Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 3

Composing the permutation cycles from right to left giveswith $6$ fixed. It therefore has order $7$ and, as a seven-cycle, sign $(-1{)}^6=+1$.

An element of $S_6$ has order $6$ precisely for cycle type $(6)$ or $(3,2,1)$. These contributeso there are $240$ elements of order $6$. Order $3$ has cycle type $(3,1,1,1)$ or $(3,3)$, contributingso there are $80$ elements of order $3$.

Inspecting the partitions of $9$ shows that the largest cycle-length least common multiple available in $S_9$ is $20$, from the partition $5+4$, but such a permutation is odd. Among integers from $16$ through $20$, orders $17$ and $19$ would require cycles of those prime lengths, order $16$ requires a cycle length divisible by $16$, and order $18$ requires disjoint cycles using at least $9+2>9$ letters. The disjoint product of a five-cycle and a three-cycle is even and has order $15$. Hence the greatest order in $A_9$ is $15$.

Every $Q\in O(3)$ with $\det Q=1$ is a rotation in three dimensions, hence is the product of two reflections in planes through its axis whose angle is half the rotation angle.

If $\det Q=-1$, its real eigenvalue is $-1$: the other two eigenvalues are either a complex-conjugate pair or two real signs, and their product is positive. Let $v$ be a corresponding unit eigenvector and let $H$ be reflection in $v^{\perp }$. Then $HQ$ fixes $v$ and has determinant $1$, so it is a rotation and hence a product of two reflections. Since $Q=H(HQ)$, every member of $O(3)$ is a product of at most three reflections.

Two reflections have determinant $1$, whereas one reflection has eigenvalues $(-1,1,1)$. The map $-I\in O(3)$ has determinant $-1$ but is not a reflection, so it cannot be a product of at most two reflections. Thus three are sometimes necessary.

For the parametrized curve $\mathbf{x}(u)=(u,f(u))$, the curvature of a plane curve isFor the upper semicircle $f(u)=\sqrt{a^2-u^2}$,and therefore $\kappa =1/a$, as required for a circle of radius $a$.

For a scalar $\varphi$, the displayed basis identities giveTaking the divergence and differentiating the basis vectors yields the Laplacian in polar coordinatesConsequently,

The regular harmonic mode with angular dependence $\cos 2\theta$ is $C{r}^2\cos 2\theta$. Its radial derivative at $r=a$ is $2Ca\cos 2\theta$, so $C=1/(2a)$. A constant is invisible to the Neumann boundary condition, givingIf two solutions existed, their difference $u$ would satisfy ${\nabla }^{2}u=0$ and ${\partial }_{n}u=0$. Green's first identity gives ${\int }_D|\nabla u{|}^2={\int }_{\partial D}u{\partial }_{n}u-{\int }_{D}u{\nabla }^{2}u=0$, so $\nabla u=0$ and $u$ is constant. Thus these are all the solutions.

For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and defineSince a transposition exchanges two columns, it has determinant $-1$. Thus any expression of $\sigma$ as $r$ transpositions has sign $(-1{)}^r$, proving that the parity is independent of the expression. Moreover, $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, so the determinant provesHence the sign homomorphism is a surjective homomorphism for $n\ge 2$.

For any homomorphism $\phi :S_n\to \{\pm 1\}$, all transpositions have the same image because they are conjugate. Since transpositions generate $S_n$, image $+1$ would make $\phi$ trivial. Surjectivity therefore forces every transposition to map to $-1$, and hence $\phi =\mathrm{sgn}$.

For fixed $g\in G$, conjugation $g_{\ast }:x\mapsto gx{g}^{-1}$ preserves multiplication:and its inverse is $(g^{-1}{)}_{\ast }$. It is therefore a group automorphism. Furthermore,so $\theta :g\mapsto g_{\ast }$ is a homomorphism $G\to \mathrm{Aut}(G)$. Its kernel consists of elements commuting with every $x$, hence

For $\alpha \in \mathrm{Aut}(G)$,so the inner automorphism group is normal in the automorphism group.

If $G=⟨x⟩$ is cyclic, every automorphism has the form $x\mapsto x^a$. Two such maps commute because $ab=ba$, so $\mathrm{Aut}(G)$ is abelian. The conclusion fails for general abelian groups: for $G=C_2\times C_2$, its three nonidentity elements may be permuted arbitrarily, givingwhich is nonabelian.

For a normal subgroup $H⊴G$, the quotient group $G/H$ is the set of cosets $gH$ with multiplication $(gH)(kH)=gkH$. If $gH=g^{\prime }H$ and $kH=k^{\prime }H$, normality moves the intervening elements of $H$ past $k$, proving that the product is independent of representatives.

If $g$ has finite order $m$, then $(gH{)}^m=H$, so $\mathrm{ord}(gH)$ divides $m$. For finite $G$, the greatest element order in $G/H$ is therefore no greater than that in $G$.

Every kernel is normal because $\phi (gh{g}^{-1})=\phi (g)\phi (h)\phi (g{)}^{-1}=1$ whenever $h\in \ker \phi$. Conversely, if $H$ is normal, the quotient map $G\to G/H$ has kernel $H$. Thus $H⊴G$ exactly when it is the kernel of a group homomorphism.

A group is metacyclic when it has a cyclic normal subgroup with cyclic quotient. A dihedral group has its cyclic rotation subgroup normal and quotient $C_2$, so every dihedral group is metacyclic.

Up to isomorphism the groups of order eight areAll except $C_2^3$ are metacyclic: use a cyclic subgroup of order four in the three noncyclic examples that have one. In $C_2^3$, every cyclic subgroup has order at most two and its quotient is not cyclic. The normal subgroups of $A_4$ are $1,V_4,A_4$; none gives the required cyclic subgroup and quotient, so $A_4$ is not metacyclic. Finally, $S_1,S_2,S_3$ are metacyclic, while $S_4$ and $S_5$ are not, by their normal-subgroup structures. Hence

The internal direct product theorem states that if $H,K⊴G$, $H\cap K=\{1\}$, and $HK=G$, then $G\cong H\times K$. Indeed, for $h\in H,k\in K$, the commutator $[h,k]$ lies in both $H$ and $K$, so it is trivial. The map $(h,k)\mapsto hk$ is consequently a homomorphism; the intersection condition makes it injective and $HK=G$ makes it surjective. Conversely, the two factors in a direct product satisfy these conditions.

In odd dimension, $-I$ is central and has determinant $-1$. Every $Q\in O(3)$ has a unique expression $Q=(-I{)}^{\epsilon }R$ with $R\in SO(3)$, soBy contrast, $SO(2)\times C_2$ is abelian, while reflections in $O(2)$ conjugate a rotation to its inverse. Thus $O(2)\ncong SO(2)\times C_2$.

The center of the unitary group consists exactly of scalar unitary matrices:The determinant gives a surjective homomorphismwhose kernel is the special unitary group $SU(2)$.

Nevertheless $U(2)$ is not isomorphic to $SU(2)\times \mathbb{T}$. The center of $U(2)$ has exactly one nonidentity element of order two, namely $-I$. The center of the proposed product is $\{\pm I\}\times \mathbb{T}$, which has three nonidentity elements of order two. Since an isomorphism preserves the center and element orders,

For a one-to-one differentiable change of variables $(x,y)\mapsto (u,v)$, the change of variables formula iswhere

Here $u=x+y$, $v=x^2+y^2$, andAt fixed $v$, write $x=\sqrt{v}\cos \theta$, $y=\sqrt{v}\sin \theta$ with $0\le \theta \le \pi$. Then $u=\sqrt{v}(\cos \theta +\sin \theta )$, soThe constant-$v$ curves are semicircles and the constant-$u$ curves are lines $x+y=u$.

The map is two-to-one where $\sqrt{v}<u<\sqrt{2v}$, because both $(x,y)$ and $(y,x)$ lie in the upper half-disc, and one-to-one elsewhere. Since $|y-x|=\sqrt{2v-u^2}$, accounting for this multiplicity givesDirectly, symmetry makes ${\iint }_{D}xdA=0$, while polar coordinates give ${\iint }_{D}ydA={\int }_0^1{r}^{2}dr{\int }_0^{\pi }\sin \theta d\theta =2/3$.

The Stokes theorem stateswhere the boundary orientation is induced by the chosen normal through the right-hand rule.

For $\mathbf{F}=(y^{2}z,xz+2xyz,0)$,and its divergence is $(-1-2y)+2y+1=0$, verifying $\nabla \cdot (\nabla \times \mathbf{F})=0$.

Orient the paraboloid $z=3-x^2-y^2$, $1\le x^2+y^2\le 4$, upward and the planar annulus $z=-1$ downward. Their outer boundary orientations cancel. The remaining boundary is the unit circle at $z=-1$ counterclockwise and the unit circle at $z=2$ clockwise, viewed from above. On a unit circle at height $z_0$, direct parametrization givesHence the boundary integral is $-\pi -2\pi =-3\pi$.

On the lower annulus, $(\nabla \times \mathbf{F})\cdot (0,0,-1)=1$, giving $3\pi$. On the paraboloid, $d\mathbf{S}=(2x,2y,1)dxdy$, so the flux integrand isThe odd terms integrate to zero over the annulus, while the rest integrates to $-6\pi$. The total surface integral is $3\pi -6\pi =-3\pi$, equal to the boundary integral.

Apply the curl of the curl identity $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-{\nabla }^{2}E$ to Faraday's law, and use the Ampère-Maxwell equation. This givesApplying the same identity to $B$, using $\nabla \cdot B=0$, gives

For a spherically symmetric charge density, Gauss's law on a sphere of radius $r$ givesFor the unit-density ball of radius $a$,The potential is continuous at $r=a$ and tends to zero at infinity.

Let $\mathbf{p}=(1,0,0)$ and write $1_{S_n}$ for the indicator of $S_n$. The layered density decomposes asThe point $\mathbf{p}$ lies on every sphere $S_n$, whose radius is $R_n=4^{-n}$ and whose outward radial direction there is $\hat{\mathbf{x}}$. A uniform ball of density $c$ contributes $c{R}_n/3$ to the field and $c{R}_n^2/3$ to the potential at its surface. Thereforeand

Taking the symmetric part givesContracting the antisymmetric part with the Levi-Civita symbol givesIndeed, $\frac{1}{2}{\epsilon }_{ijk}{\epsilon }_{k\ell m}{T}_{\ell m}=\frac{1}{2}(T_{ij}-T_{ji})$, so $S_{ij}+{\epsilon }_{ijk}{V}_k=T_{ij}$.

In three dimensions, the most general isotropic tensors of ranks zero through three areThe scalar and zero vector are unchanged by rotations. For every $R\in SO(3)$, $R_{ia}{R}_{jb}{\delta }_{ab}={\delta }_{ij}$ and $R_{ia}{R}_{jb}{R}_{kc}{\epsilon }_{abc}=(\det R){\epsilon }_{ijk}={\epsilon }_{ijk}$, proving isotropy.

The relation isTaking its trace, symmetric traceless part, and antisymmetric part separately givesEquivalently,

If $\beta =-\gamma \ne 0$, $\alpha \ne 0$, and $A_{ij}=0$, its symmetric part forces $B_{kk}=0$ and its antisymmetric part forces $B_{[ij]}=0$. Thus the solutions are exactly the symmetric traceless tensors $B_{ij}$.

Dualize the antisymmetric tensor to a vector,An isotropic symmetric rank-two tensor depending quadratically on one vector can only be a linear combination of ${\delta }_{ij}{d}_k{d}_k$ and $d_i{d}_j$: after rotating $d$ to the third axis, invariance under all rotations about that axis forces the matrix to be $\mathrm{diag}(a,a,b)$.

Since $D_{ik}{D}_{jk}={\delta }_{ij}{d}^2-d_i{d}_j$ and $D_{kl}{D}_{kl}=2d^2$, the most general answer may be writtenThe two displayed tensors are independent, for example by taking $d$ along a coordinate axis and comparing the components parallel and perpendicular to it. Hence there are exactly two independent terms.