Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2

For an integral binary quadratic form $[a,b,c]=a{x}^2+bxy+c{y}^2$ with $a>0$ and $D=b^2-4ac<0$, the reduced positive definite binary quadratic form convention is $|b|\le a\le c$, with $b\ge 0$ on either boundary $|b|=a$ or $a=c$. Since $|D|=4ac-b^2\ge 3a^2$, the enumeration of reduced binary quadratic forms for $D=-23$ has $a\le \sqrt{23/3}<3$. The integer $b$ is odd. For $a=1$, only $b=1$ is retained and $c=6$; for $a=2$, both $b=\pm 1$ give $c=3$. Thus the complete list isA proper representation of a prime uses a primitive representing vector. Completing that vector to a matrix in ${\mathrm{SL}}_2(\mathbb{Z})$ transforms a representing form to $[p,b,c]$, so its discriminant of a binary quadratic form gives $b^2\equiv -23(\mathrm{mod}p)$. Conversely, choose an odd $b$ satisfying this congruence and put $c=(b^2+23)/(4p)$. This is integral, and $[p,b,c]$ is positive definite and represents $p$ at $(1,0)$. This proves the discriminant criterion for prime representation by a binary quadratic form here.

For $p\ne 23$, quadratic reciprocity gives $(-23/p)=(p/23)$. Including the ramified prime, the answer is thereforeFor example $[23,23,6]$ supplies a proper representation of $23$. Reduction transfers every such representation to one of the three displayed forms.

The planar Brouwer fixed-point theorem says that every continuous map from the closed disk $D$ to itself has a fixed point. If a retraction $r:D\to \partial D$ existed, $x\mapsto -r(x)$ would be a continuous self-map of $D$. At a fixed point $x=-r(x)$, $x$ lies on the boundary, where $r(x)=x$; this would give $x=-x$, impossible on the unit circle.

Conversely, suppose a continuous self-map $f:D\to D$ has no fixed point. For each $x$, take the ray starting at $f(x)$ through $x$, and let $r(x)$ be its exit point on $\partial D$. The denominator $|x-f(x)|$ never vanishes, and solving the quadratic intersection with the circle shows that the exit point depends continuously on $x$. More explicitly, with $v=x-f(x)$,The ray exits beyond or at $x$, so $\lambda \ge 1$, and when $x\in \partial D$ its exit point is $x$. Hence $r$ is a retraction. Nonexistence of a boundary retraction and the planar fixed-point theorem are equivalent.

Define the continuous map $T(z)=-(z^6-z^5+2z^2+1)/6$. For $|z|\le 1$, the triangle inequality gives $|T(z)|\le (1+1+2+1)/6=5/6$. Thus $T$ maps the closed unit disk into itself. By the Brouwer fixed-point theorem, some $z$ has $T(z)=z$, which rearranges toThis supplies the requested existence and even places a root strictly inside the unit disk.

Every discrete additive subgroup of ${\mathbb{R}}^2$ has the form $\mathbb{Z}{v}_1+\cdots +\mathbb{Z}{v}_r$ for linearly independent vectors, with $r=0,1,$ or $2$. A full translation lattice has $r=2$, and conversely every such integer span is a discrete cocompact subgroup. This classifies planar lattices up to a choice of basis.

A frieze group is a discrete group of planar Euclidean isometries whose translation subgroup is infinite cyclic and has finite index. Consider the glide reflection $g(x,y)=(x+1,-y)$. Its square is translation by $(2,0)$, whereas every odd power reverses orientation. Therefore $⟨g⟩\cong \mathbb{Z}$, but none of its translation elements generates the whole group: they form the index-two subgroup $⟨g^2⟩$.

To realize exactly this frieze group, repeat a small scalene triangular motif at the points $(n,(-1{)}^n)$, reflecting the motif when $n$ is odd. Choose the motif with three distinct side lengths and no internal symmetry. The displayed pattern has glide reflection symmetry $g$ and translation period $2$, but no pure horizontal reflection or vertical mirror. Distinct side lengths force any symmetry to preserve corresponding vertices, which rules out extra motif permutations; the sequence of heights then leaves exactly the powers of $g$.

For Shannon-Fano coding, order the symbols by decreasing probability and recursively split each current list into two contiguous groups with total probabilities as close as possible. Label the two branches $0$ and $1$; the path to a symbol is its prefix code word. Here the first split is $\{a\}$ of weight $.4$ against $\{b,c,d\}$ of weight $.6$. The latter splits into $\{b\}$ and $\{c,d\}$, each of weight $.3$. One answer isFor Huffman coding, repeatedly merge the two least-probable nodes and then read the resulting binary tree from the root. Merge $c,d$ to weight $.3$, merge this node with $b$ to weight $.6$, and merge with $a$. The same displayed prefix code results. Its mean length is $.4+2(.3)+3(.2)+3(.1)=\boxed{1.9\text{ bits per symbol}}$. Choices at ties and exchange of binary labels yield other equivalent codes. Huffman coding minimizes mean length among binary prefix codes; the top-down Shannon-Fano construction does not have that general optimality guarantee.

Under independent normal distributions, maximum likelihood estimation minimizes $(y_A-\alpha {)}^2+(y_B-\beta {)}^2+(y_C-\gamma {)}^2$ subject to $\alpha +\beta +\gamma =\pi$. A Lagrange multiplier gives the same correction at every corner. Writing $s=y_A+y_B+y_C$, the interior estimates areThe result is unchanged if the common variance is unknown, since profiling it still minimizes the residual sum of squares. If physical positivity of the three angles is imposed and this projection gives a negative component, the constrained estimate over the closed simplex is instead ${\hat{\theta }}_i=\max (y_i-\lambda ,0)$, with $\lambda$ chosen to make the components sum to $\pi$. If only strictly nondegenerate triangles are admitted, a boundary optimum is merely an unattained likelihood supremum.

The normal-error model is a useful small-error approximation, but its unbounded tails assign positive probability to impossible measured angles. Periodicity or a measurement convention may matter for large errors, and equal variance and independence require experimental justification. The exact sum constraint on the true angles does not itself invalidate independence of three raw measurement errors. After imposing that constraint, however, the fitted-angle errors are correlated: the covariance of the interior estimator is ${\sigma }^2(I-11^T/3)$.

The Ricker population map has equilibria $N=0$ and $N=K$. Since $f^{\prime }(N)=e^{r(1-N/K)}(1-rN/K)$, their linearization multipliers areThus extinction is unstable and the positive equilibrium is locally asymptotically stable for $0<r<2$, becoming unstable for $r>2$.

To verify an actual period-doubling bifurcation, rather than only a multiplier crossing, put $N/K=1+x$ and $r=2+\epsilon$. The centered map isIts second iterate minus $x$ is $2\epsilon x-\frac{4}{3}{x}^3$ to leading order. After rescaling $x=\sqrt{\epsilon }y$, the limiting nonzero roots are simple, so the implicit function theorem continues them. Besides $x=0$, two nearby points therefore appear for $\epsilon >0$, with $x_{\pm }=\pm \sqrt{3\epsilon /2}+O(\epsilon )$. The map exchanges these points, and the derivative of its second iterate there is $1-4\epsilon +O({\epsilon }^{3/2})$, so the emerging two-cycle is attracting. The bifurcation at $r=2$ is supercritical period doubling.

For nonnegative populations, $f$ increases up to $N=K/r$ and decreases thereafter to zero. Hence every update after the initial time satisfiesThe bound is attained by choosing $N_0=K/r$; it is not a bound on an arbitrary initial population.

At a stationary bifurcation, a real eigenvalue of the equilibrium linearization crosses zero; the critical motion has zero frequency. At a Hopf bifurcation, a conjugate pair crosses the imaginary axis at $\pm i\omega$ with $\omega >0$, permitting a nearby limit cycle.

The Hopf normal form has a small cycle when $R^2=\mu /a+O({\mu }^2)>0$. Its radial multiplier exponent is ${\partial }_r\dot{r}{|}_R=-2\mu +O({\mu }^2)$, while the origin's radial exponent is $\mu$. Thus: for $\mu <0<a$, there is an attracting spiral and no small cycle; for $0<\mu ,a$, a repelling spiral is surrounded by an attracting cycle; for $\mu ,a<0$, an attracting spiral is surrounded by a repelling cycle; for $a<0<\mu$, there is a repelling spiral and no small cycle. Outside the repelling cycle in the third case the local radial flow points outward; a cubic normal form makes no global prediction about where it goes.

The subcritical bifurcation has $a<0$ and the unstable radius branch $R=\sqrt{\mu /a}+O(|\mu {|}^{3/2})$ on $\mu <0$, meeting the origin at $\mu =0$. On either existing branch, the angular speed is $\omega +(c-b/a)\mu +O({\mu }^2)$, so

For the beta function, split $1=t+(1-t)$ inside the integral to obtain $B(z,q)=B(z,q+1)+B(z+1,q)$. Integrating the derivative of $t^q(1-t{)}^z$ over $(0,1)$ gives $qB(z+1,q)=zB(z,q+1)$, since both endpoint terms vanish in the initial domain. Combining these identities yieldsRepeatedly apply this recurrence until $\mathrm{Re}(z+n)>0$. Each resulting expression agrees with the original integral on overlap, so it gives its unique analytic continuation. Possible singularities are only $z=0,-1,-2,\dots$.

Near $z=0$, $zB(z,q)\to qB(1,q)=1$. Continuing the recurrence downwards giveswhere the empty product is $1$. The recurrence shows that each singularity is at most a simple pole. If $q$ is not a positive integer, all these residues are nonzero and all the listed points are poles. If $q=m$ is a positive integer, poles occur only for $0\le n<m$; the later candidates are removable, as is also clear from $B(z,m)=(m-1)!/[z(z+1)\cdots (z+m-1)]$. This qualification matters when claiming a pole at every nonpositive integer.

Varying the Lagrangian gives, with the cyclic angle conventions,All successive angular gaps are $2\pi /N$ in equilibrium, so their two contributions cancel. Because the potential depends only on gaps, a uniform change of all angles costs no energy. Its Hessian therefore annihilates $(1,\dots ,1)$, giving a zero-frequency normal mode. The corresponding motion is rigid rotation, ${\theta }_i(t)={\theta }_i^{(0)}+A+Bt$, including a constant common angular shift.

For two particles write ${\theta }_2-{\theta }_1=\pi +{\delta }_2-{\delta }_1$. Both cyclic gaps contribute, so the quadratic potential is $k({\delta }_2-{\delta }_1{)}^2$. With the specified masses, the mass matrix is $km\mathrm{diag}(1,2)$ and the stiffness matrix is $2k\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$. The generalized eigenvalue problem givesThe finite-frequency normal mode keeps the mass-weighted mean angle fixed and changes the angular separation sinusoidally. Together with the two rigid-rotation constants, it spans all small motions about equilibrium.

For a regular spherical star, mass conservation gives $m^{\prime }=4\pi r^2\rho$. Multiplying the hydrostatic equilibrium equation by $r^2/\rho$ and differentiating therefore yieldsAt the center, finite density implies $m(r)=O(r^3)$, so a regular pressure satisfies $P(0)=P_c<\infty$ and $P^{\prime }(0)=0$. At a free outer boundary in vacuum, $P(R)=0$; a surrounding medium would instead prescribe its external pressure.

Differentiate the auxiliary function, using both equations:where enclosed mass is nonzero. Since $m^2/r^4\to 0$ at the regular center, $F(0)=P_c$, whereas $F(R)=G{M}^2/(8\pi R^4)$. Integrating the strictly negative derivative givesWith nonzero external pressure, the corresponding bound is $P_c>P(R)+G{M}^2/(8\pi R^4)$.

Let $1$ be the constant function and define $h=L1\in C([0,1])$. Positivity gives $h\ge 0$. Fix any $x$ and any continuous $f$. The function $g=f-f(x)1$ vanishes at $x$, so the zero-preserving assumption gives $Lg(x)=0$. By linearity this is $Lf(x)-f(x)L1(x)=0$. Since $x$ was arbitrary,This proves the zero-preserving linear operator on continuous functions representation. Zero preservation and linearity determine the multiplication operator; positivity is needed only for the multiplier's sign. The three numbered conditions in the source are hypotheses of this one request, rather than three separate subquestions.

In the Rabin cryptosystem, choose distinct large odd primes $p,q$, publish $N=pq$, and encrypt $m$ as $c=m^2\mathrm{mod}N$. Usually $p,q\equiv 3(\mathrm{mod}4)$, so a square root modulo each prime is obtained by exponentiating to $(p+1)/4$ or $(q+1)/4$. Combine the two independent choices of signs by the Chinese remainder theorem to obtain the four roots modulo $N$ for a message coprime to $N$. Redundancy or a specified encoding distinguishes the intended message. Factoring $N$ makes decryption easy; a general square-root oracle conversely factors $N$ by obtaining two roots not congruent up to sign and taking their greatest common divisor with $N$.

For the Repeated-message attack on the Rabin cryptosystem, first compute $d=\gcd (N,N^{\prime })$. If $d=1$, use the two ciphertexts to reconstruct $s\in [0,N{N}^{\prime })$ with $s\equiv m^2(\mathrm{mod}N)$ and $s\equiv m^2(\mathrm{mod}{N}^{\prime })$. Since $0<m<\min (N,N^{\prime })$, we have $m^2<N{N}^{\prime }$, so this reconstruction is the ordinary integer $m^2$. ThereforeIf distinct semiprime moduli have $d>1$, they share a prime; the gcd instead exposes their factors, allowing ordinary Rabin decryption with the usual encoding check. The two cases cover the stated protocol with distinct standard moduli.

In the coprime case, learning this message does not normally reveal the factors or a decryption key, so other messages encrypted only once remain protected by the square-root problem. Exceptional messages with $\gcd (m,N)>1$ do reveal a factor, and the shared-prime-modulus case compromises other messages too. Repetition under fresh coprime moduli is the vulnerability used here, rather than a universal attack on every ciphertext.

For spatially homogeneous solutions, the bistable cubic reaction-diffusion equation reduces to $\dot{u}=f(u)=u(1-u)(u-r)$. Its equilibria are $0,r,1$, andThus $0$ and $1$ are asymptotically stable, while $r$ is unstable. Small spatial Fourier perturbations have growth rates $f^{\prime }(u_{\ast })-m{k}^2$, so diffusion also leaves the two outer uniform states linearly stable.

At $r=1/2$, write $F(u)=-u^2(1-u{)}^2/4$, so $F^{\prime }=f$. Multiplying the stationary equation $m{U}^{\prime \prime }+F^{\prime }(U)=0$ by $U^{\prime }$ gives $m(U^{\prime }{)}^2/2+F(U)=C$. The endpoint values and vanishing endpoint derivatives make $C=0$. For the decreasing connection,Separate variables and absorb the integration constant into a translation $x_0$. The logistic front of a bistable cubic equation isIt approaches $1$ on the left and $0$ on the right and satisfies the stationary differential equation directly.

A travelling wave $u(x,t)=U(x-vt)$ obeys $m{U}^{\prime \prime }+v{U}^{\prime }+f(U)=0$, so the mechanical potential in the requested form is $V=F$, up to an additive constant. HereIts stationary points are maxima at $0,1$ and a minimum at $r$. Since $F(0)=0$ and $F(1)=(1-2r)/12$, the two maxima have equal height at $r=1/2$; the maximum at $1$ is lower for $r>1/2$ and higher for $r<1/2$.

Multiply the wave equation by $U^{\prime }$ and integrate along the heteroclinic orbit. The endpoint derivatives vanish, and $U(-\infty )=1$, $U(\infty )=0$, soConsequentlyThe integral is positive for a nonconstant front. In fact substitution of the same logistic shape from part (b) gives the exact velocity $v=\sqrt{m/2}(1-2r)$. Thus the $u=1$ phase invades rightwards when $r<1/2$, the zero phase invades leftwards when $r>1/2$, and the balanced front is stationary.

Parametrize the continued branch by $t=r{e}^{i\theta }$, $0\le \theta \le 2\pi$. Then $t^{z-1}dt=i{r}^z{e}^{iz\theta }d\theta$, givingThe apparent singularity at zero is removable, so $f$ is an entire function. The factor $r^z=e^{z\log r}$ shows its dependence on the contour radius; for instance $f(1/2)=-4\sqrt{r}$. The integrand is continued along the contour and generally returns on a different branch, so this is not an ordinary integral of a single-valued holomorphic function around a closed loop.

There is a branch-phase discrepancy in the printed identity. Set $K(z)={\int }_{-1}^1{e}^t(1-t^2{)}^{z}dt$ using the positive real base. Shrink the anticlockwise loop at $1$. The outward interval carries phase $e^{-i\pi z}$ and its return phase $e^{i\pi z}$; their difference gives $-2i\sin (\pi z){\int }_0^1{e}^t(1-t^2{)}^{z}dt$. The clockwise loop at $-1$ starts with phase $e^{i\pi z}$ and returns with $e^{-i\pi z}$, giving the same coefficient on $(-1,0)$. Small endpoint circles vanish for $\mathrm{Re}z>-1$. Thus the branch phase in a figure-eight analytic continuation integral isIf the printed interval integral uses the prescribed argument $-\pi$, then $I(z)=e^{-i\pi z}K(z)$, and the corrected formula is insteadFor example at $z=1/2$, $K>0$, $I=-iK$ and $J=-2iK$; the printed right side without the phase would be the real number $-2K$. Interpreting its interval integrand as $(1-t^2{)}^z$ restores the intended identity.

On the fixed contour away from its branch points, the continued logarithm is bounded and $e^t\exp [z\log (t^2-1)]$ is entire in $z$, uniformly on compact sets. Hence $J$ is an entire function. Division by the sine gives a meromorphic continuation of $K$, and multiplication by $e^{-i\pi z}$ gives the continuation of the specified $I$. The only candidate singularities are integers, at most simple. All $0,1,2,\dots$ are removable because the interval integral is already holomorphic on $\mathrm{Re}z>-1$.

The negative integers $z=-n$, $n\ge 1$, are genuine simple poles. At such a point the contour integrand is single-valued andThe first residue is $e$ times a rational number. The second is $e^{-1}$ times a nonzero rational number: expand $e^s(-2+s{)}^{-n}$ at $s=0$, whose degree-$n-1$ coefficient has sign $(-1{)}^n$ and nonzero magnitude. They cannot be equal because $e^2$ is irrational. Therefore $J(-n)\ne 0$, and the simple sine zero creates a simple pole. In particular ${\mathrm{Res}}_{z=-n}K=-J(-n)/[2\pi i(-1{)}^n]$.

A circle enclosing both branch points has total logarithmic change $4\pi i$, rather than zero as for the figure eight. For a circle based at $t=2$ with initial real logarithm, shrinking it leaves a connector from $1$ to $2$ as well as the two banks of the interval. Explicitly its integral isThus it is not just a known sine factor times $K$: an additional endpoint-singular integral remains. The figure-eight contour eliminates precisely that connector term. This explains why replacing it by the large circle does not give the same direct continuation method.

Put $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}/c=m\dot{\mathbf{q}}$. From Hamilton's equations,For the stated time-independent vector potential, differentiating ${\pi }_i$ gives $m{\ddot{q}}_i=(e/c){\dot{q}}_j({\partial }_i{A}_j-{\partial }_j{A}_i)$, orUse the Poisson bracket convention $[F,G]={\sum }_i(F_{q_i}{G}_{p_i}-F_{p_i}{G}_{q_i})$. Then direct differentiation givesFor $\mathbf{A}=(0,0,F(r))$ with $r=\sqrt{x_1^2+x_2^2}$, the Hamiltonian is independent of $x_3$, so ${\dot{p}}_3=0$. The radial force in the horizontal plane is ${\dot{p}}_r=e(p_3-eF/c)F^{\prime }/(mc)$. For circular motion the radial acceleration is $-r{\Omega }^2$, and $p_r$ is the mechanical radial momentum because the horizontal vector potential vanishes. ThusA circular orbit at that radius is possible when the right side is nonnegative; positive values give either sense of horizontal rotation, and the longitudinal velocity is the constant $(p_3-eF(r)/c)/m$ along that orbit.

The Godel completeness theorem for first-order logic states that $\Gamma \models \varphi$ implies $\Gamma ⊢\varphi$; together with soundness, semantic consequence and derivability coincide. Equivalently, every syntactically consistent first-order theory has a model. The compactness theorem states that a set of first-order sentences has a model if and only if every finite subset has a model. These statements use nonempty first-order structures.

For each $n$, let ${\sigma }_n$ assert the existence of $n$ distinct elements. If a theory $T$ has arbitrarily large finite models, every finite subset of $T\cup \{{\sigma }_n:n\ge 1\}$ has a model. By the compactness theorem, this extended theory has a model, necessarily infinite. Since fields of size $2^j$ are arbitrarily large, no first-order theory can have exactly the finite fields of characteristic two as its models.

Suppose instead a finite conjunction $\chi$ axiomatized exactly the infinite fields of characteristic two. Let $T_2$ be the field axioms with $1+1=0$. Then $T_2\cup \{\neg \chi \}$ would have exactly the finite characteristic-two fields as models, contradicting the preceding conclusion. Equivalently, compactness of $T_2\cup \{{\sigma }_n\}\models \chi$ would force some finite size threshold, but arbitrarily large finite fields still fail $\chi$. The infinite characteristic-two fields are not finitely axiomatizable.

The intended closed-term witness property for quantifier-free existential sentences needs at least one closed term. Under this hypothesis, suppose the set $\Gamma =\{\neg \varphi (t):t\text{ a closed term}\}$ were consistent. By the Godel completeness theorem, it has a model $M$. Take the substructure $M_0$ whose elements are the interpretations of closed terms. It is nonempty and closed under every function symbol. Atomic formulas and their negations have the same truth values in $M_0$ as in $M$, and induction extends this to every quantifier-free formula.

The provable sentence $\exists x\varphi (x)$ is true in $M_0$ by soundness. Its witness is $t^{M_0}$ for some closed term $t$, so $M\models \varphi (t)$, contradicting $M\models \Gamma$. Hence $\Gamma$ is inconsistent. The compactness theorem gives a finite inconsistent subset, so completeness and classical propositional reasoning yieldWithout a constant or nullary function, the printed assertion is false as written: in the equality-only language there are no closed terms, but $\exists x(x=x)$ is provable. Then $\Gamma$ is empty and consistent. One may repair the usual theorem by assuming a nonempty set of closed terms or adjoining a fresh constant; the latter proves a witness disjunction in the enlarged language, not the missing closed terms in the original one.

Here graphs are finite and simple. To prove Dirac theorem, take a longest path $v_1,\dots ,v_{\ell }$. All neighbours of either endpoint lie on the path. Among indices $1\le i<\ell$, let $S=\{i:v_1{v}_{i+1}\text{ is an edge}\}$ and $T=\{i:v_i{v}_{\ell }\text{ is an edge}\}$. Since $|S|+|T|=d(v_1)+d(v_{\ell })\ge n>\ell -1$, some $i$ lies in both. Following the path forward to $v_i$, jumping to $v_{\ell }$, and following it backwards to $v_{i+1}$ closes a cycle containing every vertex of the path. The degree condition makes the graph connected: two components would each have at least $\delta +1>n/2$ vertices. If $\ell <n$, connectedness gives an edge from an outside vertex to the cycle, and breaking the cycle there makes a longer path. This contradiction proves a Hamiltonian cycle.

For even $n=2m$, the complete bipartite graph $K_{m-1,m+1}$ has minimum degree $m-1$ and no Hamiltonian cycle. For odd $n=2m+1$, $K_{m,m+1}$ has minimum degree $m$ and likewise fails: a bipartite cycle uses equal numbers from its two parts. These examples include the smallest applicable $n$.

Adding one universal vertex gives $n+1$ vertices and minimum degree at least $(n+1)/2$ when the original minimum degree is at least $(n-1)/2$. Apply the proved theorem and remove the universal vertex from its cycle. The remaining sequence is a Hamiltonian path in the original graph.

A necessary condition for a Hamiltonian cycle is that deleting a nonempty vertex set $S$ leaves at most $|S|$ components, since deleting those vertices breaks a spanning cycle into at most that many path pieces. For any $k\ge 1$, take the connected star $G=K_{1,k+2}$. In $G_k$, deleting its center and the $k$ new vertices leaves $k+2$ components, more than $k+1$. Even two-connectivity does not suffice: take $G=K_{2,k+3}$, which is two-connected. Deleting its two-vertex part and the $k$ new vertices from $G_k$ leaves $k+3$ components, more than $k+2$. Both connectivity requirements admit counterexamples for every $k$.

If $\theta =0$, the splitting field is $K$ and its Galois group is trivial, hence cyclic. Otherwise choose $\alpha$ with ${\alpha }^n=\theta$. The roots are $\alpha {\zeta }_n^j$, so the splitting field is $K(\alpha )$ because ${\zeta }_n\in K$. The roots are distinct since the characteristic does not divide $n$, making this a Galois extension. Each automorphism satisfies $\sigma (\alpha )=c_{\sigma }\alpha$ with $c_{\sigma }\in ⟨{\zeta }_n⟩$. Because that root-of-unity group is fixed by all automorphisms, $c_{\sigma \tau }=c_{\sigma }{c}_{\tau }$. Thus $\sigma \mapsto c_{\sigma }$ is an injective homomorphism into a cyclic group. The splitting-field Galois group is cyclic, of order dividing $n$.

An automorphism of $L({\zeta }_n)$ fixing $K({\zeta }_n)$ fixes $K$. Because $L/K$ is a normal extension, it preserves $L$, so restriction gives a homomorphismIts kernel fixes both $L$ and ${\zeta }_n$, hence every element of their compositum. Therefore restriction is injective, proving the asserted subgroup identification. Normality is the key reason the restriction maps into automorphisms of $L$ rather than just embeddings into a larger field.

A primitive sixth root of unity satisfieswhich is irreducible over $\mathbb{Q}$ because its discriminant is $-3$. For the other polynomial choose $\alpha =i{3}^{1/6}$, so ${\alpha }^6=-3$ and ${\alpha }^3=-i\sqrt{3}$. Then ${\zeta }_6=(1-{\alpha }^3)/2\in \mathbb{Q}(\alpha )$, so this field already contains all six roots. The Eisenstein criterion at $3$ proves $x^6+3$ irreducible, giving a splitting field of degree six.

Put $r(\alpha )={\zeta }_6^2\alpha$ and $s(\alpha )={\zeta }_6\alpha$. The first automorphism fixes ${\zeta }_6$ and has order three. The second sends ${\alpha }^3$ to its negative, so sends ${\zeta }_6$ to ${\zeta }_6^{-1}$; hence $s^2=1$ and $srs=r^{-1}$. These six distinct transformations exhaust the Galois group. ThereforeThe presence of sixth roots of unity inside the splitting field is why this particular binomial has degree six rather than a generic twelve-degree splitting field.

There are seven irreducible characters, one for each conjugacy class. Besides the trivial character and the four supplied rows, let the remaining degrees be $d,e$. The sum of squared degrees gives $d^2+e^2=360-(1+25+64+64+100)=106$, whose positive integer solutions are $\{d,e\}=\{5,9\}$. Complex conjugation permutes irreducible characters while preserving degree; the known rows are real and each unknown degree occurs just once among the remaining characters, so the two missing characters are real.

Call their entries $x_j,y_j$, with degrees five and nine. Orthogonality with the regular character gives $5x_j+9y_j=-{\sum }_{\mathrm{known}}\chi (1)\chi (C_j)$ for $j>1$. Column norms give $x_j^2+y_j^2=360/|C_j|-{\sum }_{\mathrm{known}}\chi (C_j{)}^2$. These equations yield two possible rational pairs in some columns. For this integral character reconstruction from column orthogonality, a further constraint is essential: character values are algebraic integers, so rational candidates must be integers. For example at $C_6$, $5x+9y=-9$ and $x^2+y^2=1$ allow $(0,-1)$ and $(-45/53,-28/53)$; the second pair is not integral. The integral choices complete the table:Every nontrivial row has $\chi (g)\ne \chi (1)$ for $g\ne 1$. In a unitary realization, $\chi (g)=\chi (1)$ exactly when all eigenvalues of the representing matrix are one, so each nontrivial irreducible representation is faithful. If a nontrivial proper normal subgroup existed, a nontrivial irreducible representation of the nontrivial quotient would pull back to a nontrivial representation with that subgroup in its kernel. This contradicts faithfulness. The group is simple.