# Prove that multiplication by an integer $a$ that is relatively prime to $n$ defines a bijection from $\mathbb{Z}_n-\{0\}$ to itself

If gcd$(a,n)=1$, then multiplication by $a$ defines a bijection from $\mathbb{Z}_n-\{0\}$ to itself.

My working:

If $n=p$ a prime, then we can use the Fermat's Little Theorem. If $n$ is not prime in general, I tried the following:

$x\equiv0,1,2,\ldots,n-1$ (mod $n$), then $ax\equiv0,a,2a,\ldots,(n-1)a$ (mod $n$).

Since gcd$(a,n)=1$, we can cancel $a$ on both sides to get back $x\equiv0,1,2,\ldots,n-1$ (mod $n$).

My doubts:

I think my working is wrong. I know what we need to do is to show that all the elements after multiplied by $a$ have different remainders modulo $n$. But I am not sure what theorems to use, I only know some basic elementary number theory theorems. Can we solve this problem by using any simple methods?

Thanks for the help!

## 1 Answer

Sketch:

• Step $1$: Prove that the map $x\mapsto ax$ is injective. Suppose that $ax\equiv ay\pmod n$, then $ax=ay+kn$, so $a(x-y)=kn$. Therefore, $n\mid a(x-y)$. Since $a$ is relatively prime to $n$, $n\mid x-y$ and $x\equiv y\pmod n$.

• Step $2$: Prove that the map $x\mapsto ax$ is surjective. Let $z\in(\mathbb{Z}/n)^\times$ Since $a$ and $n$ are relatively prime, there exist $x$ and $y$ so that $ax+ny=1$. Consider $xz$: under this map, $zx$ maps to $axz=(1-ny)z=z-nyz\equiv z\pmod n$. Since $z$ is nonzero, $xz$ must also have been nonzero.

One could also use that $(\mathbb{Z}/n)^\times$ is a finite set and injectivity or many other techniques. Also, $(\mathbb{Z}/n)^\times$ is just another way to write the set that you describe.