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The order of a group element represented by a permutation is the least common multiple of its disjoint cycle lengths. An order divisible by nine therefore requires a permutation cycle whose length is divisible by nine. In $S_9$ that must be a nine-cycle, which uses every letter and has order nine, not eighteen. There is no element of $S_9$ of order eighteen.

In a direct product of groups, $(a,b{)}^r=(a^r,b^r)$. Thus a finite order for $(a,b)$ makes both coordinate orders finite. Conversely, if their orders are $p$ and $q$, the least positive exponent annihilating both is their least common multiple. Hencewhen both orders are finite; if either coordinate has infinite order, so does $(a,b)$.

First justify the finite bound used in the hint. Fix $k$. If no finite $T$ forced a monochromatic positive solution of ${\sum }_i{a}_i{x}_i=0$, consider the rooted tree whose level $t$ consists of the solution-free finite colorings of $[t]$, with restriction as the predecessor map. Every level is nonempty and every vertex has at most $k$ children. The König infinity lemma gives an infinite branch, hence a finite coloring of all positive integers with no such solution, contradicting the partition regular matrix hypothesis. This is the compactness bound for partition regularity.

Choose such a $T$ and let $S=\mathrm{lcm}(1,\dots ,T)$, the least common multiple. For a given finite coloring $\chi$ of the positive integers, pull it back to $[T]$ byEach $S/t$ is a positive integer in $[S]$. The defining property of $T$ gives $x_1,\dots ,x_n\in [T]$ of one color under $\tilde{\chi }$, with ${\sum }_i{a}_i{x}_i=0$. Set $y_i=S/x_i$. Their colors under $\chi$ agree, andThus reciprocal partition regularity follows. This construction is an involution on the divisors of $S$; it permits repeated coordinates and never requires a reciprocal of a positive integer to itself be integral without the common scaling factor $S$.

The distinct values in one period imply that $r$ is the least positive period, $r\mid N$, and $|Y|=r$. Write $x=j+tr$, where $0\le j<r$ and $0\le t<N/r$. Use the positive-exponent quantum Fourier transform convention already used in the concept article. The amplitude of first-register value $y$ isBy the root-of-unity filter, the inner sum vanishes unless $y=mN/r$. DefineThe distinct $f(j)$ give an orthonormal basis of the second register, and the transformed state isReindexing $j+l$ modulo $r$ showsConsequently measurement in this common eigenbasis returns each $m$ with probability $1/r$; the corresponding eigenvalue of $U_1$ is a uniformly sampled $r$th root of unity. A single measured phase $-m/r$ reveals the denominator $r/\gcd (m,r)$, which need not be $r$. Repeated samples permit exact period recovery from a Fourier sample, for example by taking the least common multiple of their reduced denominators. The simultaneous eigenspaces and their probabilities depend on $r$, not on the particular labels $f(j)$.

If a rational matrix $A$ is a partition regular matrix, every finite coloring of the positive integers admits monochromatic $y_1,\dots ,y_n$ such thatChoose a compactness bound for partition regularity $T$ for the number of colors, put $S=\mathrm{lcm}(1,\dots ,T)$, and pull back the coloring by $t\mapsto S/t$ on $[T]$. A monochromatic positive solution $x$ of $Ax=0$ gives $y_i=S/x_i$; then $(1/y_i)=x_i/S$. The least common multiple guarantees that every $y_i$ is a positive integer.

The multiplicative arithmetic function $f(n)=3^{\omega (n)}$ counts assignments of three labels to each distinct prime factor. It also counts pairs of squarefree divisors whose least common multiple is the radical of an integer $n$.

A pointwise periodic self-map has a uniform positive period exactly when its finite cycle lengths are bounded. A uniform period is divisible by every length; conversely, bounded lengths divide the least common multiple of $1,\dots ,M$ for some bound $M$. Arbitrarily many bounded-length cycles are allowed. Every such map on a finite set has a uniform period. On any infinite set containing a countably infinite subset, cycles of unbounded finite lengths and fixed points elsewhere give a counterexample. Thus, with the usual axiom of choice assumption, the sets on which every pointwise periodic self-map has a uniform period are exactly finite sets.