Prove that the set of all positive integers less than $n$ and relatively prime to n form a group under multiplication modulo n

Question

I came across the problem

Prove that the set of all positive integers less than $n$ and relatively prime to n form a group under multiplication modulo n.

Proving the associativity of multiplication modulo n, closure and existence of identity are fairly easy.

But how would we prove that there is inverse for all elements i.e.$\forall$ $a \in U(n),\space \exists b \in U(n)$ such that $ab(modn) = 1$?

MY TRY:
I know that if $gcd(a,n) = 1$ $\exists x$ such that $ax(modn) = 1$. But $x$ should be in U(n) in order to complete the proof.
Also, from the theory of diophantine equations such $x$ is not unique rather if $x_{0}$ is a particular solution then, $x_{0}+nt$ for $t\in \Bbb Z$ is also a solution.
So, we can find some x such that $0\le x \le n-1$ and $ax(modn) = 1$. But, how would we prove that such $x$ is relative prime to n i.e. $gcd(x,n) = 1$?
I got stuck here. Any hint in that direction would be a great help Other ways of solving the problem are also welcomed.

How did you prove the closure property? Those ideas might be helpful for answering your question about the inverse. — Jens Schwaiger, Jul 01 '20 at 04:25
@JensSchwaiger I used fact that if a and b are co-prime to n then gcd(ab,n) = $1$ so, by euclidean algorithm gcd(ab(modn),n) = $1$. — ImBatman, Jul 01 '20 at 04:30
Yes, Euclidian algorithm is the key. It is also the key for solving your problem. If $\text{gcd}(a,n)=1$ there are integers $x,y$ such that $ax+ny=1$. But then $\text{gcd}(x,n)=1$, too. — Jens Schwaiger, Jul 01 '20 at 04:41
@Kabir FYI, a very similar question is Proving that a set Rn (relatively prime with respect to any n) is a group. — John Omielan, Jul 01 '20 at 04:50

score 2 · Accepted Answer · answered Jul 01 '20 at 04:33

Hint:

You have shown for each $a$ coprime to $n$ there is an $x$ so that $ax=1$ mod $n$.

Towards a contradiction, assume $\gcd(x,n) = d \neq 1$. Then any linear combination $rx+sn$ must be divisible by $d$... What happens if we take $r=a$? Do you see how to finish the proof from here?

I hope this helps ^_^

Prove that the set of all positive integers less than $n$ and relatively prime to n form a group under multiplication modulo n

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