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Every identity-preserving biholomorphism of $\mathbb{C}/\Lambda$ lifts to multiplication by a nonzero complex number $\zeta$ satisfying $\zeta \Lambda =\Lambda$. Relative to a $\mathbb{Z}$-basis of $\Lambda$, multiplication by $\zeta$ is represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. Its real determinant is $|\zeta {|}^2$, while complex multiplication preserves orientation, so $\det A=1$ and $|\zeta |=1$. The Cayley-Hamilton theorem givesConsequently $2\mathrm{Re}\zeta =\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$, and $\zeta$ is a root of unity of order $1$, $2$, $3$, $4$, or $6$.

For $C_n=⟨g:g^n=e⟩$, the complex irreducible representations are precisely the $n$ one-dimensional maps $g\mapsto e^{2\pi ij/n}$, with $0\le j<n$. Their characters have the displayed values, forming the discrete Fourier character table. Indeed a representing matrix $A$ satisfies $A^n=I$. Applying the fundamental theorem of algebra to its characteristic polynomial gives an eigenvalue and an eigenvector over the complex numbers. Its line is an invariant subspace under every power of $A$; irreducibility therefore forces the entire space to be that line. Its eigenvalue is an $n$th root of unity. Different roots give inequivalent one-dimensional representations, and there are exactly $n$ possibilities.

All conjugacy classes are singletons because the group is Abelian. The number of the displayed irreducibles equals the group order, and their squared dimensions sum to $n$, in agreement with the sum of squares of irreducible representation dimensions. Over the real numbers, conjugate nonreal characters instead combine into the planes described by the irreducible real representations of a finite cyclic group.

At a primitive cube root of unity, the cubic homogeneous module has basis $x^3,x^{2}y,x{y}^2,y^3$. Its only simple modules occurring as submodules are the one-dimensional subspaces of $⟨x^3,y^3⟩$, on which $K=1$ and $E=F=0$. Its quotient by this two-dimensional submodule is simple, but it has no simple complement: any lift of the highest $K$ weight space contains $x^{2}y$, whose $E$ image is $x^3\ne 0$. Thus the cubic module is not a semisimple module.

For a primitive $m$th root of unity ${\zeta }_m$, every Galois automorphism sends ${\zeta }_m$ to a unique ${\zeta }_m^a$ with $a$ invertible modulo $m$. The assignment $\sigma \mapsto a$ is a group homomorphism, since composition multiplies the exponents, and is injective because ${\zeta }_m$ generates the cyclotomic field. The irreducibility of cyclotomic polynomials gives degree $\phi (m)$, equal to the order of the target, so the map is an isomorphism. The cyclotomic character on an absolute Galois group factors through this finite quotient and generally has a nontrivial kernel.

Let $q\nmid m$ be outside the excluded primes, let $\zeta$ have order $m$, and let $\rho$ be a primitive qth root of unity. Galois equivariance identifies the conjugates with $\Phi ({\rho }^i\zeta )$ for $1\le i\le q-1$. The distribution axiom includes the missing $i=0$ factor, so the field norm is $\Phi ({\zeta }^q)/\Phi (\zeta )$. Arithmetic Frobenius sends $\zeta$ to ${\zeta }^q$, giving the displayed factor. At higher powers of an already ramified prime all q translates occur instead. These are different norm relations; norm compatibility in just a p-power tower does not supply the auxiliary-prime relation by itself.

A cyclotomic integer is an element of $\mathbb{Z}[{\zeta }_n]$, where ${\zeta }_n$ is a root of unity. For a primitive $n$-th root, this is the full ring of integers of the cyclotomic field $\mathbb{Q}({\zeta }_n)$.

Suppose fields with integer U(1) charges $Q_i$ acquire nonzero vacuum expectation values, with at least one $Q_i\ne 0$. The stabilizer in the circle group is exactly ${\mathbb{Z}}_d$, where $d=\gcd (|Q_1|,\dots ,|Q_r|)$. Indeed, $e^{i\theta }$ preserves every condensate if and only if $e^{i{Q}_i\theta }=1$ for every $i$. The Bezout identity implies $e^{id\theta }=1$, and conversely $d\mid Q_i$ proves every $d$th root of unity preserves every condensate. Neutral fields impose no constraint; if all condensing fields are neutral, the whole circle group survives.

The complex algebraic numbers form a field, the algebraic closure of $\mathbb{Q}$ inside $\mathbb{C}$. To see closure under the field operations, if $\alpha ,\beta$ are algebraic numbers, the tower law makes $\mathbb{Q}(\alpha ,\beta )$ a finite-dimensional field extension of $\mathbb{Q}$. Every element of this extension is an algebraic element, because its successive powers are linearly dependent. In particular it contains $\alpha \pm \beta$, $\alpha \beta$ and $\alpha /\beta$ when $\beta \ne 0$.

It is an infinite-degree algebraic extension. A primitive $p$th root of unity has minimal polynomial of an algebraic element $1+X+\cdots +X^{p-1}$ of degree $p-1$ for every prime $p$, so the degrees of its finite subextensions are unbounded. Finally a complex root of a polynomial whose coefficients are algebraic numbers is itself algebraic over their finite-degree coefficient field and hence over $\mathbb{Q}$, again by the tower law. Thus every such polynomial splits in this field, establishing the algebraic closure assertion.

A nonidentity Möbius transformation of finite order $n$ is conjugate to multiplication by a primitive $n$th root of unity. Zero and infinity are fixed; for any other point, a return after $k$ iterates requires ${\mu }^k=1$, whose least positive solution is $n$. Thus its only orbit sizes are one and $n$. For the identity, every orbit has size one. This is a direct orbit computation, independent of an appeal to the orbit-stabilizer theorem.

If $Ev=0$ and $Kv=\lambda v$, induction using the defining commutator gives the displayed identity. When $q$ is not a root of unity, a finite string of length $n+1$ in a simple module has $\lambda =ϵq^n$, $ϵ=\pm 1$. Its distinct $K$ eigenvalues make the string vectors independent, and the nonzero raising coefficients prove simplicity.

If a cap set $A\subseteq {\mathbb{F}}_3^d$ has subset density $\alpha >0$ and $3^d{\alpha }^2\ge 2$, some affine subspace of codimension one has relative subset density at least $\alpha +{\alpha }^2/2$. The zero-sum count and the Parseval identity on a finite group give a nonzero finite abelian Fourier coefficient of magnitude at least ${\alpha }^2/2$. The three slice densities are $\alpha +2\mathrm{Re}(z{\omega }^j)$, where $z$ is this finite abelian Fourier coefficient and $\omega$ a primitive cube root of unity; one slice has increment at least $|z|$. Translation preserves the zero-sum condition because the characteristic is three.

Forthere are $2n$ linear characters, given by $a\mapsto {\xi }^j$, $b\mapsto 1$ for a primitive $(2n)$th root of unity $\xi$, and $n$ two-dimensional irreducible representationswhere ${\omega }^3=1$ and $\omega \ne 1$. The two-dimensional representations parametrized by $\epsilon$ and $-\epsilon$ are isomorphic. The sum of squares of irreducible degrees is $2n+4n=6n$, so this list is complete.

A nonzero algebraic integer whose conjugates all have modulus at most one is a root of unity. Its norm forces every conjugate to have modulus one, and bounded integral coefficient vectors show that two positive powers coincide.

The curve has good reduction at five and seven: its discriminant is $-432D^2$, and the hypotheses make $D$ a unit at those primes. Over ${\mathbb{F}}_5$ the cubing map is a bijection, soThe character sum vanishes after replacing $x^3+D$ by a variable ranging over the whole field. Over ${\mathbb{F}}_7$, where $D=2$, the possible cubes are zero once, one three times, and minus one three times. The corresponding right sides are two, three and one. Two and one are nonzero squares while three is not, so the count is $1+2+0+6=9$.

Rational torsion injects into the reduction group at a good odd prime, a standard reduction theorem. Thus its order divides $\gcd (6,9)=3$. If a point of order three existed, its coordinates would be integral by the Nagell–Lutz theorem. Since $2P=-P$, the duplication formula givesThus either $x=0$ and $y^2=D$, or $x^3=-4D$ and $y^2=-3D$.

At the prime in the hypothesis, $D$ is a nonsquare unit. Since $p\equiv 1(\mathrm{mod}3)$, there is a nontrivial cube root of unity $\zeta \in {\mathbb{F}}_p$, and $(2\zeta +1{)}^2=-3$. Hence $-3$ is a nonzero square, so both $D$ and $-3D$ are nonsquares. Neither ordinate equation is possible even modulo $p$. There is no rational point of order three, proving

For $K=\mathbb{Q}(\zeta )$, every Galois automorphism sends the primitive root $\zeta$ to another primitive $m$th root of unity, uniquely ${\zeta }^a$ for $a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }$. The Cyclotomic character of a finite cyclotomic extension isIf ${\chi }_m(\sigma )=a$ and ${\chi }_m(\tau )=b$, then $(\sigma \tau )(\zeta )=\sigma ({\zeta }^b)={\zeta }^{ab}$, proving the group homomorphism property. Its kernel is trivial: an automorphism fixing $\zeta$ fixes the field it generates. Thus the character is injective. Here its domain is the finite cyclotomic Galois group; the corresponding map from an absolute Galois group need not be injective.

The additive cyclic group ${\mathbb{Z}}_n$ is generated by $1$, with $n\cdot 1=0$. Every complex irreducible representation is one-dimensional by a. If $R(1)=z$, the defining relation requires $z^n=1$. Conversely each such root of unity defines a group homomorphism. Hence the complete list isChanging the representative of $g$ changes the exponent by an integer multiple of $2\pi i$, so the formula is well-defined. These representations are pairwise inequivalent because their values on the generator differ. The case $k=0$ is the trivial representation.

The matrices obey $Q^5=P^2=I$ and $PQP=Q^{-1}$. Moving every $P$ to the right therefore reduces every product to $Q^k$ or $Q^{k}P$, with $0\le k<5$. The five diagonal matrices $Q^k$ are distinct because $\alpha$ is a primitive fifth root of unity; the five $Q^{k}P$ are distinct and off-diagonal. ThusThis is the dihedral group of the regular pentagon, with $Q$ a rotation and $P$ a reflection.

Conjugation by $P$ interchanges $Q^k$ and $Q^{-k}$, while conjugation by a power of $Q$ fixes rotations. AlsoSince multiplication by $2$ permutes the residues modulo $5$, all five reflections form one conjugacy class. The full list of conjugacy classes isTo prove the representation is irreducible over $\mathbb{C}$, any proper nonzero invariant subspace would be a line invariant under $Q$. Its two distinct eigenvalues force such a line to be one of its coordinate eigenspaces. But $P$ exchanges those two lines, so neither is $G$-invariant. Hence the given complex representation is irreducible.

Choose a normalization in which the circle group has period $2\pi$ and the U(1) charges ${\hat{Q}}_i$ of the complex scalar fields are integers. By restriction of a representation,Write $v_i=⟨{\varphi }_i⟩$. A vacuum expectation value $v_i\ne 0$ is preserved by $e^{i\theta }$ precisely when $e^{i{\hat{Q}}_i\theta }=1$. Therefore the specified finite cyclic group survives spontaneous symmetry breaking exactly when every condensing complex scalar field obeysOnly U(1) charges of fields with nonzero vacuum expectation values enter this test; in the paper all $N$ fields condense.

Suppose at least one condensing U(1) charge is nonzero and put $d=\gcd (|{\hat{Q}}_i|:v_i\ne 0)$. Since each such U(1) charge is a multiple of $d$, every $d$th root of unity preserves the vacuum expectation values. Conversely, the Bezout identity expresses $d$ as an integer combination of these U(1) charges, so a common stabilizer obeys $e^{id\theta }=1$. The discrete remnant of a circle symmetry is consequentlyA nontrivial retained finite cyclic group exists exactly when $d>1$, and its possible orders are the divisors $n>1$ of $d$. For example, condensing U(1) charges $6$ and $10$ leave ${\mathbb{Z}}_2$. If $d=1$, the stabilizer is trivial. If all condensing U(1) charges vanish, the entire circle group is unbroken and there is no largest finite $n$; this case does not break the symmetry assumed in the question.

Let $\omega =e^{i\pi /7}$, a primitive fourteenth root of unity. The geometric series gives ${\sum }_{k=1}^{14}{\omega }^k=0$. Taking the real part and adding the remaining fifteenth term therefore yieldsThe upper summation limit matters: the first fourteen terms complete one full cycle, and the fifteenth supplies one additional term.

A root of unity satisfies $|z{|}^3=1$, hence has modulus one. Writing $z=e^{i\varphi }$, its argument satisfies $3\varphi =2\pi k$ for some integer $k$. The three distinct complex roots are thereforeThey are equally spaced on the unit circle. Taking $k=0,1,2$ exhausts them, since adding three to $k$ repeats a root and a nonzero polynomial of degree three has at most three distinct roots.