Reference for this theorem: $a, b$ coprime, $f(k) := ka \bmod b$, then $f$ is bijection on $\lbrace 0, …, b−1 \rbrace$.

Question

I need to use the following theorem in a paper but have to expect that some of the audience (physicists) is not familiar with it, so I would like to reference it:

Let $a$ and $b$ be two coprime integers. Let $f(k) := ka \bmod b$. Then $$k≠j ⇒ f(k) ≠ f(j) \quad ∀ k,j ∈ \left\lbrace 0, …, b-1 \right\rbrace$$ or with other words: $f$ is a bijection on $\left\lbrace 0, …, b-1 \right\rbrace$.

What is the name of this theorem and if it does not have one, what is a sligthly more general theorem to reference instead? If even this does not exist, I am interested in a citable reference for this.

I have gone through Wikipedia’s lists of theorems and lemmas and checked everything named by less than two persons and did not find anything. I also checked a few books on number theory and did not find this statement.

Note that I am not looking for a proof (I can do that myself).

Answer 1

Proposition 2.1.13 from Elementary Number Theory: Primes, Congruences, and Secrets by William Stein (freely available on his site) is:

If $\text{gcd}(a,n) = 1$, then the equation $ax ≡ b ~(\text{mod } n)$ has a solution, and that solution is unique modulo $n$.

From there, it is only a small step to the required statement.

Answer 2

I don't think it has a name, but it follows from this fact:

If $X$ is finite then $f: X \to X$ is a bijection iff it is an injection iff it is a surjection.

You could also phrase the specific instance of that fact as follows:

Since $a$ is a unit mod $b$, the map $k \mapsto ka$ is a bijection.

If you want to get down to the details, you could say that $f$ is injective because

If $\gcd(a,b)=1$ and $b$ divides $ac$, then $b$ divides $c$.

This is an extension of Euclid's lemma and, like it, follows from Bézout's identity.