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For each $n \in \mathbb{N}$, let $F_n$ be a finite set with $n$ elements. For any function $f : F_n \to F_n$ and $k \in \mathbb{N}$, $f^k$ is the result of composing $f$ with itself $k$ times.

Say that $n$ distinguishes powers $i$ and $j$ iff there is some function $f : F_n \to F_n$ such that $f^i \neq f^j$.

For example, 2 distinguishes each odd power from each even power (because of the order-2 rotation), but does not distinguish any odd powers. But every number greater than 2 distinguishes 1 and 3.

Conjecture. For all numbers $0 < m < n$, there exist powers $i$ and $j$ that $m$ does not distinguish but $n$ does distinguish.

Is this conjecture true? How can I prove it?

Jeff Russell
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1 Answers1

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Let us first investigate what powers a function can distinguish. Fix $f:F_m\to F_m$, and let $R_i$ denote the image of $f^i$. Note that the sets $R_i$ form a decreasing sequence of subsets of $F_m$ which eventually stabilizes. Write $R=\bigcap_i R_i$, the eventual value on which the sequence stabilizes. Note that $R=R_{m-1}$, since the cardinality of $R_i$ can only decrease up to $m-1$ times before stabilizing. Note that $f$ maps $R$ to itself surjectively, and hence bijectively since $R$ is finite. Let $d$ be the order of $f$ as a permutation of the set $R$. We then have that for $i,j\geq m-1$, $f^i=f^j$ iff $i-j$ is divisible by $d$.

Thus for $i,j\geq m-1$, if $m$ distinguishes $i$ and $j$ then $i-j$ is not divisible by the order of some permutation of a subset $R\subseteq F_m$.

Now fix $0<m<n$. Note that $m$ does not distinguish $n-2$ and $n-2+m!$, since $n-2\geq m-1$ and $m!$ is divisible by the order of any permutation of any subset of $F_m$. On the other hand, $n$ does distinguish $n-2$ and $n-2+m!$: let $f:\{1,\dots,n\}\to\{1,\dots,n\}$ be defined by $f(x)=\min(x+1,n)$. Then $f^{n-2}(1)=n-1$ but $f^{n-2+m!}(1)=n$, so $f^{n-2}\neq f^{n-2+m!}$. This proves your conjecture.

Eric Wofsey
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