Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 3

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

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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

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SetBecause the region lies in the positive quadrant, its four inequalities becomeThus this change of variables sends $D$ to a rectangle in the $uv$-plane. Its Jacobian determinant issoSince $x^2=uv$, the double integral is

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Use $x>0$ as the parameter of the smooth curve:Its velocity vector and speed areThe unit tangent vector is therefore

The parameter-independent formula for the curvature of a space curve givesDifferentiating,Hence $\kappa$ increases for $0<x<1/\sqrt{2}$ and decreases for $x>1/\sqrt{2}$. Its global maximum occurs at $x=1/\sqrt{2}$ and equalsThe corresponding point is

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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 $p$.

Now let $|G|=33$. Cauchy's theorem gives a subgroup $H$ of order $11$. Let $H$ act by the conjugation action on the set $\mathcal{X}$ of subgroups of order $11$. It fixes $H$. If it also fixed $K\ne H$, then $H$ would normalize $K$; since $H\cap K=\{e\}$, the product $HK$ would be a subgroup of order $121$, which is impossible. Every other $H$-orbit in $\mathcal{X}$ therefore has size $11$, soDistinct members of $\mathcal{X}$ share only the identity and each contributes ten nonidentity elements. Hence $1+10|\mathcal{X}|\le 33$, forcing $|\mathcal{X}|=1$. Thus $H$ is a normal subgroup.

Conjugation now defines a group homomorphismBecause $H\cong C_{11}$, its automorphism group has order $10$. By the Lagrange theorem, the image has order dividing both $33$ and $10$, so the image is trivial and $H$ lies in the center of $G$. Cauchy's theorem also supplies $x$ of order $3$. If $h$ generates $H$, then $h$ and $x$ commute and $hx$ has order $\mathrm{lcm}(11,3)=33$. Therefore

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A Möbius transformation of the Riemann sphere is a mapwith its natural values at the pole and at infinity.

For three distinct points $z_1,z_2,z_3$, defineThe usual limiting conventions cover an infinite $z_i$. This Möbius transformation sends $(z_1,z_2,z_3)$ to $(1,0,\infty )$. Defining $T_w$ similarly, the mapsends $z_i$ to $w_i$. If two Möbius transformations do so, their quotient fixes $0,1,\infty$. A Möbius transformation fixing infinity is affine, and fixing zero and one then makes it the identity. This proves uniqueness.

With this convention, the cross-ratio isSubstitution shows that translations, nonzero scalings, and inversion preserve it; since these generate the Möbius group, every Möbius transformation preserves cross-ratios.

Conversely, suppose a bijection $f$ of the Riemann sphere preserves every cross-ratio. Let $m$ be the unique Möbius transformation agreeing with $f$ at three chosen points $z_1,z_2,z_3$. For any other $z$, preservation by $f$ and $m$ givesThe last coordinate in a cross-ratio with three fixed distinct entries is injective, so $f(z)=m(z)$. The equality already holds at the three base points, hence $f=m$ everywhere and $f$ is Möbius.

Finally, the map $z\mapsto a\overline{z}+b$ is constant when $a=0$ and therefore is not Möbius. If $a\ne 0$, it is bijective and fixes infinity. Were it Möbius, it would have the affine form $\alpha z+\beta$. Equality on real $z$ forces $\alpha =a$ and $\beta =b$, whereas equality at $z=i$ would require $ai=-ai$, contradicting $a\ne 0$. Thus

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A subgroup $N\le G$ is normal whenIts left and right cosets then agree, and multiplicationis well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the quotient group $G/N$.

For a group homomorphism $\theta :G\to H$, its kernel and image areThe kernel is normal becausewhenever $k\in \ker \theta$. Conversely, if $K⊴G$, the quotient mapis a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the mapis well defined and bijective and preserves multiplication. This is the first isomorphism theorem.

DefineEuler's formula shows that $\Phi$ is a homomorphism, its image is the complex unit circle, and its kernel is $\mathbb{Z}$. The first isomorphism theorem therefore givesThe image of $\mathbb{Q}/\mathbb{Z}$ consists exactly of the roots of unity: if $t=p/q$, then $\Phi (t{)}^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$.

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The set $S(\mathbb{N})$ consists of all bijections $\mathbb{N}\to \mathbb{N}$. The composition of two bijections is a bijection, composition is associative, the identity map is an identity, and every bijection has an inverse, so $S(\mathbb{N})$ is a group.

An element of $S_{\mathrm{fin}}(\mathbb{N})$ has finite support. The support of a composition is contained in the union of the two supports, and a permutation and its inverse have the same support. The identity has empty support. Hence $S_{\mathrm{fin}}(\mathbb{N})$ is a subgroup.

Let $\sigma$ be a cycle and choose $n$ that it moves. Because $\sigma$ has finite support, the sequencemust repeat. Since $\sigma$ is invertible, its first repetition returns to $n$; let the least positive return time be $l$. The cycle condition says that every moved point occurs in this orbit, so ${\sigma }^l$ fixes every point. No smaller positive power fixes $n$. Thereforewhich is finite.

For $\tau \in S_{\mathrm{fin}}(\mathbb{N})$, partition its finite support into the orbits of the cyclic group $⟨\tau ⟩$. On each orbit, let ${\sigma }_i$ agree with $\tau$ and fix every point outside that orbit. Then each ${\sigma }_i$ is a permutation cycle, their supports are pairwise disjoint, andDisjoint cycles commute because at every point at most one of them acts nontrivially.

Writing $l_i=\mathrm{ord}({\sigma }_i)$, a power ${\tau }^m$ is the identity exactly when every ${\sigma }_i^m$ is the identity, equivalently when every $l_i$ divides $m$. Thus

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Since $r=(x_j{x}_j{)}^{1/2}$, the chain rule givesFor the radial vector field $g(r)\mathbf{x}$, the product rule and Einstein summation convention give

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Differentiating the spherical coordinate systemgives the orthogonal decompositionwhereThus the scale factors of orthogonal coordinates are

Since the total differential satisfiescomparison of coefficients gives the gradient

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Each field has only an azimuthal component and is independent of $r$ and $\varphi$. The curl in spherical coordinates therefore reduces toand its $\varphi$ component vanishes. The half-angle identitiesshow in both cases that

Apply part (a) with $g(r)=r^{-3}$. Thenaway from the origin, explicitly confirming that each resulting curl has zero divergence.

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Part (c) givesThe half-space $x_1>0$ is a simply connected domain, so the Poincare lemma makes this curl-free field a conservative vector field: it is the gradient of a single-valued scalar potential.

Using $\tan u+\cot u=2/\sin (2u)$,The spherical-coordinate formula from part (b) shows that a potential must satisfy $f_{\varphi }=2$ and may be independent of $r,\theta$. On $x_1>0$ the azimuthal angle has the single-valued branchHence one solution, up to an additive constant, is

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On the part with $1\le z\le 2$, writeThis is the upper portion of a hyperboloid of one sheet. Choosing the outward orientation, whose normal points radially away from the $z$-axis, the vector area element isIts radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below.

Forthe curl isOn the surface, $z^2={\rho }^2-1$, and thereforeDirect integration gives

To verify this with the Stokes theorem, note that on a circle of fixed $\rho ,z$,The induced boundary orientation runs in the negative $\varphi$ direction on the top circle and the positive $\varphi$ direction on the bottom circle. Hence

For $S^{\prime }$, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive $\varphi$ direction at $z=-1$, where ${\rho }^2=2$, and negative $\varphi$ direction at $z=\sqrt{2}$, where ${\rho }^2=3$. Thus

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Let ${\varphi }_1,{\varphi }_2$ be two solutions with the same Neumann boundary condition, and set $u={\varphi }_1-{\varphi }_2$. ThenMultiplying by $u$ and applying Green's first identity givesThe integrand is nonnegative. If $m>0$, both terms can vanish only when $u=0$, so the solution of the modified Helmholtz equation is unique.

If $m=0$, the identity only forces $\nabla u=0$. On a connected region, $u$ may be any constant. Thus solutions of the Laplace equation with prescribed normal derivative are unique only up to an additive constant, so

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Put $\eta =\psi -\varphi$. The common Dirichlet boundary condition gives $\eta =0$ on $S$. Expanding the energy functional,Integration by parts and the field equation make the cross term zero:Consequently

For $m>0$, equality holds exactly when $\eta =0$. For $m=0$, equality holds exactly when $\nabla \eta =0$, so $\eta$ is constant on the connected region; its zero boundary value then again forces $\eta =0$. The minimizing property therefore proves uniqueness for the Dirichlet problem in both cases:

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Let a change of orthonormal coordinates be represented by a rotation matrix $R$. Since both angular momentum and angular velocity are vectors,Using $L_i=I_{ij}{\omega }_j$ in both frames givesfor every vector $\mathbf{\omega }$. Multiplication by $R^{-1}=R^T$ yieldswhich is exactly the transformation law for a rank-two tensor. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf{r}=\mathbf{x}-\mathbf{a}$. The vector triple product givesThereforeAt the centre of mass,Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho {\int }_{B}dV=M$. Hence the parallel axis theorem in tensor form is

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Reflection symmetry of the cube in each coordinate plane makes every off-diagonal integralvanish because its integrand is odd in one coordinate. Permuting the three coordinate axes leaves the cube unchanged, so its three diagonal moments are equal. Thus its inertia tensor at the centre has the isotropic form

Forpart (a) and $\lambda =M{\ell }^2/6$ giveThe displayed symmetric matrix has orthogonal eigenvectorswith corresponding eigenvaluesFor a unit angular velocity, $|\mathbf{L}|=|I\mathbf{\omega }|$ is minimized and maximized along eigenvectors for the smallest and largest eigenvalues. Their ratio is

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