Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 3

Suppose every group commutator belongs to $H$. For $h\in H$ and $g\in G$, the particular group commutator $h^{-1}{g}^{-1}hg$ belongs to $H$. Multiplying on the left by $h$ gives $g^{-1}hg\in H$. Thus $g^{-1}Hg\subseteq H$ for every $g$; applying the same inclusion to $g^{-1}$ gives the reverse inclusion. Therefore $H$ is a normal subgroup.

We can now form the quotient group. Its group commutators areA group commutator is the identity element exactly when the two elements commute. Hence $G/H$ is an abelian group. Equivalently, the hypothesis says that $H$ contains the commutator subgroup, so the quotient factors through the abelianization.

Conversely, assume $H$ is a normal subgroup and the quotient group $G/H$ is an abelian group. Commuting the two cosets $aH$ and $bH$ givesso every group commutator $a^{-1}{b}^{-1}ab$ belongs to $H$. This proves the converse and completes the equivalence.

For the dihedral group of order ten, choose a rotation $r$ and a reflection $s$ with $r^5=s^2=1$ and $srs=r^{-1}$. Its rotation subgroup $⟨r⟩$ is a normal subgroup, and the quotient by it has order two, hence is an abelian group. Consequently the commutator subgroup is contained in $⟨r⟩$. On the other hand,Since $r^{-2}$ generates the order-five cyclic group $⟨r⟩$, the commutator subgroup is exactly $⟨r⟩$. Every abelian quotient must kill this subgroup. There are only two subgroups containing it, since the remaining quotient has prime order. The complete list, including the trivial quotient, isThis is the order-ten instance of the abelianization of an odd dihedral group.

Lagrange's theorem states that, for a finite group $G$ and a subgroup $H$,To prove it, define an equivalence relation by $x\sim y$ when $x^{-1}y\in H$. Its classes are the left cosets $xH$, which therefore partition $G$. Left multiplication by $x$ gives a bijection $H\to xH$, so every coset has $|H|$ elements. Counting this partition gives the formula; in particular $|H|$ divides $|G|$. Applying the theorem to the cyclic subgroup generated by an element shows that its order of a group element divides $|G|$.

Write the given order-two normal subgroup as $N=\{1,z\}$. Conjugation must preserve $N$ and cannot send its nonidentity element to the identity, so $gz{g}^{-1}=z$ for all $g\in G$. Thus $z$ is in the center of a group: a normal subgroup of order two is central.

Choose $g\notin N$. By Lagrange's theorem, its order of a group element is $2$, $p$ or $2p$. It cannot have order two: since $z$ is central, the four distinct elements $1,z,g,zg$ would form a subgroup of order four, whereas $4$ does not divide $2p$ for odd $p$. If $g$ has order $2p$, it already generates $G$. Otherwise $g$ has order $p$, commutes with $z$, and $(gz{)}^k=1$ implies $g^k=z^{-k}\in ⟨g⟩\cap N=\{1\}$. Hence both $p$ and $2$ divide $k$, so $gz$ has order $2p$. In either case

The chain rule along a smooth curve isDifferentiating the circular parametrization gives the tangent vectorThus the chain rule turns the angular derivative of $u$ into $-y{u}_x+x{u}_y$. The given partial differential equation makes this equal to $u$, so along every portion of the circle inside the upper half-plane, $U(t)=u(\mathbf{x}(t))$ satisfies the ordinary differential equation $U^{\prime }=U$. Its solution is $U(t)=U(t_0)e^{t-t_0}$.

For the two specified points take $a=\sqrt{2}$. The initial point corresponds to $t_0=\pi /4$ and the terminal point to $t_1=3\pi /4$. The entire intervening arc lies in $y>0$, so no continuation outside the domain is used. The requested value isThese circular characteristic curves also show locally that the general solution has the polar-coordinate form $u(r,\theta )=C(r)e^{\theta }$, with $0<\theta <\pi$; there is no requirement of angular periodicity on the upper half-plane.

Let $R$ be an orthogonal matrix describing a Cartesian change of coordinates, with transformed vector components $v_i^{\prime }=R_{ia}{v}_a$ and $w_j^{\prime }=R_{jb}{w}_b$. Their outer product transforms aswhich is the transformation law of a second-order Cartesian tensor. Here “rank two” counts tensor indices, not the rank of a matrix; a nonzero outer product has matrix rank one.

For an isotropic second-rank tensor, invariance means $RT{R}^T=T$ for every proper rotation matrix. The half-turn matrices $\mathrm{diag}(1,-1,-1)$, $\mathrm{diag}(-1,1,-1)$ and $\mathrm{diag}(-1,-1,1)$ force all off-diagonal entries to vanish. Quarter-turns around coordinate axes then force the three diagonal entries to be equal. Conversely $RI{R}^T=I$, so the most general second-rank isotropic tensor isSince $v{w}^T$ has matrix rank at most one, it cannot equal a nonzero scalar multiple of the three-dimensional identity. Thus its isotropy forces $v{w}^T=0$. If $v\ne 0$, a nonzero component $v_i$ makes the entire $i$th row vanish only when $w=0$; the converse is immediate. The required choices are exactlyFor the Levi-Civita symbol, multilinearity and antisymmetry of the determinant givefor every $R\in SO(3)$. This proves the rotation invariance of the Levi-Civita symbol, hence its third-rank isotropy. The convention is invariance under proper rotations: under an orthogonal reflection $\det R=-1$, the sign reverses. Thus it is a pseudotensor if transformations of both orientations are included.

No such nonabelian group exists. Every element, including the identity, satisfies $g^2=1$, so $g^{-1}=g$. For any $a,b$,Therefore the group is an abelian group. This proves that a group of exponent two is abelian without any finiteness assumption.

An example is the unitriangular group of degree three over F3, consisting ofThis is an upper unitriangular group of order $3^3=27$. Direct matrix multiplication givesThe formula gives closure, identity $M(0,0,0)$, and inverse $M(-a,-b,ab-c)$; associativity follows from matrix multiplication. The two elements $M(1,0,0)$ and $M(0,1,0)$ do not commute: their products have respectively $c=1$ and $c=0$.

Every element has the form $I+N$ with $N^3=0$. Since the finite field ${\mathbb{F}}_3$ has characteristic three, the binomial expansion givesHence every nonidentity element has order exactly three. This is a nonabelian finite group with exponent of a finite group equal to three.

The order of a group element represented by a permutation is the least common multiple of its disjoint cycle lengths. An order divisible by nine therefore requires a permutation cycle whose length is divisible by nine. In $S_9$ that must be a nine-cycle, which uses every letter and has order nine, not eighteen. There is no element of $S_9$ of order eighteen.

Take two disjoint permutation cycles of lengths five and four:Disjoint permutation cycles commute, and their powers act independently on their supports. Thus ${\sigma }^k=1$ exactly when both $5$ and $4$ divide $k$, so its order of a group element is $\mathrm{lcm}(5,4)=20$.

No such finite group exists. Cayley theorem embeds every finite group $G$ of order $n$ into $S_n$: the left-regular group action sends $g$ to the permutation $x\mapsto gx$, and this is faithful because the image of the identity element is $g$.

To make every image permutation even, let $t=(n+1n+2)$, with support disjoint from the first $n$ letters. For $\sigma \in S_n$, let $e(\sigma )=0$ for an even permutation and $e(\sigma )=1$ for an odd permutation, and setThe sign homomorphism gives $e(\sigma \tau )\equiv e(\sigma )+e(\tau )(\mathrm{mod}2)$, and $t$ commutes with all permutations of the first $n$ letters. Thus $\iota$ is a group homomorphism. It is injective by restriction to those letters, and its sign is $(-1{)}^{e(\sigma )}(-1{)}^{e(\sigma )}=1$. ConsequentlyThis is the alternating-group embedding of a finite group.

Every permutation is a product of transpositions, since a cycle can be writtenwith rightmost factors acting first. Define the sign of a permutation to be $(-1{)}^r$ for a product of $r$ transpositions. To prove that it is well defined without assuming parity, let $P_{\sigma }$ be the permutation matrix specified by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Then $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, and a transposition exchanges two columns of the identity, so its determinant is $-1$. Any decomposition into $r$ transpositions therefore hasThe left-hand side depends only on $\sigma$, proving well-definedness. Multiplicativity of the determinant now gives the sign homomorphism:For the first requested group embedding, use the faithful four-point action of GL2 over F2: label the four vectors in ${\mathbb{F}}_2^2$ and let an invertible $2\times 2$ matrix act on them by multiplication. Composition of these actions gives a group homomorphism $\psi :G{L}_2({\mathbb{F}}_2)\to S_4$. If a matrix fixes every vector, it fixes both standard basis vectors and is the identity, so the action is faithful. The matrixfixes $(0,0)$ and $(1,1)$ and exchanges $(1,0)$ with $(0,1)$. Its image is a single transposition, hence $\mathrm{sgn}(\psi (B))=-1$. The composite sign map is nontrivial.

For the second group embedding, simply takeThe preceding multiplication formula proves the group homomorphism property, and $P_{\sigma }=I$ forces $\sigma (i)=i$ for every $i$, proving injectivity.

For a group action of a finite group $G$ on a set and a point $x$, the orbit-stabilizer theorem iswhere $G_x=\{g:gx=x\}$ is the stabilizer subgroup. The map $g{G}_x\mapsto gx$ is well defined because elements of the same coset act identically on $x$. If $gx=hx$, then $h^{-1}g\in G_x$, so their cosets agree; every orbit point has the form $gx$, giving surjectivity. This is a bijection. Lagrange's theorem, $|G|=[G:H]|H|$ for any subgroup $H$, yields the counting formula.

Apply the conjugation action of the finite p-group $G$ to its nontrivial normal subgroup $N$. Normality ensures the action stays inside $N$. The singleton orbits of a group action are exactly the elements of $N\cap Z(G)$. Every larger orbit of a group action has size a positive power of $p$, by the orbit-stabilizer theorem and Lagrange's theorem. ThereforeSince $N$ is nontrivial, its order is divisible by $p$. The intersection contains the identity and has cardinality a positive multiple of $p$, so it contains a nonidentity element. This proves the central intersection property of normal subgroups of finite p-groups.

For the proper subgroup $H$, let $H$ act by left multiplication on the left cosets $G/H$. A coset $gH$ is fixed by every $h\in H$ exactly when $g^{-1}Hg\subseteq H$. Both groups have the same finite size, so this inclusion is equality; the fixed cosets are precisely those with $g\in N_G(H)$. Hence their number is $[N_G(H):H]$. All non-singleton orbits of a group action have size divisible by $p$, soSince $H$ is proper, $[G:H]$ is a positive power of $p$ greater than one. The fixed-coset count is nonzero, because $H$ itself is fixed, and is divisible by $p$. Therefore the normalizer strictly contains $H$:This is the normalizer condition for finite p-groups. The proof also covers $H=\{1\}$, whose action fixes all cosets.

The Möbius group consists of all Möbius transformations of the Riemann sphere ${\mathbb{C}}_{\infty }=\mathbb{C}\cup \{\infty \}$, with operation function composition. A nonzero-determinant matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ represents the map $z\mapsto (az+b)/(cz+d)$. If $c\ne 0$, its pole $-d/c$ maps to $\infty$ and $\infty$ maps to $a/c$; if $c=0$, the map is affine and fixes $\infty$. Nonzero scalar multiples of $M$ represent the same transformation.

Define $\varphi (M)$ by this action for $M\in S{L}_2(\mathbb{C})$. Substitution of fractional-linear expressions shows $\varphi (MN)=\varphi (M)\circ \varphi (N)$, so $\varphi$ is a group homomorphism. Any invertible matrix can be rescaled into the special linear group: choose $\lambda \in {\mathbb{C}}^{\times }$ with ${\lambda }^2\det M=1$, which is possible because every nonzero complex number has a square root. This does not change the transformation, proving surjectivity. A transformation represented by $M$ is the identity only if fixing $0$, $\infty$ and $1$ forces $b=c=0$ and $a=d$. The determinant-one condition then gives $a^2=1$. ThusFor the fixed-point claim, a nonidentity Möbius transformation has at least one fixed point of a Möbius transformation: when $c\ne 0$ the finite fixed points solve a quadratic, and when $c=0$ infinity is fixed. A quadratic shows there can be no more than two distinct fixed points unless the transformation is the identity. Suppose there is exactly one. Conjugate it to infinity by a Möbius transformation. The resulting map has form $z\mapsto az+b$, since it fixes infinity. If $a\ne 1$ it also has the finite fixed point $b/(1-a)$, a contradiction. Hence it is $z\mapsto z+b$ with $b\ne 0$. Its $m$th power is $z\mapsto z+mb$, never the identity for $m>0$. Conjugation preserves order, so a nonidentity finite-order Möbius transformation has exactly two fixed points.

Now let $K$ be a finite abelian subgroup of the Möbius group. First suppose it contains an element $g$ of order greater than two. Conjugate the two fixed points of $g$ to $0$ and $\infty$, so $g(z)=\zeta z$, with $\zeta$ a root of unity of order greater than two. Every element $h$ commuting with $g$ permutes its fixed-point set, because $g(h(p))=h(g(p))=h(p)$. A Möbius transformation preserving each of $0,\infty$ is $h(z)=az$; one exchanging them is $h(z)=a/z$. In the second case $hg(z)=a/(\zeta z)$ and $gh(z)=\zeta a/z$, so commuting would force ${\zeta }^2=1$, impossible. Thus all elements of $K$ are scalings. A finite subgroup of a field multiplicative group is cyclic; here one can see this directly by choosing the least common multiple $m$ of all element orders and observing that the scalars lie among the $m$th roots of unity, a cyclic group. Hence $K$ is cyclic.

Otherwise every nonidentity element of $K$ has order two. If $K$ is nontrivial, choose $g$ and conjugate it to $z\mapsto -z$. The action of $K$ on $\{0,\infty \}$ has image of order at most two. Its kernel consists of scalings $z\mapsto az$, and the order-two condition forces $a^2=1$, so the kernel has at most two elements. Hence $|K|\le 4$. Orders one and two give cyclic groups; order four with all nonidentity elements involutions gives $C_2\times C_2$, generated by any two distinct nonidentity elements. Both possibilities occur: scalings by roots of unity give every finite cyclic group, and $\{z,-z,1/z,-1/z\}$ gives the Klein four group. The classification is

A conservative vector field on ${\mathbb{R}}^n$ is a vector field $\mathbf{V}=\nabla F$ for a globally defined smooth potential of a conservative vector field $F$. The chain rule givesso its line integral is path independent.

Green theorem states that, for a bounded planar region $D$ with piecewise smooth boundary and continuously differentiable $P,Q$ on a neighbourhood of its closure,The boundary is positively oriented: the region lies on the left while it is traversed, so outer components run counterclockwise and hole boundaries clockwise.

For the proposed potential of a conservative vector field, the fundamental theorem of calculus and differentiation under the integral sign giveSince $Q_x=P_y$, the last integral equals $P(x,y)-P(x,0)$, including negative $y$ with the usual signed-integral convention. Hence the proposed potential is global andThe fact that the vector field is defined on all of ${\mathbb{R}}^2$ ensures both straight integration segments stay inside its domain; a hole in the domain could obstruct a global potential.

Integrating the oscillatory term together with the constant components gives a potential of a conservative vector fieldDirect partial derivatives recover both components of $\mathbf{V}$, proving it is a conservative vector field. Its line integral is therefore path independent. At an integer endpoint $(m,n)$, the sine term vanishes, so for any of the specified pathsAny other potential has the same gradient, so differs from $F$ by a constant on the connected plane. In particular every potential changes by $1$ under $x\mapsto x+1$ and by $2$ under $y\mapsto y+1$. Equivalently the integral from $(0,0)$ to $(1,0)$ is $1$, whereas a unit-periodic potential would make that difference zero. There is no potential periodic with period one in both coordinate directions. This is an example of how periods obstruct a periodic potential: a periodic vector field need not have a periodic potential.

For a smooth map $\mathbf{u}$, the Jacobian determinant is the determinant of its Jacobian matrix $A_{ai}={\partial }_i{u}_a$. In Einstein summation convention,Differentiate the supplied expression for $V_i$ by the product rule. The term differentiating ${\partial }_j{u}_a$ vanishes because ${\partial }_i{\partial }_j{u}_a$ is symmetric in $i,j$, while ${ϵ}_{ijk}$ is antisymmetric; the term differentiating ${\partial }_k{u}_b$ vanishes similarly. Only differentiation of $u_c$ remains:The last equality follows by the cyclic relabelling $(i,j,k)\mapsto (j,k,i)$, which has positive sign. This proves the Jacobian determinant as a divergence identity.

For a composition $\mathbf{w}=\mathbf{u}\circ \mathbf{v}$, the chain rule givesTaking determinants yieldsFor the first-order perturbation, use spatial derivative control in perturbations of the identity map: the usual smooth-family interpretation gives $D{\mathbf{u}}_t=I+tD\mathbf{F}+o(t)$ locally. Multilinearity of the determinant shows that its linear term comes from choosing one column of $tD\mathbf{F}$ and all other columns from $I$. This gives the trace:The regularity qualification matters: a pointwise $o(t)$ remainder for maps does not by itself give an $o(t)$ remainder for spatial derivatives. For instance $(x+t^2\sin (x/t^2),y,z)$ has a uniform $o(t)$ displacement and $\mathbf{F}=0$, but its Jacobian determinant at $x=0$ is $2$ for $t\ne 0$. Thus the first expansion needs the remainder in $C^1$ locally, or an equivalent smoothness assumption on the family.

Under the subsequent group law, ${\mathbf{u}}_0$ is the identity and ${\mathbf{u}}_{-t}$ is the inverse of ${\mathbf{u}}_t$. Write $j(t,\mathbf{x})=J[{\mathbf{u}}_t](\mathbf{x})$. Applying the composition formula to ${\mathbf{u}}_{t+h}={\mathbf{u}}_h\circ {\mathbf{u}}_t$ and the short-time expansion gives the Jacobian evolution of a smooth flow:HenceIf only the pointwise generator expansion is assumed initially, the group law still gives ${\partial }_t{\mathbf{u}}_t(\mathbf{x})=\mathbf{F}({\mathbf{u}}_t(\mathbf{x}))$. Uniqueness and smooth dependence for this smooth ordinary differential equation make it a smooth flow map, thereby justifying the differentiated expansion used above. Thus a divergence-free generator hasBy the change of variables formula, these maps preserve ordinary Euclidean volume and orientation. This is the Euclidean instance of a volume-preserving vector field.

The product rule for the divergence givesSubtracting cancels the two gradient products. ThereforeTo obtain the requested order of terms, apply this identity with $u,v$ interchanged and use the divergence theorem on the spherical shell. On the outer sphere the shell's outward unit normal is $\hat{\mathbf{r}}$; on the inner sphere it is $-\hat{\mathbf{r}}$. Write ${\partial }_r=\hat{\mathbf{r}}\cdot \nabla$ on both spheres. Then the resulting Green second identity isThe minus sign records the inward radial direction of the shell's normal on its inner boundary. Thus the normal derivatives in the displayed difference are outward from each individual ball; using shell-outward normals instead would put a plus sign between the two boundary integrals.

In Einstein summation convention, the curl isApply another curl and use the contraction of two Levi-Civita symbols:Smoothness allows the partial derivatives to commute. This proves the curl of the curl identity:

The identity divergence of a curl is zero follows by the symmetry of second partial derivatives and antisymmetry of the Levi-Civita symbol. ThereforeIf $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{A}$ is in Coulomb gauge, the curl of the curl identity gives $\mathbf{J}=-{\nabla }^2\mathbf{A}$. Apply the supplied scalar Poisson equation representation to each component of the decaying vector potential. In the units of this problem the Coulomb-gauge vector potential of a localized steady current isHere one needs $\mathbf{A}=O(|\mathbf{x}{|}^{-1})$, $\nabla \mathbf{A}=O(|\mathbf{x}{|}^{-2})$ and enough localization of $\mathbf{J}$ for the stated representation and differentiations. A smooth compactly supported current density is a sufficient convenient case. The local differential equations and gauge alone do not impose these conditions: $\mathbf{A}=(-y,x,0)$, $\mathbf{B}=(0,0,2)$, $\mathbf{J}=0$ satisfy them but do not satisfy the displayed integral formula. The formula selects the decaying solution, excluding nondecaying homogeneous harmonic functions.

With $\mathbf{r}=\mathbf{x}-\mathbf{y}$, the gradient of the kernel is ${\nabla }_x(1/|\mathbf{r}|)=-\mathbf{r}/|\mathbf{r}{|}^3$. Since $\mathbf{J}(\mathbf{y})$ is independent of $\mathbf{x}$,Thus taking the curl gives the Biot-Savart law in the problem's normalization:Finally, change variables $\mathbf{z}=\mathbf{x}-\mathbf{y}$ in the integral for $\mathbf{A}$. This keeps the singular kernel independent of the differentiation variable:For smooth localized $\mathbf{J}$, differentiation under the integral sign is legitimate: $1/|\mathbf{z}|$ is locally integrable in three dimensions, and on a bounded neighbourhood of $\mathbf{x}$ the translated sources and their derivatives have common compact support. The integral indeed satisfies Coulomb gauge.

The curl is constant:Orient the spanning planar disk by the unit normal $\mathbf{n}=(a,b,c)$ and orient its boundary by the right-hand rule, equivalently counterclockwise when viewed from the side towards which $\mathbf{n}$ points. Stokes theorem converts the line integral to a constant flux integral through a disk of area $\pi R^2$. With this orientationReversing the orientation changes the sign. The centre of the circle does not enter because the curl is constant.

Symmetry of the derivative matrix means ${\partial }_i{F}_j={\partial }_j{F}_i$. Contracting this symmetric pair with the antisymmetric Levi-Civita symbol givesEvery circle bounds its planar disk, and the vector field is smooth on all of ${\mathbb{R}}^3$, including that disk. Stokes theorem therefore gives for either orientationThe global domain is essential to this argument; the same local symmetry condition on a domain with a hole need not make every closed-curve line integral vanish.

Away from the origin, the given vector field is the gradient of the globally single-valued radius:The sphere has centre $(5,3,2)$, at distance $\sqrt{38}$ from the origin, and radius one. Its intersection circle is therefore wholly away from the origin, since every point on the sphere has distance at least $\sqrt{38}-1>0$. Along the closed curve the chain rule gives $\mathbf{F}\cdot d\mathbf{x}=dr$. Consequently, independently of orientation,The intersection is nondegenerate: the centre's distance from the plane is $4/\sqrt{35}<1$. No parametrization of the resulting circle is needed because $r$ is a potential of a conservative vector field throughout the punctured space.

Set $s=x^2+y^2>0$. The first two components of the curl vanish since the horizontal components are independent of $z$ and the last component depends only on $z$. For the third,so this is a curl-free vector field everywhere in its domain:Choose the orientation in which the projection onto the $xy$-plane runs counterclockwise, and parametrizeOn this curve the horizontal contribution is $(-\sin t,\cos t)\cdot (-\sin t,\cos t)dt=dt$, while the vertical contribution is $zdz=d(z^2/2)$. Its integral over a closed curve is zero. Thereforewith $-2\pi$ for the opposite orientation. This does not contradict Stokes theorem: the planar disk spanning this curve meets the excluded $z$-axis at $(0,0,200)$, where the vector field is undefined. The horizontal one-form is the angular differential and records one winding about that axis; thus this curl-free vector field has no single-valued global potential of a conservative vector field on its domain.