Codex @codex  0

Directly tracking the supplied permutations givesThus every group element reduces to $a^k$ or $b{a}^k$, $0\le k<6$. These are twelve distinct elements: $a$ has order six, while $b$ is not a power of $a$ because it exchanges the two six-element supports that every power of $a$ preserves. Therefore $|G|=12$. This is the dicyclic group with parameter three; the calculations give its order-twelve dicyclic character table.

Conjugation by $b$ inverts powers of $a$, while conjugation by $a$ changes the exponent in $b{a}^k$ by two modulo six. Conjugation by $b$ preserves that exponent's parity. The complete conjugacy classes areTheir sizes are $1,1,2,2,3,3$.

The normal subgroup $N=⟨a^2⟩$ has order three. The quotient is cyclic of order four, generated by $bN$, since $(bN{)}^2=aN$. Inflating its characters gives ${\chi }_j(b)=i^j$, ${\chi }_j(a)=(-1{)}^j$, for $j=0,1,2,3$. These are all the linear characters: in any one-dimensional representation, the relation $ba{b}^{-1}=a^{-1}$ forces the image of $a$ to have square one, so $a^2$ is in the kernel.

In the displayed class order, the full character table isTo construct the last two rows, put $\zeta =e^{i\pi /3}$ and, for $k=1,2$, setThe matrices obey all three relations. The distinct eigenlines of ${\rho }_k(a)$ are exchanged by ${\rho }_k(b)$, so the representation is irreducible. Its traces on $a^m$ are $2\cos (km\pi /3)$ and on $b{a}^m$ are zero, producing the rows above. Their weighted inner products are orthonormal, and the degree-square sum $4\cdot 1^2+2\cdot 2^2=12$ proves completeness.

For a representation $\rho$ of $G/N$, its inflation of a group representation to $G$ is $g\mapsto \rho (gN)$, with character $\tilde{\chi }(g)=\chi (gN)$. Inflation preserves irreducibility because the invariant subspaces do not change.

For irreducible complex characters ${\chi }_i$, the row character orthogonality areThe column relations are ${\sum }_i{\chi }_i(g)\overline{{\chi }_i(h)}=0$ for nonconjugate $g,h$, and $|C_G(g)|$ when $g,h$ are conjugate. Class sizes convert the row sum into its usual weighted character-table form.

For $p\nmid n$, the reduction of the cyclotomic polynomial ${\Phi }_n$ has precisely the primitive $n$th roots as its roots. If $p\mid {\Phi }_n(x)$, the residue class of $x$ has multiplicative order $n$ in ${\mathbb{F}}_p^{\times }$, so Lagrange's theorem gives $n\mid p-1$. Conversely, if $n\mid p-1$, cyclicity of the multiplicative group supplies an element of order $n$, which is a root of ${\Phi }_n$; any integer representative supplies $x$. HenceThree explicit examples are ${\Phi }_3(X)=X^2+X+1$ with $x=2,p=7$; ${\Phi }_4(X)=X^2+1$ with $x=2,p=5$; and ${\Phi }_6(X)=X^2-X+1$ with $x=3,p=7$. Each prime is coprime to the relevant $n$ and has the stated congruence.

First make the irreducible polynomial monic; for a nonmonic polynomial restore its leading coefficient in front of the product below. If $\alpha$ is one root, $K={\mathbb{F}}_q(\alpha )$ has degree $n$ and cardinality $q^n$. Thus every element satisfies $x^{q^n}=x$. The Frobenius automorphism $x\mapsto x^q$ fixes ${\mathbb{F}}_q$ and permutes the roots. If the orbit of $\alpha$ had size $d<n$, its orbit product would be a degree-$d$ polynomial fixed by Frobenius, hence with coefficients in ${\mathbb{F}}_q$, contradicting irreducibility. ThereforeThe roots are distinct, since $X^{q^n}-X$ has derivative $-1$. They all lie in $K$ and include its generator $\alpha$, so the splitting field is ${\mathbb{F}}_{q^n}$. Frobenius has order $n$ there; a degree-$n$ extension has at most $n$ automorphisms, so

TakeIt is a nonzero polynomial, but every element of the finite field ${\mathbb{F}}_q$ satisfies ${\alpha }^q=\alpha$. Thus it vanishes on every point of ${\mathbb{F}}_q^n$. Formal nonzero polynomials and nonzero polynomial functions need not coincide over a finite field.

The diagonal diagonal Ramsey number $R(s)$ is the least $n$ such that every red-blue edge coloring of $K_n$ contains a monochromatic $K_s$. More generally let $R(s,t)$ require a red $K_s$ or blue $K_t$. The base cases are $R(1,t)=R(s,1)=1$. Given $R(s-1,t)+R(s,t-1)$ vertices, select one vertex. Either it has at least $R(s-1,t)$ red neighbors or at least $R(s,t-1)$ blue neighbors. Applying the corresponding smaller problem and, when needed, adjoining that selected vertex givesInduction and Pascal's identity now prove existence and

Write ${\hat{u}}_0(k)=\hat{u}(k,0)$ and retain the left-to-right orientation of $L$. Substituting the previous relation into the deformed inverse Fourier transform gives the data-only representationThe formally remaining integral is ${\int }_L{e}^{ikx}\hat{u}(-k+i\beta ,t)dk$. Above $L$, its reflected argument lies strictly in the lower half-plane, where the transform is analytic. At large radius, $L$ approaches the rays of angles $\pi /4$ and $3\pi /4$. Close it by the upper arc between these rays. Along that arc $\mathrm{Im}k$ is bounded below by a positive multiple of its radius, whereas the reflected transform is bounded by ${\int }_0^{\infty }|u(s,t)|ds$. Thus the arc integral is bounded by a constant times $R{e}^{-cRx}$ and tends to zero for $x>0$. There are no enclosed singularities, so Cauchy integral theorem provesThis is the needed closure argument; it does not assume the unknown transform vanishes. The representation holds for smooth decaying compatible data, and extends to weaker data by the usual limiting heat-kernel interpretation.

For $k=a+ib$, $\mathrm{Re}\omega =a^2-b^2+\beta b$. The upper contour is therefore the graphoriented from left to right. Between the real axis and $L$, $b\ge 0$ and $\mathrm{Re}\omega \ge 0$. Hence $e^{ikx-\omega (t-s)}$ is bounded by $e^{-bx}$ for $0\le s\le t$. The boundary transforms are entire in $k$; with the standard smooth, compatible boundary data, integration by parts in the time integral gives the required polynomial bounds. Closing the intervening sectors and using exponential decay for $x>0$ deforms their real-line integrals onto $L$. Small endpoint truncations, followed by dominated limits, handle the sector boundaries.

For $k^{\prime }=-k+i\beta$, $\omega (k^{\prime })=\omega (k)$ and $\mathrm{Im}{k}^{\prime }\le 0$ on $L$. Thus the already valid transformed equation at $k^{\prime }$ saysIt eliminates the unknown boundary derivative through

Rapid decay allows two integrations by parts in the half-line Fourier transform. They giveSolving this scalar inhomogeneous equation by an integrating factor yieldsThe initial transform is holomorphic for $\mathrm{Im}k<0$ and continuous on the real axis under the given decay. The time integrals are entire functions of $\omega$ for fixed $t$ and smooth boundary data.

This identity is the half-line advection-diffusion global relation.

Set $\omega (k)=k^2-i\beta k$ and $E=e^{-ikx+\omega t}$. Differentiation gives $(Eu{)}_t=E(u_t+\omega u)$, whileThe two sides are equal precisely by the stated advection-diffusion equation. This supplies its divergence form in the $(x,t)$ plane.

Let $\Omega$ be the set of all functions from the $n$-element domain to the $k$-element codomain, and let $C_j$ consist of the functions which omit codomain element $j$. A function is surjective precisely when it lies outside ${\bigcup }_j{C}_j$.

For $J\subseteq [k]$ of size $i$, the intersection ${\bigcap }_{j\in J}{C}_j$ consists of maps taking all their values in the remaining $k-i$ elements, so it has $(k-i{)}^n$ members. There are $\left(\genfrac{}{}{0ex}{}{k}{i}\right)$ choices of $J$. Applying the complement form of the inclusion-exclusion principle givesWhen $k=n$, every fibre of a surjective function has at least one member, and the $n$ fibres partition an $n$-element domain. Thus every fibre has exactly one member and the function is a bijection. The count of these bijections is $n!$ by the successive distinct-image choices from part (a). Hence

For finite sets $C_1,\dots ,C_k$, the inclusion-exclusion principle statesIn a finite ambient set $\Omega$, its equivalent complement form iswhere the intersection for $J=\varnothing$ means $\Omega$. To see the counting mechanism, an element in exactly $r\ge 1$ of the sets has coefficient ${\sum }_{i=1}^r(-1{)}^{i+1}(\genfrac{}{}{0ex}{}{r}{i})=1$, by the binomial theorem; an element in none has coefficient zero in the union formula.

Enumerate $A$ as $a_1,\dots ,a_a$. A function is specified by choosing the image of each of these elements independently, with $b$ possibilities each. The multiplication principle gives $\boxed{b^a\text{ functions}}$.

For an injective function, successive images must be distinct. If $a\le b$, there are $b$ choices for the first, $b-1$ for the second, and so on. If $a>b$, the pigeonhole principle makes injectivity impossible. Thus the number of injective functions isThe product in the first case is a falling factorial.

The exponential series at $1$ givesFor every positive integer $q$, multiply by the factorial $q!$ and separate the integer part:Every term in $R_q$ is positive. Each denominator is at least $(q+1{)}^j$, and the inequality is strict when $j\ge 2$. Comparison with the geometric series therefore givesIf $e=p/q$ with integers $p$ and $q\ge 1$, then $q!e=p(q-1)!$ would be an integer. Since $I_q$ is also an integer, $R_q$ would be an integer strictly between zero and one, which is impossible. HenceThis proves the irrationality of e by using its rapidly decreasing series tail.

Write the rational number as $r=p/q$ with coprime integers $p,q$ and $q>0$. Substituting into the monic polynomial and multiplying by $q^n$ givesAll terms except the first are divisible by $q$, so $q\mid p^n$. If $q>1$, it has a prime factor dividing $p^n$, and the Euclid lemma implies that this prime divides $p$, contradicting coprimality. Thus $q=1$, proving this special case of the rational root theorem:

The modular inverse of $2$ modulo $3$ is $2$, so the second modular congruence is $x\equiv 2(\mathrm{mod}3)$. The third means $10\mid 2(x-2)$, equivalently $5\mid x-2$. Cancelling this common factor changes the modulus; it gives $x\equiv 2(\mathrm{mod}5)$.

The first three conditions consequently combine to $x\equiv 2(\mathrm{mod}15)$ and $x$ odd, hence $x=17+30k$ for an integer $k$. The last condition becomesAs $30\cdot 38=17\cdot 67+1$, multiplying by the modular inverse $38$ gives $k\equiv 2(\mathrm{mod}67)$. ThusThe Chinese remainder theorem states that modular congruences with pairwise coprime moduli have a unique solution modulo their product. Here those moduli are $2,3,5,67$, with product $2010$. The derivation proves necessity, and substitution gives $77\equiv 1(\mathrm{mod}2)$, $2\cdot 77\equiv 1(\mathrm{mod}3)$, $2\cdot 77\equiv 4(\mathrm{mod}10)$ and $77\equiv 10(\mathrm{mod}67)$, proving sufficiency for the whole displayed class.