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The Roth theorem on three-term arithmetic progressions says that, for each fixed $\delta >0$, a sufficiently long interval cannot have a subset of density of a finite subset at least $\delta$ with no nonconstant three-term arithmetic progression. Equivalently, the maximum size of a three-term-progression-free subset of $[N]=\{1,\dots ,N\}$ is $o(N)$. Here is a density increment proof, including its analytic ingredients.

Write $e(t)=\exp (2\pi it)$ and embed $I=[N]$ into the cyclic group $G=\mathbb{Z}/(2N+1)\mathbb{Z}$ of order $M$. For normalized Fourier analysis on a finite abelian group take $\hat{h}(r)=M^{-1}{\sum }_{x}h(x)e(-rx/M)$. The orthogonality of complex exponentials follows from a finite geometric series: its average is zero for a nonzero frequency and one for frequency zero. Expanding in these linear phases consequently proves both Fourier inversion on a finite group and the Parseval identity on a finite group. In particular,The second identity follows by summing the frequency expansion over $x,d$: the two constraints leave precisely the frequencies $(r,-2r,r)$. Since $M$ is odd, multiplication by two permutes the frequencies. Also a congruence $x+z=2y$ with $x,y,z\in I$ is an equality in the integers, since $|x+z-2y|<M$.

Let $A\subseteq I$ have subset density $\alpha =|A|/N\ge \delta$, and suppose it contains no nonconstant three-term arithmetic progression. Its balanced indicator function of a finite subset is $f=1_A-\alpha 1_I$, extended by zero outside $I$. Then $\hat{f}(0)=0$. Counting endpoints with the same parity givesIf $N\ge 4{\alpha }^{-2}$, their difference in absolute value is at least ${\alpha }^3{N}^2/(4M^2)$. Telescope the difference of the products into three terms, each containing $f$ and two factors chosen from $1_A,\alpha 1_I$. Put $s={\max }_r|\hat{f}(r)|$. The Cauchy-Schwarz inequality and the Parseval identity on a finite group bound each term by $s\alpha N/M$, since the squared normalized $L^2$ norm of either other factor is at most $\alpha N/M$. Therefore some nonzero frequency $r$ satisfiesThis is the Fourier detection of a progression-free subset of an interval step; it has just been proved rather than assumed.

We now turn this Fourier coefficient on a finite abelian group into a density increment. Set $Q=\lfloor \sqrt{N}\rfloor$. Partition the unit interval into $Q$ equal pieces and place the $Q+1$ fractional parts of $0,r/M,\dots ,Qr/M$ in them. The pigeonhole principle gives $1\le q\le Q$ with $\Vert qr/M\Vert \le 1/Q$. This proves the particular Dirichlet approximation theorem needed here. Put $\kappa ={\alpha }^2/(1000\pi )$ and $L=\lfloor \kappa Q\rfloor$. For sufficiently large $N$ depending only on $\delta$, $L\ge \kappa Q/2$ and $2L\le Q$. Each residue chain modulo $q$ in $I$ has at least $Q$ elements. Split each such chain into arithmetic progressions with between $L$ and $2L-1$ elements, absorbing its final short remainder into the previous block.

The linear phase changes by at most $4\pi L/Q\le {\alpha }^2/250$ within any block. Freezing the linear phase at the first element of each block therefore changes the preceding correlation by at most ${\alpha }^{2}N/250$, because $|f|\le 1$. The triangle inequality givesThe block sums of the balanced indicator function of a finite subset add to zero, so their positive total is half their absolute total. Some block consequently satisfiesHere $c>0$ is absolute, and shrinking it absorbs the floors. An affine parametrization of $P$ identifies $A\cap P$ with a subset of $[|P|]$ and preserves arithmetic progressions.

Repeat this Roth density-increment step. Every step raises the subset density by at least ${\delta }^2/144$, so fewer than $T=\lceil 144/{\delta }^2\rceil +1$ steps are possible. All the lengths satisfy $N_{j+1}\ge c{\delta }^2\sqrt{N_j}$ while above a threshold depending only on $\delta$. Choose the initial $N$ so large that this lower recursion stays above that threshold for $T$ steps; this is possible by working backwards through finitely many squarings. We would then force a subset density greater than one. Thus every fixed positive density eventually forces a nonconstant three-term arithmetic progression. For an infinite subset of the positive integers with positive upper asymptotic density, apply the finite conclusion along initial intervals whose subset density is bounded away from zero; this gives the usual infinite formulation as well.

A clean proof outline uses hypergraph removal rather than the quadratic density increment route. The structure is: establish removal for the three-uniform tetrahedron, encode four-term arithmetic progressions as its copies, and use the many edge-disjoint copies coming from constant progressions to contradict removal. We give the technical regularity stage in outline, as requested, and prove the counting inequality and the encoding that make the argument work.

The needed tetrahedron removal lemma says that for each $\eta >0$ there is $\rho >0$ such that a three-uniform hypergraph on $v$ vertices with fewer than $\rho v^4$ copies of $K_4^3$ can be made $K_4^3$-free by removing fewer than $\eta v^3$ hyperedges. Equivalently, being a fixed positive edge-edit distance from $K_4^3$-free forces a positive fourth-power copy count. Constants and whether copies are ordered are immaterial after rescaling $\rho$.

Here is the global proof of this hypergraph removal input. Strong regularity for three-uniform hypergraphs partitions both vertices and the bipartite pair sets between vertex classes. The resulting three-class cells, called triads in a hypergraph regularity partition, are supported on three pair cells. On all but a prescribed small total weight, the triple-edge function has approximately constant relative density and a small relative three-dimensional box norm. This two-level refinement is essential: a vertex partition alone does not capture correlations carried by pairs. The regularity construction uses the squared L2 norms of conditional expectations of edge indicators as bounded energies. Refinement increases this energy by the squared L2 norm of the difference between the new and old conditional expectations, by orthogonality. Thus a nonuniformity witness causing a definite discrepancy causes a definite energy increment. A witness to nonuniformity refines the relevant pair or triple partition and raises the appropriate energy. A hierarchy of tolerances, with sufficiently strong refinement at the pair level, makes these increases terminate after a bounded number of stages. Bounds may depend very badly on the requested tolerance, which is harmless for a qualitative theorem.

Delete hyperedges with two vertices in the same vertex class, hyperedges meeting exceptional classes or irregular cells, hyperedges supported on pair cells of very small density, and hyperedges in triads of very small triple density. Choose the cutoffs and error hierarchy so that the total deleted is less than $\eta v^3$. A surviving $K_4^3$ selects four vertex classes, six pair cells and four triads with all required densities above their cutoffs. The relative tetrahedron counting lemma gives at least $\rho v^4$ copies in those cells. This contradicts the assumed sparse copy count, and proves the tetrahedron removal lemma. The strong regularity and relative counting estimates are the technical portion outlined here; the dependence of their constants is not needed.

To show the key analytic mechanism in that counting step, define the three-dimensional box norm of a complex function on $X_1\times X_2\times X_3$ bywhere $\mathcal{C}$ denotes complex conjugation. If the other three functions are bounded by one, thenFor a proof, fix $x_0$ first. Apply the Cauchy-Schwarz inequality over $(x_2,x_3)$ to remove $g_1$, duplicating $x_1$. Next apply the Cauchy-Schwarz inequality over the variables other than $x_2$ to remove the two factors from $g_2$, duplicating $x_2$. A third application removes the four factors from $g_3$ and duplicates $x_3$. The resulting eighth power is exactly the cube average displayed above. Averaging over $x_0$ proves the inequality. This also shows nonnegativity of the cube average by its successive squared-sum form.

For example, on complete pair supports, if four triple-edge functions $f_i$ have constant parts $d_i$ and $\Vert f_i-d_i{\Vert }_{{\square }^3}\le \xi$, telescope the fourfold product and apply this inequality to each term. Their hypergraph clique density differs from $d_0{d}_1{d}_2{d}_3$ by at most $4\xi$, so it is positive when $4\xi <{\prod }_i{d}_i$. On general pair supports the same successive squaring argument is combined with sufficiently regular pair cells; choosing relative errors small compared with the pair-density product gives the relative tetrahedron counting lemma. This explains both the counting step and why its support must be regularized before the triple-edge densities. It supplies a proved important step without pretending that the strong regularity theorem is a one-line assertion.

We now prove the four-term progression hypergraph encoding completely. Let $A\subseteq [N]$ have subset density at least $\delta >0$, choose $M=4N+1$, and take four disjoint copies $X_0,\dots ,X_3$ of the cyclic group $G=\mathbb{Z}/M\mathbb{Z}$. Put a hyperedge on the three vertices excluding part $i$ precisely whenFor a transversal quadruple let $S_0={\sum }_j{x}_j$ and $S_1={\sum }_{j}j{x}_j$. Its four edge conditions are $S_1-i{S}_0\in A$, $i=0,1,2,3$. Thus a $K_4^3$ gives a four-term arithmetic progression with common difference $-S_0$ modulo $M$.

Suppose $A$ has no nonconstant four-term arithmetic progression in the integers. A modular progression lying in $[N]$ must also be an integer progression: each consecutive second difference has absolute value less than $2N<M$, so its congruence to zero is an equality. Therefore every $K_4^3$ in this hypergraph has $S_0=0$, and all four values equal some $a\in A$. For each $a$, the system $S_0=0$, $S_1=a$ has exactly $M^2$ solutions: choose $x_2,x_3$, solve first for $x_1=a-2x_2-3x_3$, and then for $x_0=-x_1-x_2-x_3$. There are exactly $|A|M^2=O(M^3)$ copies of $K_4^3$.

These copies are edge-disjoint. Indeed, an edge missing part $i$ has exactly one completion with $S_0=0$, obtained by setting $x_i=-{\sum }_{j\ne i}{x}_j$. Its common value is the already prescribed $L_i\in A$. Hence destroying all copies requires at least $|A|M^2$ edge deletions. Since $M\le 5N$, this is at least $\delta M^3/5$. Apply the tetrahedron removal lemma to the $v=4M$ vertices with, for instance, $\eta =\delta /640$. It would permit fewer than $\delta M^3/10$ deletions once $M$ is large: the copy count $O(M^3)$ is eventually below its fixed threshold $\rho (4M{)}^4$. This is impossible.

Every fixed positive density therefore forces a nonconstant four-term arithmetic progression in sufficiently long intervals. This proves the length-four case of the Szemerédi theorem. Applying it to sufficiently large initial intervals along a subsequence witnessing positive upper asymptotic density also gives the usual infinite-set formulation.

Use normalized Fourier analysis on a finite abelian group, identifying $A$ with its indicator function $a=1_A$ on the cyclic group ${\mathbb{Z}}_n$. The conventions areThe last operation is normalized convolution on a finite group. With these conventions the Fourier coefficients on a finite abelian group use a normalized average, while sums over frequencies use counting measure.

Expanding the normalized convolution on a finite group and putting $z=x-y$ givesFor the translation of a function $g(x)=f(x-u)$, putting $z=x-u$ yieldsThe two transforms are thereforeThe negative sign in the translation factor follows from the negative sign in our Fourier coefficient on a finite abelian group convention.

Conversely, assume $H$ is a normal subgroup and the quotient group $G/H$ is an abelian group. Commuting the two cosets $aH$ and $bH$ givesso every group commutator $a^{-1}{b}^{-1}ab$ belongs to $H$. This proves the converse and completes the equivalence.

For the dihedral group of order ten, choose a rotation $r$ and a reflection $s$ with $r^5=s^2=1$ and $srs=r^{-1}$. Its rotation subgroup $⟨r⟩$ is a normal subgroup, and the quotient by it has order two, hence is an abelian group. Consequently the commutator subgroup is contained in $⟨r⟩$. On the other hand,Since $r^{-2}$ generates the order-five cyclic group $⟨r⟩$, the commutator subgroup is exactly $⟨r⟩$. Every abelian quotient must kill this subgroup. There are only two subgroups containing it, since the remaining quotient has prime order. The complete list, including the trivial quotient, isThis is the order-ten instance of the abelianization of an odd dihedral group.

The Möbius group consists of all Möbius transformations of the Riemann sphere ${\mathbb{C}}_{\infty }=\mathbb{C}\cup \{\infty \}$, with operation function composition. A nonzero-determinant matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ represents the map $z\mapsto (az+b)/(cz+d)$. If $c\ne 0$, its pole $-d/c$ maps to $\infty$ and $\infty$ maps to $a/c$; if $c=0$, the map is affine and fixes $\infty$. Nonzero scalar multiples of $M$ represent the same transformation.

Define $\varphi (M)$ by this action for $M\in S{L}_2(\mathbb{C})$. Substitution of fractional-linear expressions shows $\varphi (MN)=\varphi (M)\circ \varphi (N)$, so $\varphi$ is a group homomorphism. Any invertible matrix can be rescaled into the special linear group: choose $\lambda \in {\mathbb{C}}^{\times }$ with ${\lambda }^2\det M=1$, which is possible because every nonzero complex number has a square root. This does not change the transformation, proving surjectivity. A transformation represented by $M$ is the identity only if fixing $0$, $\infty$ and $1$ forces $b=c=0$ and $a=d$. The determinant-one condition then gives $a^2=1$. ThusFor the fixed-point claim, a nonidentity Möbius transformation has at least one fixed point of a Möbius transformation: when $c\ne 0$ the finite fixed points solve a quadratic, and when $c=0$ infinity is fixed. A quadratic shows there can be no more than two distinct fixed points unless the transformation is the identity. Suppose there is exactly one. Conjugate it to infinity by a Möbius transformation. The resulting map has form $z\mapsto az+b$, since it fixes infinity. If $a\ne 1$ it also has the finite fixed point $b/(1-a)$, a contradiction. Hence it is $z\mapsto z+b$ with $b\ne 0$. Its $m$th power is $z\mapsto z+mb$, never the identity for $m>0$. Conjugation preserves order, so a nonidentity finite-order Möbius transformation has exactly two fixed points.

Now let $K$ be a finite abelian subgroup of the Möbius group. First suppose it contains an element $g$ of order greater than two. Conjugate the two fixed points of $g$ to $0$ and $\infty$, so $g(z)=\zeta z$, with $\zeta$ a root of unity of order greater than two. Every element $h$ commuting with $g$ permutes its fixed-point set, because $g(h(p))=h(g(p))=h(p)$. A Möbius transformation preserving each of $0,\infty$ is $h(z)=az$; one exchanging them is $h(z)=a/z$. In the second case $hg(z)=a/(\zeta z)$ and $gh(z)=\zeta a/z$, so commuting would force ${\zeta }^2=1$, impossible. Thus all elements of $K$ are scalings. A finite subgroup of a field multiplicative group is cyclic; here one can see this directly by choosing the least common multiple $m$ of all element orders and observing that the scalars lie among the $m$th roots of unity, a cyclic group. Hence $K$ is cyclic.

Otherwise every nonidentity element of $K$ has order two. If $K$ is nontrivial, choose $g$ and conjugate it to $z\mapsto -z$. The action of $K$ on $\{0,\infty \}$ has image of order at most two. Its kernel consists of scalings $z\mapsto az$, and the order-two condition forces $a^2=1$, so the kernel has at most two elements. Hence $|K|\le 4$. Orders one and two give cyclic groups; order four with all nonidentity elements involutions gives $C_2\times C_2$, generated by any two distinct nonidentity elements. Both possibilities occur: scalings by roots of unity give every finite cyclic group, and $\{z,-z,1/z,-1/z\}$ gives the Klein four group. The classification is

For every real $\alpha$, direct substitution gives the axially equivariant monomial rational map identityThe first is a spatial rotation through $\alpha$ about the third axis compensated by a target rotation through $n\alpha$, hence by an isospin rotation of the Skyrmion. The second pairs the spatial half-turn about the first axis with a target half-turn about that axis. For $n>1$, these generate the continuous axial rotation group together with perpendicular half-turns, often written $D_{\infty }$; it is the group of rotations preserving an axis in the three-dimensional special orthogonal group. Under rotations about the axis, the pure spatial stabilizer of the map itself includes the cyclic group $C_n$, whereas the larger symmetry just described is combined spatial-target equivariance. Also $R(\overline{z})=\overline{R(z)}$ gives a paired reflection symmetry of its angular density.

The angular Jacobian of a rational map isIt depends only on latitude. Put $u=\log |z|$; then $J_R=n^2{\cosh }^{2}u/{\cosh }^2(nu)$. Since $\cosh (nu)\ge \cosh u$ for $n\ge 1$, the maximum is $n^2$ at $u=0$, the equator. For $n>1$ it vanishes at the two poles, producing an axially symmetric ring of angular Skyrme baryon density. For $n=1$, $R$ is the identity: every spatial rotation is compensated by the same target rotation, and the rational map approximation for Skyrmions becomes the fully spherically symmetric Skyrmion hedgehog ansatz. The charge and uses areThe degree-one construction captures the symmetry of the unit Skyrmion exactly; its radial profile still has to be solved. The degree-two map supplies the appropriate toroidal two-Skyrmion symmetry and useful initial data. Higher monomials give axial charge-$n$ competitors, but axial symmetry need not minimize the energy: the usual lowest-charge examples already include the tetrahedral three-Skyrmion and the cubic four-Skyrmion. Topological charge determines neither shape nor energy optimality by itself.

Use the class equation for the action of $G$ on itself by conjugation:where the sum runs over noncentral conjugacy classes. By Lagrange's theorem, each noncentral class has size a positive power of $p$ greater than one. Thus $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. Since the identity element belongs to the center of a group,For $|G|=p^3$, the possible orders of the center of a group are $p,p^2,p^3$. Order $p^3$ would make $G$ an abelian group. Order $p^2$ would make the quotient group $G/Z(G)$ have prime order and hence be a cyclic group. But a cyclic quotient by the center forces $G$ to be Abelian: if $gZ(G)$ generates the quotient, any two elements are $g^{a}z$ and $g^{b}w$ with $z,w\in Z(G)$, and their products commute. Both larger orders contradict the hypothesis. Consequently

Use the usual convention that a three-term arithmetic progression must be nonconstant. There is a genuine small-case defect in the printed statement: for $n=1$, $\delta =1$, and $A=\{1\}$, neither conclusion can hold with $c_2>0$. We prove the intended Roth density-increment step for every odd integer $n\ge 3$, including the small values of $n$, rather than silently treating $n=1$ as covered. Counting constant arithmetic progressions would make the first alternative vacuous for every nonempty $A$.

Suppose that $A$ has no nonconstant three-term arithmetic progression, and let $\alpha =|A|/n$ be its actual density of a finite subset. Thus $0<\delta \le \alpha \le 1$. Put $I=[n]$, $N=2n+1$, and embed $I$ in the cyclic group $G=\mathbb{Z}/N\mathbb{Z}$. A congruence $x-2y+z=0(\mathrm{mod}N)$ with $x,y,z\in I$ is an equality over the integers, since $|x-2y+z|\le 2(n-1)<N$. Thus this embedding creates no extra three-term arithmetic progressions. Since $N$ is odd, multiplication by two permutes $G$.

Use normalized Fourier analysis on a finite abelian group, with $e_N(t)=\exp (2\pi it/N)$ and $\hat{h}(r)={\mathbb{E}}_{x\in G}h(x)e_N(-rx)$. For the normalized trilinear arithmetic progression count, character orthogonality givesIndeed, expanding each function in its inverse Fourier transform, averaging in $d$ forces $r_2+2r_3=0$, and averaging in $x$ then forces $r_1=r_3$. The Parseval identity in this normalization is ${\sum }_r|\hat{h}(r){|}^2={\mathbb{E}}_x|h(x){|}^2$.

Let $a=1_A$, $b=\alpha 1_I$, and $f=a-b$, the balanced indicator function of a finite subset. Then $\hat{f}(0)=0$. Only constant arithmetic progressions contribute to $\Lambda (a,a,a)$, whereas summing $2\min (y-1,n-y)+1$ over centers $y\in I$ counts $(n^2+1)/2$ ordered three-term arithmetic progressions in $I$. HenceSet $M={\max }_r|\hat{f}(r)|$. Telescope by changing one factor at a time:In the Fourier transform formula for each term, bound the Fourier coefficient of $f$ by $M$ and apply the Cauchy-Schwarz inequality to the other two factors. The permutation $r\mapsto -2r$ preserves the required squared sums. Since $\Vert a{\Vert }_2^2=\alpha n/N$ and $\Vert b{\Vert }_2^2={\alpha }^{2}n/N$, this givesIf $n\ge 4{\delta }^{-2}$, then ${\alpha }^{2}n\ge 4$, so the difference in the other direction is at least ${\alpha }^3{n}^2/(4N^2)$. Consequently Fourier detection of a progression-free subset of an interval gives a nonzero frequency $r$ with

To turn this into a density increment, take $Q=\lfloor \sqrt{n}\rfloor$. The Dirichlet approximation theorem, proved here by placing $0,r/N,\dots ,Qr/N$ into $Q$ equal intervals modulo one, gives $1\le q\le Q$ such that $\Vert qr/N{\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. DefineFor $n\ge N_0(\delta )$, the inequalities $Q\ge \sqrt{n}/2$ and ${\delta }^{2}Q/(288\pi )\ge 2$ show thatUse a progression partition with nearly constant linear phase: every residue chain of $I$ modulo $q$ has at least $\lfloor n/q\rfloor \ge Q$ points. Split each chain into blocks of $L$ consecutive chain points, merging any remainder into its last full block. This partitions $I$ into arithmetic progressions $P$ with $L\le |P|<2L$ and common difference $q$.

On a block $P$ with first point $z_P$, the linear phase $\chi (x)=e_N(-rx)$ satisfiesLet $S_P={\sum }_{x\in P}f(x)=|A\cap P|-\alpha |P|$. Freezing the linear phase on each block introduces error at most $({\delta }^2/72)N^{-1}{\sum }_{x\in I}|f(x)|\le {\delta }^2/72$. ThereforeThe block sums total zero. Their positive part consequently totals at least $N{\delta }^2/144$. Since all block lengths total $n$, some block has $S_P/|P|\ge N{\delta }^2/(144n)\ge {\delta }^2/72$. Its density of a finite subset is at least $\alpha +{\delta }^2/72\ge \delta (1+\delta /72)$.

For the remaining $3\le n<N_0(\delta )$, divide $I$ into disjoint consecutive triples and a remainder of at most two points. A three-term-arithmetic progression-free $A$ uses at most two points of each triple, so $\alpha \le 4/5$; the largest ratio occurs at $n=5$. Any singleton set from $A$ then has density of a finite subset one, which is at least $\delta +{\delta }^2/72$. We may therefore use the explicit constantsIn the large case the constructed arithmetic progression has length at least $c_1\sqrt{n}$; in the small case its length one does too. Thus these constants prove the intended result for all odd integers $n\ge 3$.

Use the positive-exponent quantum Fourier transform convention $F_N|x⟩=N^{-1/2}{\sum }_t{e}^{2\pi ixt/N}|t⟩$. The period of a function here is the least positive translation leaving the function unchanged on the cyclic group ${\mathbb{Z}}_N$. The injective promise means $f(x)=f(x^{\prime })$ exactly when $x\equiv x^{\prime }(\mathrm{mod}r)$. In particular $r\mid N$: the subgroup of period translations is generated by $r$, and wrapping the argument by $N$ preserves its value. This is exact quantum period finding on a finite cyclic group, rather than the truncated, nondividing-period version requiring continued fractions.

Prepare a uniform superposition state in the first quantum register using $F_N|0⟩$, and perform a reversible computation of the function into a second quantum register, cleaning its workspace by uncomputation:A quantum measurement in the computational basis of the second quantum register leaves a periodic coset state in the first:For the quantum Fourier transform of this periodic coset state, the finite geometric series vanishes unless the measured label is $t=mN/r$. On those labels its normalized amplitude is $r^{-1/2}{e}^{2\pi iam/r}$. Thus $m$ has the uniform distribution on a finite set on $\{0,\dots ,r-1\}$, irrespective of $a$.

Compute the greatest common divisor by the Euclidean algorithm, and set $q=N/\gcd (t,N)$, with $\gcd (0,N)=N$. Since $t=mN/r$,The exact period recovery from a Fourier sample is therefore certified by testing $f(q\mathrm{mod}N)=f(0)$. Every candidate $q$ divides $r$ and lies between $1$ and $r$, so the injective promise makes that equality equivalent to $q=r$. This includes $r=N$, where the test uses input $0$, and $r=1$, which always succeeds. The test does not assert that an arbitrary equality of function values certifies an arbitrary period; the Fourier-derived divisor restriction is essential.

The concise result isHere $\phi$ is the Euler totient function, with $\phi (1)=1$. The totient lower bound from the Mertens product gives $\phi (r)/r=\Omega (1/\max \{1,\log \log N\})$ for $r\le N$. Each trial uses polynomial time quantum Fourier transforms, function evaluations and Euclidean algorithm arithmetic in $\log N$. Repeating $O(\max \{1,\log \log N\})$ independent trials gives constant success probability; the herald accepts only correct answers.

The printed asymptotic bound has the wrong direction for this argument. In Big O notation, the stated upper bound on the number of coprime integers supplies no success lower bound and is itself false uniformly: for prime $N$, $\phi (N)=N-1$, not $O(N/\log \log N)$. Likewise the single-trial success need not be $O(1/\log \log N)$, since it tends to one for $r=N$ prime. The meaningful corrected guarantee is the lower bound above. If an upper-bounded success rate is literally required, one may deliberately retain an accepted result with an independent probability proportional to $1/\max \{1,\log \log N\}$; this discards useful successes and is unnecessary for efficient recovery.

In an irreducible complex group representation $\rho$, every $\rho (z)$ for $z$ in the center of a group commutes with the entire representation. Schur lemma therefore gives $\rho (z)={\lambda }_{z}I$. Faithfulness makes $z\mapsto {\lambda }_z$ an injective homomorphism $Z(G)\to {\mathbb{C}}^{\times }$.

Its image is a finite subgroup. If its order is $m$, Lagrange's theorem for finite groups shows that every element is an $m$th root of unity. These roots form a cyclic group, and every subgroup of a cyclic group is cyclic. HenceBoth irreducibility over the algebraically closed field and faithfulness are used: scalar action alone would not embed the centre.

Write $x=g^{-1}h$ and $S(x)={\sum }_{\chi \in \hat{G}}\chi (x)$, where $\hat{G}$ is the character group of a finite abelian group. The Fundamental theorem of finitely generated abelian groups writes the finite group $G$ as a product of cyclic groups. A character of a cyclic group factor of order $r$ is determined by an arbitrary $r$th root of unity, so $|\hat{G}|=|G|$. This description also shows that the characters separate points: if $x\ne 1$, a nonzero coordinate of $x$ is detected by a character $\psi$ with $\psi (x)\ne 1$.

Multiplication by $\psi$ permutes $\hat{G}$, soFor $x\ne 1$ this forces $S(x)=0$. For $x=1$ every summand is one. Since $\chi (g{)}^{-1}\chi (h)=\chi (g^{-1}h)$, the character-sum cancellation lemma givesAll character values are roots of unity, so inversion here is also complex conjugation. This is the finite-abelian version of character orthogonality.

The elliptization theorem says that a closed oriented three-manifold with finite fundamental group has spherical geometry. When the fundamental group is a cyclic group, it is a lens space. Consequently the candidates are $L(5,q)$ with $q\in \{1,2,3,4\}$, using the same surgery orientation convention throughout.

The oriented classification of lens spaces says that $L(p,q)$ and $L(p,q^{\prime })$ are related by an orientation-preserving homeomorphism exactly when $q^{\prime }\equiv q^{\pm 1}(\mathrm{mod}p)$. Since $1^{-1}=1$, $4^{-1}=4$, and $2^{-1}=3$ modulo five, the complete list isThese are three distinct oriented homeomorphism classes. Reversing orientation interchanges $L(5,1)$ and $L(5,4)$; the class containing $L(5,2)$ and $L(5,3)$ admits an orientation-reversing homeomorphism. The elliptization theorem and the stated oriented classification of lens spaces are the general classification results used here.

The splitting field of $f\in K[t]$ is the smallest extension $L/K$ in which $f$ is a product of linear factors; equivalently, $L$ is generated over $K$ by all its roots of a polynomial. To prove existence and uniqueness of splitting fields, adjoin a root of an irreducible polynomial dividing $f$ and repeat until $f$ splits. If $L$ and $L^{\prime }$ are two resulting fields, an embedding between the fields generated so far extends by sending each newly adjoined root to a root of its transformed minimal polynomial. Repetition gives a $K$-embedding $L\to L^{\prime }$, whose image contains every root of $f$ and is therefore all of $L^{\prime }$. Thus the splitting field exists and is unique up to $K$-isomorphism.

Now write $K={\mathbb{F}}_q$, and let the distinct irreducible factors of $f$ have degrees $d_1,\dots ,d_r$. The irreducible factors of a finite-field Frobenius polynomial show that a root of the degree-$d_i$ factor lies in ${\mathbb{F}}_{q^n}$ exactly when $d_i\mid n$. Hence the splitting field over a finite field isEvery finite field is a perfect field, so $L/K$ is separable; it is normal because it is a splitting field. It is consequently a Galois extension. More explicitly, the Galois group of a finite field extension is the cyclic group of order $m$ generated by the finite-field Frobenius automorphism $x\mapsto x^q$.

Finally, let $N_2(n)$ denote the number of monic irreducible polynomials over a finite field of degree $n$ over ${\mathbb{F}}_2$. Sinceand $\mu (1)=1$, $\mu (3)=-1$, $\mu (9)=0$, we obtain

Cauchy's theorem for finite groups states that if a prime number $p$ divides the order of a finite group $G$, then $G$ contains an element of order $p$.

ConsiderThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$ is divisible by $p$. The cyclic group $C_p$ acts on $X$ by cyclically shifting the coordinates; the shifted tuple remains in $X$ because $x_2\cdots x_p{x}_1=x_1^{-1}(x_1\cdots x_p)x_1=e$. Every orbit has size one or $p$. The fixed points are exactly $(x,\dots ,x)$ with $x^p=e$. Their number is therefore divisible by $p$. Since the identity is one fixed point, another exists; its order is $p$.

Now let the abelian group $G$ have order $pq$ for distinct primes $p,q$. Cauchy's theorem supplies $x,y$ of orders $p,q$. Their cyclic subgroups intersect trivially, and commutativity shows that $xy$ has order $pq$. Thus $G$ is cyclic andThe corresponding assertion for order $p^2$ is false: $C_{p^2}$ is not isomorphic to $C_p\times C_p$, since the former has an element of order $p^2$ while every nonidentity element of the latter has order $p$.

Every finite quotient of the abelian group $A_{\neg p}$ is abelian. Part iii excludes elements of prime order $q\ne p$ by the Cauchy theorem for groups, so any finite quotient is a finite abelian $p$-group. If its exponent divides $p^n$, the quotient map kills $p^n{A}_{\neg p}$ and therefore factors throughEvery quotient of a cyclic group is cyclic, so the finite quotient is isomorphic to $\mathbb{Z}/p^k\mathbb{Z}$ for some $k\le n$. Conversely, reduction modulo $p^k$ gives a surjection $A_{\neg p}\to \mathbb{Z}/p^k\mathbb{Z}$. Thus these are exactly the nontrivial finite quotients, together with the trivial case $k=0$.

No. The center of a group satisfiesbecause $Z(D_8)=\{1,r^2\}\cong C_2$. This centre is not cyclic. The cyclic-center obstruction to a faithful irreducible representation therefore rules out a faithful irreducible complex representation of $C_2\times D_8$.

Yes. If $g$ generates the cyclic group $C_n$, define the one-dimensional complex representationIt is irreducible because it is one-dimensional, and it is faithful because the displayed scalar has multiplicative order $n$. Thus every $C_n$, including the trivial case $n=1$, has a faithful irreducible representation.

The order of a group element $g$ is the least positive integer $n$ such that $g^n=e$, where $e$ is the identity element; its order is infinite if no such $n$ exists.

Let $g$ have finite order $n$. A group homomorphism preserves the group operation and the identity, soThe order of an element divides every positive exponent that gives the identity. Hence

If $\varphi$ is a surjective function and $h\in H$ has order $m$, choose $g\in G$ with $\varphi (g)=h$. The first result gives $m\mid n$, where $n=\mathrm{ord}(g)$. The elementthen has order $m$, because the order of $g^k$ is $n/\gcd (n,k)$.

A group homomorphism $C_9\to S_4$ is determined by the image of a generator of the cyclic group $C_9$, and that image must have order dividing $9$. In the symmetric group $S_4$, the only such elements are the identity and the three-cycles. There arethree-cycles. Therefore the number of homomorphisms is

A group $G$ is abelian when $xy=yx$ for every $x,y\in G$. It is cyclic when some generator of a group $g$ satisfiesIf $x=g^r$ and $y=g^s$ belong to a cyclic group, thenso every cyclic group is abelian.

The Klein four-group $C_2\times C_2$ is abelian, but every nonidentity element has order two, so no element generates all four elements. Thus an abelian group need not be cyclic.

The condition on proper subgroups does not force $G$ to be abelian. The quaternion groupis nonabelian because $ij=k$ whereas $ji=-k$. Its proper subgroups are the trivial subgroup, $\{\pm 1\}$, and the three cyclic subgroupseach of order four. They are all cyclic, so

A group action of $G$ on $X$ is a map $(g,x)\mapsto gx$ such that $ex=x$ and $(gh)x=g(hx)$. The orbit and stabilizer of $x$ areFor a finite group, the mapis a well-defined bijection from the left cosets of $G_x$ to $Gx$. Each coset has $|G_x|$ elements, so the orbit-stabilizer theorem is

The Cauchy theorem for groups says that if a prime number $p$ divides $|G|$, then $G$ has an element of order $p$. To prove it, letThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$, which is divisible by $p$. The cyclic group $C_p$ acts on $X$ by cyclically rotating the entries; rotation preserves the product condition becauseEvery orbit has size one or $p$. The fixed points are exactly the tuples $(g,\dots ,g)$ with $g^p=e$. Their number is therefore divisible by $p$. Since the identity gives one fixed point, there is another, and its entry has order $p$.