Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 2

Write $p=8k+7$. In Gauss's lemma, for the multiplier 2, precisely the integers $j$ with $p/4<j\le (p-1)/2$ give residues $2j>p/2$. Their number isIt is even, so the Legendre symbol $(2/p)=1$: 2 is a quadratic residue modulo $p$. For completeness, the sign rule follows by replacing those large residues by their negatives; the resulting absolute residues permute $1,\dots ,(p-1)/2$. Multiplying them and cancelling their nonzero product gives $2^{(p-1)/2}\equiv (-1{)}^{2k+2}=1(\mathrm{mod}p)$, equivalent to the residue assertion by Euler's criterion.

Now take the prime $p=2q+1$. The condition $q\equiv 3(\mathrm{mod}4)$ gives $p\equiv 7(\mathrm{mod}8)$, so $2^q\equiv 1(\mathrm{mod}p)$. Thus $p$ divides $2^q-1$. For integers $q>3$, $2^q-1>2q+1$ (true at 4 and preserved on increasing $q$), so this divisor is proper and $2^q-1$ is composite. At $q=3$, the divisor is the number itself: $2^3-1=2\cdot 3+1=7$. The congruence still holds, but it does not establish compositeness.

Choose a point of $K$ with both coordinates positive, so its product is positive. The set $K\cap [0,\infty {)}^2$ is nonempty and compact; the continuous function $xy$ therefore attains its maximum there. A maximizer cannot lie on either axis, where the product is zero. It consequently belongs to $K\cap P^{+}$ and also maximizes the product on that set.

For uniqueness, the function $\Phi (x,y)=\log x+\log y$ is strictly concave on the positive quadrant: its Hessian is $\mathrm{diag}(-1/x^2,-1/y^2)$, negative definite. If two distinct positive points attained the same maximal product, their midpoint would belong to $K\cap P^{+}$ by convexity and strict concavity would give a larger value of $\Phi$, hence a larger product. This contradiction proves existence and uniqueness of the positive product maximizer. This argument is positive-product maximization on a convex set; compactness should be applied to the closed quadrant intersection, not directly to the potentially noncompact set $K\cap P^{+}$.

Here a planar lattice means a discrete subgroup of the additive Euclidean plane; full rank is not required. Discreteness at zero gives an $ϵ>0$ such that distinct lattice points are at least $ϵ$ apart. Consequently every bounded set contains only finitely many lattice points, by placing disjoint small disks around them. If the group is zero, it has the stated one-generator form with $w=0$.

Otherwise choose a shortest nonzero lattice vector $w_1$. Every lattice point on its line is an integer multiple: subtract the nearest integer multiple of $w_1$ from a putative noninteger multiple to obtain a nonzero shorter vector, a contradiction. This already proves the one-generator case when all lattice points lie on that line.

For the remaining case choose the perpendicular orientation so that some point has positive height above the line of $w_1$. Reduce its parallel component modulo $w_1$ into $[0,|w_1|)$. Among reduced points of positive height there is a least height $d>0$. Indeed, heights tending to zero would give infinitely many points in a bounded strip, contrary to the finiteness just proved; bounded sets also ensure that a positive infimum is attained. Choose $w_2$ at height $d$.

Given any lattice vector $v$, subtract an integer multiple of $w_2$ so that its height lies in $[0,d)$, then reduce its parallel component modulo $w_1$. Minimality of $d$ forces the residual height to be zero, and the one-dimensional argument forces the residual to be an integer multiple of $w_1$. ThereforeTogether with the rank-one and zero cases this proves the planar discrete-subgroup basis assertion. Under the alternative convention that a Euclidean lattice must have full rank, only the two-generator case is called a lattice.

A digital signature lets the holder of a private key authenticate a message so that anyone with the corresponding trusted public key can verify its origin and integrity. The signature does not conceal the message. Public-key certification, a suitable message hash and protection of the signing key are needed for that interpretation.

For the ElGamal signature scheme, choose a large prime $p$ and a primitive root $g$ modulo $p$. The secret key is $x$, and the public key is $(p,g,y)$ with $y=g^x\mathrm{mod}p$. Represent the message by a hash $m$ modulo $p-1$. Choose a fresh secret nonce $k$ coprime to $p-1$ and setThe signature is $(r,s)$. After checking the prescribed ranges, the verifier accepts whenA genuine signature passes because $xr+ks\equiv m(\mathrm{mod}p-1)$. The security aim is that someone without $x$ cannot efficiently produce a valid signature on a new authenticated message. Nonce secrecy and freshness matter: reuse gives linear congruences relating two message hashes, the nonce and the secret key. Signing unhashed arbitrary messages also permits algebraic forgeries, so the hash is part of the practical scheme.

The independent Laplace distribution responses have log likelihoodFor every fixed positive $\sigma$, maximizing this expression means minimizing the sum of absolute residuals. Thus $\hat{\beta }$ is a least absolute deviations estimate, not a least-squares estimate. Such a minimizer exists: minimize the distance in the finite-dimensional closed column space of the design matrix. It need not be unique if the data or design allow multiple minimizers.

Write $s=S_1(\hat{\beta })$. For $s>0$, differentiating the profiled log likelihood gives $-n/\sigma +s/{\sigma }^2=0$. Its unique maximum isIf the fitted residual sum is zero, the likelihood instead increases without bound as $\sigma \downarrow 0$: there is no joint maximum with the stated positive scale parameter. The displayed estimate then denotes the boundary limit, not an admissible positive-scale MLE.

An ordinary exponential dispersion family has log density $[y\theta -b(\theta )]/a(\sigma )+c(y,\sigma )$, so changing the mean changes the log density by an affine function of $y$. For two distinct Laplace locations ${\mu }_1,{\mu }_2$, their log-density difference is $(|y-{\mu }_2|-|y-{\mu }_1|)/\sigma$, with different slopes on the intervals separated by the two locations. It is not affine in $y$. Equivalently the kink at $y=\mu$ moves with the parameter. The Laplace location family is not an ordinary exponential dispersion family.

First-season recruits in year $n+1$ contribute $\alpha \sigma \gamma X_n$. Seeds produced in year $n-1$ that survive two winters, fail first-season germination and germinate in their second season contribute $\beta {\sigma }^2(1-\alpha )\gamma X_{n-1}$. HenceThis treats recruitment from older seeds as absent; otherwise the two-season recurrence would need further terms. The characteristic polynomial is $r^2-ar-b$. For distinct roots,with constants determined by two initial values. For $b=0,a>0$, the recurrence is first order after the initial step; for $a=b=0$ it gives $X_{n+1}=0$. These describe the zero-root degeneracies without interpreting $0^0$ in the formula.

Since $a,b\ge 0$, $|r_{-}|\le r_{+}$. The positive root is less than one exactly when $a+b<1$: the polynomial is nonpositive at zero, has only one positive root, and its value at one is $1-a-b$. ThereforeThis is the decay condition for the two-season seed-bank recurrence. The deterministic model describes expected or averaged abundance; literal eventual integer extinction requires that interpretation. A corresponding branching model with geometrically decaying expected abundance becomes extinct almost surely, since the probability of any surviving plants is at most the expected number.

The simultaneous zeros are the origin and the four corners $(\pm 1,\pm 1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, with eigenvectors $(1,1),(1,-1)$, so the origin is a saddle equilibrium of Poincaré index $-1$. At a corner $J=-2xyI$. Thus $(1,1)$ and $(-1,-1)$ are stable nodes; $(1,-1)$ and $(-1,1)$ are unstable nodes, all of index $+1$. Although the node eigenvalues coincide, the equilibria are hyperbolic and their stability follows from their strictly signed real eigenvalues.

The lines $x=\pm 1$ and $y=\pm 1$ are invariant and divide the plane into nine convex regions. A nonconstant periodic trajectory cannot cross any of these lines by uniqueness, and cannot be contained in one of them because one-dimensional autonomous flows have no nonconstant periodic solutions. It must therefore lie inside one region. The central region contains only the origin, whose index is $-1$; all other regions contain no equilibria. But a simple periodic orbit has index $+1$, equal to the sum of the enclosed equilibrium indices. Its interior stays in the same convex region, giving a contradiction in either case. There are no periodic orbits.

Away from the invariant lines, trajectory curves have the first integralsince both logarithmic derivatives equal $-2xy$. The diagonal $y=x$ is the unstable separatrix of the origin, leading toward the two sinks; $y=-x$ is its stable separatrix, coming from the two sources. The diagram shows these separatrices, the invariant lines and the direction field.

Near zero the integrand is asymptotic to $t^{z-1}/2$, and at infinity it decays exponentially. The defining integral therefore converges and is analytic for $\mathrm{Re}z>0$, by locally uniform domination, also allowing logarithmic factors for differentiation. Multiplication by the entire function $1/\Gamma (z)$ does not enlarge the domain of the unevaluated integral itself.

On $\mathrm{Re}z>1$, subtract the two convergent integrals and useThe substitution $s=2t$ gives $I(z)-\zeta (z)=-2^{1-z}\zeta (z)$. Thusfirst there and then wherever the two analytic continuations agree. The Riemann zeta function is meromorphic on the plane with its only pole a simple pole at one. The prefactor has a simple zero there, $1-2^{1-z}=(z-1)\log 2+O((z-1{)}^2)$. It cancels the pole, giving $I(1)=\log 2$. Elsewhere both factors are holomorphic. The analytic continuation of $I$ is entire; it is the Dirichlet eta function.

The Euler-Lagrange equations areUsing time independence of the vector potential gives $m{\ddot{r}}_i=(e/c){\dot{r}}_j({\partial }_i{A}_j-{\partial }_j{A}_i)$, which is the component form ofTaking its scalar product with the velocity shows $\dot{T}=m\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}=0$: the kinetic energy is constant, since the magnetic force does no work.

For $\mathbf{B}=(0,0,F^{\prime }(x))$, the equations give $m\ddot{y}=-(e/c)F^{\prime }(x)\dot{x}$ and $\ddot{z}=0$. Thus $mc\dot{y}+eF(x)=-C$ and $\dot{z}=v_z$ are constants of motion. Substitution into $2T/m={\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2$ yieldsThe constant $D$ in this representation is nonpositive. The gauge choice $\mathbf{A}=(0,F(x),0)$ also makes the conserved transverse canonical momenta explicit.

Set $s=a(t)/a(t_{\mathrm{dec}})$. Free propagation in the homogeneous expanding universe redshifts frequencies by $\nu ={\nu }_{\mathrm{dec}}/s$ and dilutes physical number density by $s^{-3}$. Number conservation in corresponding frequency intervals givesInserting the initial Planck spectrum, the factor $s^{-2}(s\nu {)}^2$ becomes ${\nu }^2$, while its exponent is $hs\nu /(k{T}_{\mathrm{dec}})$. HenceThe shape is preserved even without collisions maintaining thermal equilibrium; it follows from redshift of an initially Planckian distribution.

For the total photon number density put $x=h\nu /(kT)$. For the energy density include the energy $h\nu$ of each photon. Thenwhere $\alpha =(8\pi /c^3)(k/h{)}^3{\int }_0^{\infty }{x}^2/(e^x-1)dx$ and $\xi =(8\pi h/c^3)(k/h{)}^4{\int }_0^{\infty }{x}^3/(e^x-1)dx$. Both integrals are finite positive constants. Thus $n\propto a^{-3}$ and $ϵ\propto a^{-4}$, with the extra power supplied by photon-energy redshift.

For a discrete memoryless channel, its channel capacity in bits per use is $C={\max }_{P_X}I(X;Y)$. The Shannon second coding theorem identifies this with the supremum of rates achievable by arbitrarily long block codes with error probability tending to zero: every rate below $C$ is achievable, and no rate above $C$ can have vanishing error. The channel is used independently at each transmission, and coding rate is measured using base-two logarithms.

Let $q$ be the probability of using the second input symbol. The second output symbol occurs with probability $q/2$, so $H(Y)=h_2(q/2)$. The first input has deterministic output and the second has one bit of conditional output Shannon entropy; hence $H(Y\mid X)=q$ andFor $0<q<1$, its derivative is $\frac{1}{2}{\log }_2[(2-q)/q]-1$, which vanishes at $q=2/5$. Strict concavity of binary entropy makes this the unique global maximum. Substituting givesThus Shannon's theorem supplies operational coding rates arbitrarily close to this value from below, and rules out reliable transmission above it.

For fixed $t$, let $B$ have the binomial distribution with parameters $m,t$. The Bernstein polynomial is $f_m(t)=\mathbb{E}[f(B/m)]$. Continuity on the compact interval gives uniform continuity and a finite bound $M=\Vert f{\Vert }_{\infty }$.

Given $ϵ>0$, choose $\delta >0$ so $|f(s)-f(t)|<ϵ$ whenever $|s-t|<\delta$. Since $\mathbb{E}[B/m]=t$ and $\mathrm{Var}(B/m)=t(1-t)/m\le 1/(4m)$, Chebyshev's inequality givesThe bound is independent of $t$, including both endpoints. Taking the supremum and then letting $m$ tend to infinity proves $f_m\to f$ uniformly on $[0,1]$. This provides the required polynomial-approximation theorem with its proof.

For each node define the Lagrange interpolation polynomialThe nodes are distinct, so this is well defined, has degree $n-1$, and satisfies $L_j(x_{k,n})={\delta }_{jk}$. Its square has degree $2n-2$, within the assumed exactness range. ThereforeThe strict inequality follows because $L_j$ is not the zero polynomial. Exactness for the constant polynomial also gives ${\sum }_{j=1}^n{a}_{j,n}=1$. Thus the quadrature is a positive weighted average; the requested nonnegativity follows in stronger form.

Choose a polynomial $P$ with $\Vert f-P{\Vert }_{\infty }<ϵ$, using part (i). For all sufficiently large $n$, its degree is at most $2n-1$, so $I_n(P)={\int }_0^{1}P$. Positive weights summing to one give $|I_n(f-P)|\le ϵ$, and $|{\int }_0^1(f-P)|\le ϵ$. HenceSince $ϵ$ is arbitrary, the quadratures converge to the integral of every continuous $f$. The positivity established in part (ii), not just polynomial exactness on its own, controls the approximation error uniformly in $n$.

The diffusive flux is $\mathbf{J}=-D\nabla C$ by Fick's first law. Conservation in any fixed control volume $V$ gives $\frac{d}{dt}{\int }_{V}C=-{\int }_{\partial V}\mathbf{J}\cdot \mathbf{n}+{\int }_{V}F$. The divergence theorem and arbitrariness of $V$ yieldAfter the initial point injection, $F=0$ and $D=kC$ with $k>0$, so radial symmetry gives $C_t=k{r}^{-2}(r^{2}C{C}_r{)}_r$. Total mass is $4\pi {\int }_0^{\infty }{r}^{2}Cdr=4\pi M$.

A spreading radius $R$ and typical concentration obey $C{R}^3\sim M$ and $R^2\sim kCt$. Thus $R^5\sim Mkt$. In the proposed ansatz this givesDirect substitution, including the time derivative of $\xi$, givesMultiplying by ${\xi }^2$ makes the left side $-({\xi }^{3}f{)}^{\prime }/5$. Regularity and zero radial flux at the origin set the integration constant to zero, so $f{f}^{\prime }=-\xi f/5$. On the positive region $f^{\prime }=-\xi /5$. Nonnegativity, compact support and continuity at its edge giveIts normalization is $1={\int }_0^{{\xi }_0}{\xi }^{2}fd\xi ={\xi }_0^5/75$. Therefore the spherical nonlinear-diffusion source profile isThe concentration and flux vanish at the free boundary, so there is no distributional flux jump despite the derivative kink. The solution satisfies the diffusion equation weakly across that boundary and classically inside its positive region. Its support shrinks to the origin as $t\downarrow 0$, and testing against a continuous compactly supported function gives the limit $4\pi M$ times its value at zero. This verifies the prescribed point injection as well as the mass normalization.

For the Laplace-type contour ansatz, differentiating under the integral and putting $P(t)=t^2-t-2$ gives the differential-equation residualIntegration by parts turns this into $[e^{zt}Pf{]}_{\partial \gamma }+{\int }_{\gamma }{e}^{zt}[(2-t)f-(Pf{)}^{\prime }]dt$. Thus it vanishes whenever the endpoint contribution vanishes and $(Pf{)}^{\prime }=(2-t)f$. The latter condition is $P{f}^{\prime }=3(1-t)f$, solved byup to a constant multiplier. Closed contours avoiding the poles make the endpoint contribution zero.

Choose separate positively oriented small circles around 2 and $-1$. The residue theorem gives, after division by $2\pi i$,ThereforeThe Wronskian of the two basis functions is $-9z{e}^z$, nonzero for $z\ne 0$. They span the two-dimensional solution space on every regular domain, and both extend entire across the apparent singular point at zero. Their initial values satisfy $w(0)=A+B$ and $w^{\prime }(0)=2(A+B)$. Thus every solution has $w^{\prime }(0)=2w(0)$, and the requested combination $w(0)=0$, $w^{\prime }(0)\ne 0$ is impossible. No additional logarithmic solution is lost: the two independent functions already span the solution space away from zero.

Hamilton's equations generate the phase-space vector field $(H_{p_1},\dots ,H_{p_n},-H_{q_1},\dots ,-H_{q_n})$. Its divergence vanishes:by equality of mixed derivatives. The Jacobian $J(t)$ of the flow obeys $\dot{J}=(\mathrm{div}V)J=0$ with $J(0)=1$. Consequently $d\tau =d{q}_1\cdots d{q}_{n}d{p}_1\cdots d{p}_n$ is preserved for every time on which the smooth flow exists. This proves Liouville theorem in Hamiltonian mechanics directly; explicit time dependence of the Hamiltonian would not change the divergence argument.

For a time-independent function $F$, Hamilton's equations giveThus a constant of motion has $[F,H]=0$. For $F={\sum }_k{p}_k$ and the given Hamiltonian, $[F,H]=-{\sum }_k{V}_{q_k}$. Simultaneous translation $q_k\mapsto q_k+s$ leaves every difference $q_i-q_j$, and hence $V$, unchanged. Differentiation with respect to $s$ gives ${\sum }_k{V}_{q_k}=0$, so $[F,H]=0$ also in this example. This is conservation of total momentum from translational symmetry.

A canonical transformation preserves the symplectic form and therefore the form of Hamilton's equations for arbitrary Hamiltonians. In one degree of freedom,Thus the transformation is canonical exactly when $[Q,P]=1$. Equivalently the chain rule gives $[F,G{]}_{q,p}=[Q,P{]}_{q,p}[F,G{]}_{Q,P}$ for arbitrary transformed functions, so the bracket and Hamiltonian dynamics are preserved precisely with this factor one. The condition also makes the Jacobian nonzero and hence supplies local invertibility; a global canonical change of variables additionally requires the stipulated global bijectivity on its domain.

A deduction from premises $\Gamma$ in propositional logic is a finite sequence of formulas, each an allowed logical axiom instance, a member of $\Gamma$, or obtained from earlier formulas by an inference rule, such as modus ponens. For example a Hilbert system can use the schemas $p\to (q\to p)$, $[p\to (q\to r)]\to [(p\to q)\to (p\to r)]$, and $(\neg q\to \neg p)\to (p\to q)$, with modus ponens as its rule. Its last formula is the conclusion. A finite deduction uses only finitely many premises. The completeness theorem for propositional logic says $\Gamma \models v$ implies $\Gamma ⊢v$; together with soundness, syntactic consistency is equivalent to existence of a satisfying valuation.

For compactness, if $\Gamma$ had no model, completeness would give a deduction of a contradiction. That deduction uses a finite subset of $\Gamma$, which therefore also has no model. Contrapositively, if every finite subset has a model, so does $\Gamma$.

Fix the given valuation on $Q$ and let $L_v$ be its collection of true literals. Suppose $\{s\}\cup L_v$ were inconsistent. A finite subset of its literals, with conjunction $\ell \in L(Q)$, would satisfy $s⊢\neg \ell$ by the deduction theorem and classical logic. But $\neg \ell \in U$ whereas the given valuation makes $\ell$ true, contradicting that it satisfies all of $U$. Hence $\{s\}\cup L_v$ is consistent.

If $U\cup \{\neg t\}$ had a model, restrict that valuation to $Q$ and apply the preceding consistency result. Completeness supplies an extension to $P\cup Q$ making $s$ true and preserving those $Q$ values. Combine its $P$ values with the original model's $Q\cup R$ values. Disjointness of $P,Q,R$ makes this a well-defined valuation satisfying $s$ and $\neg t$, contradicting soundness of $s⊢t$. Thus $U\cup \{\neg t\}$ is inconsistent.

Compactness now supplies $u_1,\dots ,u_k\in U$ with $u_1,\dots ,u_k⊢t$. Put $u={\bigwedge }_j{u}_j$. ThenThis proves propositional interpolation. The empty conjunction is the truth constant; if the language omits truth and falsehood symbols, use an equivalent tautology when $Q$ is nonempty, or adjoin those constants for the empty-common-language case.

Let $R(s,t)$ be the two-colour clique Ramsey number. At a vertex of $K_{R(s-1,t)+R(s,t-1)}$, either its red neighbourhood has at least $R(s-1,t)$ vertices or its blue neighbourhood has at least $R(s,t-1)$. In the first case a red $(s-1)$-clique extends through the vertex, or there is a blue $t$-clique; the second case is symmetric. HenceWith $R(1,t)=R(s,1)=1$, induction and Pascal's identity give $R(s,t)\le \left(\genfrac{}{}{0ex}{}{s+t-2}{s-1}\right)$. Taking $s=t$ provesThe four-vertex graph is the paw graph, a triangle with a pendant edge; the copy need not be induced. For the lower bound, colour two disjoint triangles in $K_6$ red and all edges between them blue. Red components have only three vertices and blue is bipartite, so neither colour contains a paw. Thus $R(H)\ge 7$.

For the upper bound, $R(3,3)\le 6$ guarantees a monochromatic triangle in $K_7$, say red on a vertex set $T$. A red edge from $T$ to any outside vertex immediately supplies a red paw. Otherwise all edges from $T$ to the remaining four vertices are blue. If those four vertices contain a blue edge $ab$, then $a,b$ and any vertex of $T$ form a blue triangle, and another vertex of $T$ joined to $a$ supplies its pendant edge. If they contain no blue edge, they are a red $K_4$, which contains a red paw. Therefore

An automorphism sends the primitive root of unity ${\xi }_n$ to another primitive root of unity, hence uniquely to ${\xi }_n^a$ with $a\in (\mathbb{Z}/n\mathbb{Z}{)}^{\ast }$. Define $\chi (\sigma )=a$. Composition multiplies exponents, and an automorphism with exponent one fixes the generator and thus all of $L$. Therefore $\chi$ is an injective group homomorphism and $G$ is abelian.

The extension is finite Galois: it is the splitting field of $X^n-1$, whose roots are all present and distinct. Existence of a root of order $n$ in positive characteristic implies that the characteristic does not divide $n$. The Fundamental theorem of Galois theory identifies an intermediate field with $L^H$. Since $G$ is abelian, $H$ is normal; the normal-subgroup criterion gives $L^H/K$ Galois, with group $G/H$.

For a non-surjective nontrivial example, take $K=\mathbb{Q}(i)$ and $n=8$. Its cyclotomic extension has degree two, and its image is $\{1,5\}$ inside the four-element group $(\mathbb{Z}/8\mathbb{Z}{)}^{\ast }$. For $K=\mathbb{Q}$ and prime $p$, ${\Phi }_p(X)=1+\cdots +X^{p-1}$ is irreducible: ${\Phi }_p(X+1)$ is Eisenstein at $p$. Thus the extension has degree $p-1$, equal to the unit group's order, proving surjectivity.

For $n=7$, let $\zeta ={\xi }_7$. The cyclic Galois group of order six has exactly one subgroup of each order dividing six, so there are exactly four fields:Here ${\sigma }_a(\zeta )={\zeta }^a$. To identify the quadratic field without guessing, put $A=\zeta +{\zeta }^2+{\zeta }^4$ and $B={\zeta }^3+{\zeta }^5+{\zeta }^6$. Multiplication gives $A+B=-1$ and $AB=2$, so $(A-B{)}^2=-7$. The order-three subgroup fixes $A$, and $A-B\notin \mathbb{Q}$, proving the quadratic identification. The real field is cubic because complex conjugation is ${\sigma }_6$, with fixed subgroup of order two.

For a complex irreducible representation, every central element acts as a scalar by Schur lemma. If the representation is faithful, these scalars embed $Z$ in ${\mathbb{C}}^{\ast }$. Every finite subgroup of ${\mathbb{C}}^{\ast }$ is cyclic: if its order is $m$, all its elements satisfy $z^m=1$, and the $m$th roots of unity form a cyclic group. Thus $Z$ is cyclic.

The trivial group satisfies the conclusion immediately. Otherwise let $G$ be a finite $p$-group with cyclic centre, and decompose a faithful representation as a direct sum of irreducibles using Maschke's theorem. The nontrivial cyclic $p$-group $Z$ has a unique subgroup $C$ of order $p$. Some component does not kill $C$, since otherwise the whole representation would kill it. If that component had a nontrivial kernel $N$, the kernel would be a normal subgroup and the allowed intersection theorem gives $N\cap Z\ne 1$. Every nontrivial subgroup of cyclic $Z$ contains $C$, contradicting the choice of component. That irreducible component is faithful.

Finally the regular representation is faithful, since left multiplication by a nonidentity element does not fix the basis vector at the identity. A cyclic centre therefore supplies a faithful irreducible component of it. The first argument proves the converse, giving a finite $p$-group has a faithful irreducible complex representation exactly when its centre is cyclic.

Since $10\equiv 2(\mathrm{mod}4)$, the ring of integers is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{10}]$, with field norm $N(x+y\sqrt{10})=x^2-10y^2$. Norms $\pm 2$ and $\pm 3$ are impossible: modulo five they would require a square congruent to 2 or 3. An algebraic integer is a unit exactly when its norm is $\pm 1$.

If 2 factored into two nonunits, their absolute norms would both be 2, since $N(2)=4$; this is impossible. The same argument for $N(3)=9$ excludes a factorization of 3. Finally $ϵ+1=4+\sqrt{10}$ has norm 6, so a nonunit factorization would require absolute norms 2 and 3, also impossible. Hence 2, 3 and $ϵ+1$ are irreducible. The conjugate $\overline{ϵ}+1$ is irreducible by the same argument.

The equality $6=2\cdot 3=(4+\sqrt{10})(4-\sqrt{10})$ gives inequivalent factorizations into irreducibles: no norm-six factor is associated to either 2 or 3, whose norms are 4 and 9. A principal ideal domain is a unique factorization domain. Consequently ${\mathcal{O}}_K$ is not a principal ideal domain.

Write $r=\sqrt{10}$. The ideals are ${\mathfrak{p}}_2=(2,r)$, ${\mathfrak{p}}_3=(3,r+1)$ and ${\overline{\mathfrak{p}}}_3=(3,r-1)$, with residue fields of orders 2, 3 and 3 respectively. Thus their ideal norms are 2, 3 and 3.

The generators of ${\mathfrak{p}}_2^2$ are $4,2r,10$. They are multiples of 2, and the integer combination of 4 and 10 giving 2 proves ${\mathfrak{p}}_2^2=(2)$. Similarly ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ is generated by $9,3r+3,3r-3,r^2-1=9$. The difference of the middle generators is 6; together with 9 this generates 3. Therefore the product is $(3)$.

The ideals ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ are coprime, so their product equals their intersection. The element $4+r$ belongs to both, hence $(4+r)\subseteq {\mathfrak{p}}_2{\mathfrak{p}}_3$. Both ideals have norm 6, so equality holds. Likewise $2+r$ belongs to ${\mathfrak{p}}_2$ and ${\overline{\mathfrak{p}}}_3$, and its norm is $-6$. ThusThe discriminant is 40. The Minkowski bound for ideal classes in a real quadratic field is $\sqrt{40}/2=\sqrt{10}<4$, so every ideal class has an integral representative of norm 1, 2 or 3. The only possible prime ideals with those norms are the ones just listed. Their classes satisfy $[{\mathfrak{p}}_2{]}^2=1$ and $[{\mathfrak{p}}_3]=[{\overline{\mathfrak{p}}}_3]=[{\mathfrak{p}}_2]$. Thus there are at most two classes. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm $\pm 2$, already ruled out. Therefore the class number is exactly two.

The element $ϵ=3+\sqrt{10}>1$ has norm $-1$. To prove that it is a fundamental unit, let $u=x+y\sqrt{10}$ be a positive real unit with $1<u<ϵ$. Its conjugate is $\pm 1/u$, so $y=(u-\overline{u})/(2\sqrt{10})>0$. For norm $-1$, the increasing function $u+1/u$ gives $y<(ϵ+1/ϵ)/(2\sqrt{10})=1$; for norm $+1$, $y<(ϵ-1/ϵ)/(2\sqrt{10})=3/\sqrt{10}<1$. Either contradicts integer $y$. There is no positive unit between one and $ϵ$. For any positive unit, multiply by a suitable integer power of $ϵ$ to place it in $[1,ϵ)$; it must then equal one. HenceTo classify the norm-six elements, a principal ideal generated by such an element has norm 6. Unique prime-ideal factorization and part (b) force it to be either ${\mathfrak{p}}_2{\mathfrak{p}}_3=(ϵ+1)$ or ${\mathfrak{p}}_2{\overline{\mathfrak{p}}}_3=(ϵ-1)$. Therefore its generator is an associate of one of these two elements. Since their norms are respectively 6 and $-6$, the unit multiplying the first must have even exponent and the unit multiplying the second must have odd exponent. Combining the two cases gives all integer solutions:Conversely every displayed element is integral, since $ϵ$ is a unit, and has norm 6. This proves completeness of the norm-six equation in the quadratic field of square root ten.