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I noticed an interesting result when playing around with mods. If I take all the residue classes of $n$ and multiply each of them by an integer $q$ such that $n$ and $q$ are coprime, then each residue class maps to a unique new residue class.

For example, if $n=7$ and $q=3$ then the residue classes ${0,1,2,3,4,5,6}$ get mapped to ${(0\cdot3), (1\cdot3), (2\cdot3),(3\cdot3),(4\cdot3),(5\cdot3),(6\cdot3)}$ which in turn becomes ${0,3,6,2,5,1,4}$. This is simply a permutation of $0,1,2,3,4,5,6$ which are the residue classes we started with.

In other words, is there a way to prove that when you multiply $\{0,1,2,\dots, (n-1)\}$ by $q$, no two products will be equivalent mod n?

Alejandro Breen
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1 Answers1

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Let $gcd(q,n)=1$. Then the mapping $\Bbb Z_n\rightarrow\Bbb Z_n:x\mapsto qx$ is bijective and so permutes the residues $0,1,\ldots,n-1$.

The reason is that in the above case, $q$ is invertible in $\Bbb Z_n$. Then $qx=qy$ implies by multiplying with $q^{-1}$, $x=y$ and so the mapping is injective.

But a mapping $X\rightarrow X$, where $X$ is finite, is already bijective if it is injective (or surjective).

Wuestenfux
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