Proving Inverses Exist Within Set of Units of $\Bbb Z/n\Bbb Z$ I'm doing some practice problems for abstract algebra and I came across this problem:
Let $\mathbb{Z}_n^*$ be the set of all elements with a multiplicative inverse from $\mathbb{Z}_n$, and $n > 1$.
For instance, $\mathbb{Z}_6^* = \{1, 5\}$. Prove that $\mathbb{Z}_n^*$ is a group.
I understand how to show the associativity and identity property, but I'm stuck on the inverse part. How do we know that any $a$ in $\mathbb{Z}_n^*$ has an inverse $a'$ that exists within $\mathbb{Z}_n^*$?
Intuitively it kinda makes sense to me, but I can't put it into a concrete proof. Thanks for the help!
 A: Let $M$ be a monoid, that is, a set endowed with an operation that is associative and has a neutral element $1$; I'll denote the operation by simple juxtaposition, you can use the symbol you prefer.
Since the operation is associative, the inverse of an element, if it exists, is unique; denote the inverse of $a$ (if it exists) by $a^{-1}$.
Denote by $U(M)$ the set of elements of $M$ that have an inverse. Clearly $1\in U(M)$, because $1^{-1}=1$.

Theorem. $U(M)$ is a submonoid of $M$, that is, if $a,b\in U(M)$, then $ab\in U(M)$.

The proof is just checking that $(ab)^{-1}$ exists, because $b^{-1}a^{-1}$ is the required element.

Corollary. $U(M)$, with the operation induced by $M$, is a group.

In your case $M$ is $\mathbb{Z}/n\mathbb{Z}$ with multiplication.
A: This is much more general: 

In any commutative unital ring $R$, the set of units $R^\times$ is a
  (multiplicative) group.

$a\in R^\times$ means there exists an element $a'\in R$ (the inverse of $a$) such that $aa'=a'a=1$. But this can be read as $a'a=aa'=1$, which means $a'$ is a unit, with inverse $a$. Hence $a'\in R^\times$.
