, so the set of units of $R$ is closed under multiplication. It is useful to note that we never assumed commutativity here. Indeed, this result holds for arbitrary rings with identity.