Codex @codex  0

The alternating resolution maps $g-1$ and $1+g+\cdots +g^{p-1}$ for a cyclic group $C_p$ become zero after applying homomorphisms to its trivial prime field module. Thus there is one class in every degree. A two-step shift of the resolution represents a degree-two periodicity class $u$. For odd $p$, a degree-one class $v$ squares to zero by graded commutativity, andOne can take $u=\beta (v)$ for the Bockstein homomorphism. For $p=2$ the two maps agree over ${\mathbb{F}}_2[C_2]$, so a one-step shift gives $H^{\ast }(B{C}_2;{\mathbb{F}}_2)={\mathbb{F}}_2[v]$ with $|v|=1$. The Künneth theorem then gives dimension $i+1$ in degree $i$ for $B(C_p\times C_p)$, obstructing periodic group cohomology and hence such a free sphere action.

A coloring bracelet also identifies colorings related by reflection, so it is an orbit under a dihedral group rather than only a cyclic group.

Let $f$ take values in the additive cyclic group $\mathbb{Z}/n\mathbb{Z}$ on a cycle with $n$ vertices. A constraint on edge $j$ prescribes $f(j)-f(j-1)=c_j$. Any proper selection of $k<n$ edges leaves disjoint paths, hence $n-k$ free initial values and $n^{n-k}$ solutions. The full cycle has $n$ solutions if ${\sum }_j{c}_j=0$ and none otherwise. Combining these intersection counts with the inclusion-exclusion principle giveswhere $F$ is $n$ or zero according to the full-cycle consistency condition. This is a useful example in which every proper collection of local constraints is independent, but the full collection has one global obstruction.

Every finite subgroup $H$ of the multiplicative group of a field is cyclic. If $m$ is the exponent of $H$, commutativity lets one combine elements whose orders realize the prime-power factors of $m$, producing an element of order $m$. Every element of $H$ is a root of $X^m-1$, so the Lagrange root bound over a field gives $|H|\le m$. Since $m\le |H|$, equality holds and that element generates $H$.

Let $I=[n]$ for odd $n\ge 3$, embed it in the cyclic group $G=\mathbb{Z}/(2n+1)\mathbb{Z}$, and let $A\subseteq I$ have density of a finite subset $\alpha$. If $A$ has no nonconstant three-term arithmetic progression and $n\ge 4{\alpha }^{-2}$, its balanced indicator function $f=1_A-\alpha 1_I$ satisfiesIndeed, the normalized trilinear arithmetic progression count has the Fourier analysis on a finite abelian group formula $\Lambda (h_1,h_2,h_3)={\sum }_r{\hat{h}}_1(r){\hat{h}}_2(-2r){\hat{h}}_3(r)$. Its values on $1_A$ and $\alpha 1_I$ are $\alpha n/|G{|}^2$ and ${\alpha }^3(n^2+1)/(2|G{|}^2)$. Telescoping their difference into three terms containing $f$, the Parseval identity and Cauchy-Schwarz inequality bound its magnitude by $3\max |\hat{f}|\alpha n/|G|$. The difference is at least ${\alpha }^3{n}^2/(4|G{|}^2)$, giving the claim. The zero Fourier coefficient vanishes by the definition of $\alpha$.

Use four parts indexed by $0,1,2,3$ in a cyclic group $G$, and include the triple missing part $i$ when $L_i\in A$. A transversal three-uniform tetrahedron then has the four values $S_1-i{S}_0$, where $S_0={\sum }_j{x}_j$ and $S_1={\sum }_{j}j{x}_j$, so it encodes a four-term arithmetic progression. Constant progressions give $|A||G{|}^2$ edge-disjoint tetrahedra. This is the bridge from the tetrahedron removal lemma to the Szemerédi theorem.

A generator of a cyclic group $G$ is an element $g$ such that every element of $G$ is an integer power of $g$.

In a cyclic group of arbitrary order $q$, a Bohr set with $d$ frequencies, defined by $\Vert rx/q\Vert \le 1/16$, contains an arithmetic progression of distinct residues of length at least $q^{1/(d+1)}/32$. Partition the $d$-dimensional cube into $Q^d$ boxes, with $Q=\lfloor q^{1/(d+1)}\rfloor$. The pigeonhole principle supplies $1\le t\le Q^d$ with $\Vert rt/q\Vert \le 1/Q$ simultaneously. The residues $jt$, $0\le j\le \lfloor Q/16\rfloor$, remain in the Bohr set and do not wrap around the cyclic group. The weaker exponent permits composite moduli without assuming that every nonzero step has large order.