- #1

mathmari

Gold Member

MHB

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Let $n\in \mathbb{N}$ and $M=\{1, 2, \ldots , n\}\subset \mathbb{N}$. Let $d:M\times M\rightarrow \mathbb{R}$ a map with the property $$\forall x, y\in M : d(x,y)=0\iff x=y$$

Let \begin{equation*}G=\{f: M\rightarrow M \mid \forall x,y\in M : d(x,y)=d\left (f(x), f(y)\right )\}\end{equation*} and $S_n$ the symmetric group, i.e. the set of all bijective maps of $M$ to itself with the composition.

Each $f\in G$ ist injective.

I want to show that $G\subseteq S_n$. We have to show that $f\in S_n$, i.e. that $f$ is surjective, or not? Could you give me a hint how we could show that? (Wondering)