# Multiplication in the division algorithm for groups abstract algebra

For $n \in \mathbb{N}$, $n>1$, let $\mathbb{Z}_n^\times := \{1, \dots, n-1\}$

For positive integers $a$ and $n$, show that $ax \bmod{n} = 1$ has a solution if and only if $\gcd(a,n)=1$.

I have this part of the proof solved.

Using the above show that $(\mathbb{Z}_n^\times,\cdot)$, where $a\cdot b := (ab) \bmod{n}$, is a group if $n$ is a prime.

This is the part I am having trouble solving. I am using Euclids lemma and that multiplication is associative in the DA. I think I have to show closure, but I am not sure how using the above proof.

• It's saying "use the above proof" just to help you realize that $$\{ 1, 2, \dots, p-1\} = \mathbb Z_p^\times$$ when $p$ is prime. To prove that what you have is a group, show that it satisfies the group axioms. – Andrew Tawfeek Oct 22 '17 at 23:31

## 1 Answer

Closure is actually enough in a finite group. Not sure if you have proven that or not yet. Otherwise, the key is that every element of $\{1, 2, \ldots, p-1\}$ is relatively prime to $p$. Suppose $d$ was a common divisor with $d>1$, then $d \mid p$, a prime, which is impossible. This means from what you have shown above it has an inverse under multiplication.