Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3

Suppose $G/Z(G)$ is generated by $gZ(G)$. Every element of $G$ can then be written $g^{a}z$ with $z\in Z(G)$. HenceThus the cyclic quotient by the center condition forces $G$ to be abelian.

If $|H/Z(H)|=7$, the quotient has prime order and is therefore cyclic. The result just proved would make $H$ abelian, so $Z(H)=H$ and the quotient would have order one, a contradiction. Therefore

Such a group does exist. Take $H=S_3$. Its center is trivial, sohas order six. Therefore

Elements $g,h\in G$ are conjugate whenfor some $x\in G$. Thenfor every integer $n$, so $h^n=e$ exactly when $g^n=e$. They therefore have the same order.

The Möbius group consists of the Möbius transformationsof the Riemann sphere, under composition. If $h=xg{x}^{-1}$, thenThus $p$ is fixed by $g$ exactly when $x(p)$ is fixed by $h$. Conjugation gives a bijection between their fixed-point sets, so conjugate elements have the same number of fixed points.

A nonidentity Möbius transformation has at most two fixed points because its fixed-point equation is quadratic on the Riemann sphere. If it has only one repeated fixed point, conjugate that point to infinity; the transformation becomes a nontrivial translation $z\mapsto z+c$, which has infinite order. Consequently every nontrivial finite-order element has two distinct fixed points:This is fixed points of a finite-order Möbius transformation.

In Einstein notation,This is

Similarly,after cyclically relabelling the first term and swapping two indices in the second. Hence

The data are rotationally symmetric, so seek a radial solution. The radial Laplacian in two dimensions givesTwice integrating,Regularity at the center forces $C=0$, and $u(1)=1$ gives $D=0$. Thus

On the annulus the logarithmic term is allowed, soThe condition $u(1)=1$ gives $D=0$, while $u(2)=1$ givesTherefore

A second-rank tensor is antisymmetric whenUnder an orthogonal Cartesian coordinate change with matrix $R$,Thereforeafter relabelling the dummy indices. Hence the antisymmetric second-rank tensor property is independent of Cartesian coordinates.

For $F=(y,z,x)$,Its antisymmetric part isAlso,For example, $-(1/2){ϵ}_{12k}{B}_k=1/2$, and the same componentwise calculation gives

The order of an element $x$ is the least positive integer $d$ such that $x^d=e$, and the order of a finite group is its number of elements.

Lagrange theorem states that if $H\le G$ and $G$ is finite, thenIndeed, the left cosets of $H$ partition $G$. Multiplication by a coset representative is a bijection from $H$ to each coset, so every coset has $|H|$ elements. Summing over the $[G:H]$ cosets proves the formula.

The cyclic subgroup $⟨x⟩$ has exactly $d$ elements when $x$ has order $d$. Applying Lagrange to $⟨x⟩\le G$ gives

No. The alternating group $A_4$ has order twelve, and six is a proper divisor of twelve, but $A_4$ has no element of order six. Its nonidentity elements are eight 3-cycles of order three and three double transpositions of order two.

Let $a$ and $b$ generate $C_m$ and $C_n$. The order of $(a,b)$ in the direct product of groups isWhen $\gcd (m,n)=1$, this is $mn=|C_m\times C_n|$. Hence $(a,b)$ generates the whole product and

Every element satisfies $x^2=e$, including the identity. For any $x,y\in G$,Since $x^{-1}=x$ and $y^{-1}=y$, this implies $xy=yx$, so $G$ is abelian. It is therefore a vector space over the two-element field, with group operation as vector addition and scalar multiplication by zero or one. A basis of this finite vector space givesfor some $r\ge 0$.

Write $d=qr+s$ with $0\le s<r$. Since $x^d=e\in H$ and $x^r\in H$,The minimality of the positive integer $r$ forces $s=0$. Therefore

Among the $n+1$ right cosetstwo must coincide because $[G:H]=n$. If $H{x}^i=H{x}^j$ with $0\le i<j\le n$, right multiplication by $x^{-i}$ gives $H=H{x}^{j-i}$, so

The least such exponent need not divide $n$. For example, take $G=S_3$, $H=⟨(12)⟩$, and $x=(13)$. Then $[G:H]=3$, while $x\notin H$ and $x^2=e\in H$, so the least exponent is two, which does not divide three.

For $x\in X$,The stabilizer contains the identity; if $g,h$ fix $x$, thenso it is a subgroup.

The orbit-stabilizer theorem statesDefineThe map is well defined and bijective: two elements give the same image exactly when they differ by an element of the stabilizer. Lagrange's theorem then proves the formula.

The orbit of $(1,\dots ,k)$ consists precisely of the ordered $k$-tuples of distinct elements of $\{1,\dots ,n\}$:It has sizeThe stabilizer consists of permutations fixing $1,\dots ,k$ pointwise, while freely permuting the remaining letters:Thusverifying orbit-stabilizer.

Choose $h\in H\setminus \mathrm{Stab}(x)$ and put $y=hx\ne x$. For any $z\ne x$, double transitivity supplies an element $s\in G$ taking the ordered pair $(x,y)$ to $(x,z)$. Since $sx=x$, one hasThen $sh\in H$ sends $x$ to $z$. Together with the identity for $z=x$, this proves that $H$ acts transitively on $X$.

Now let $g\in G$. Transitivity of $H$ gives $h_0\in H$ with $h_{0}x=gx$. Henceso $g\in H$. ThereforeThis is the point-stabilizer maximality property of a doubly transitive group action.

Let $G$ act on its underlying set by left multiplication. The resulting homomorphismis injective because ${\lambda }_g(e)=g$. This is Cayley theorem. If $g\ne e$, then $gx=x$ would imply $g=e$ after right cancellation, so every nonidentity permutation in the image is fixed-point free.

If $|G|$ is even, pair every element with its inverse. Elements not equal to their inverses occur in pairs. Since the identity is self-inverse and the group has even size, there must be another self-inverse element $g\ne e$. It has order two.

A permutation is odd when its sign is $-1$. If a subgroup $H\le S_m$ contains an odd element, the restrictionis surjective. Its kernel is the set of even elements and has index two. Each coset has the same size, so precisely half of $H$ is odd.

Now let $n=4k+2$. An element of order two in the fixed-point-free regular representation is a product ofdisjoint transpositions, so it is odd. The image $H\cong G$ therefore has an index-two normal subgroup of even permutations. Its order is $2k+1>1$, so it is nontrivial and proper. Hence

Nonabelian simple groups of even order do exist; the smallest example is the alternating group $A_5$, of order sixty.

Identify $A$ with $\{(a,1):a\in A\}$. Componentwise use of the stated multiplication shows that this set contains the identity and is closed under products and inverses, so it is a subgroup. It has the two cosetsand therefore has index two. Every index-two subgroup is normal, so

For $ϵ=-1$,so every element outside $A$ has order two.

The element $(e,-1)$ commutes with $(a,1)$ exactly whenor $a^2=e$. Since $A$ itself is abelian, $D(A)$ is abelian exactly when every $a\in A$ satisfies $a^2=e$.

Suppose $D(A)$ is nonabelian. An element $(a,1)$ is central exactly when $a=a^{-1}$, equivalently $a^2=e$. No element $(a,-1)$ is central, because commuting with every $(b,1)$ would require $b^2=e$ for every $b$, which would make the whole group abelian. Hence

If $Q\in O(2)$, then $Q^{T}Q=I$, soand $\det Q=\pm 1$. If $\det Q=1$, its first column is $(\cos \theta ,\sin \theta {)}^T$ for some $\theta$, and orthonormality plus positive orientation forceThus every element of $SO(2)$ is a rotation.

Let $J=\mathrm{diag}(1,-1)$. If $\det Q=-1$, then $QJ\in SO(2)$ and $Q=(QJ)J\in SO(2)J$. Conversely every matrix in $SO(2)J$ has determinant $-1$. Therefore the union is disjoint and

The standard isomorphisms areand

For $z\in B$, let $R_z$ be the corresponding rotation. The identityshows thatrespects the multiplication of the generalized dihedral group. The decomposition in part (b) makes it bijective. Hence

For a regular parametrized curve $r(t)$, the Frenet frame isEquivalently, wherever $dT/dt\ne 0$, one may write $N=(dT/dt)/|dT/dt|$ and $B=T\times N$.

Differentiation givesand $|r^{\prime }|=6|\sin t\cos t|$. ThusThe plus sign applies on $(0,\pi /2)\cup (\pi ,3\pi /2)$ and the minus sign on $(\pi /2,\pi )\cup (3\pi /2,2\pi )$.

On each open quadrant,The constant binormal proves that the curve is planar; indeed $z=\sqrt{3}y$.

For $n=(a,b,c)$ and $r=(x,y,z)$, the integrand is $\frac{1}{2}(n\times r)\cdot dr$. Sincethe Stokes theorem gives the signed planar area, and its absolute value gives the ordinary area.

The $xy$ projection of the given curve is the astroid $x=2{\cos }^{3}t$, $y={\sin }^{3}t$, whose area isThe plane $z=\sqrt{3}y$ stretches area from its $xy$ projection by $\sqrt{1+3}=2$. Hence

LetThenUsing polar coordinates on the first quadrant,Since $I>0$, the Gaussian integral is

For fixed $x>0$, $x\ne 1$,with limiting value one at $x=1$. Integrating over $0<x<1$ in the opposite order givesOn the other hand, after setting $x=e^{-u}$,Consequently

In polar coordinates, the circle is $r=2$, the line $y=\sqrt{3}x$ is $\theta =\pi /3$, and $y=1$ is $r\sin \theta =1$. The inequalities selectThe surface height isThus the required volume is

SetThe ellipsoid becomes the unit ball and the Jacobian determinant is $abc$. Thereforeassuming $a,b,c>0$.

Fix the first index $i$ and form the vector field with components $F_j=T_{ij}$. The divergence theorem givesSubstitution yields the tensor divergence theorem

For $u=(-y,x,\alpha )$, one has $\nabla \cdot u=0$, and therefore

Close the upper half-ellipsoid with its unit-disk base $D$ in the plane $z=0$. By symmetry,On the base, the outward normal is $n=(0,0,-1)$, soThe first two components integrate to zero over the disk, while the third integrates to $-\pi {\alpha }^2$. The closed-surface tensor identity therefore givesMultiplying by ${\beta }^2$,

The vector identity from Question 3 givesIntegrating over $V$ and applying the divergence theorem proves the Poynting theorem

Both divergences vanish, so $\rho =0$. Faraday's and Ampere's curl equations respectively requireFor positive constants these imply

The Poynting vector isOnly the two $x$-faces of the box contribute to its outward flux. With $a=\pi /(2k)$,The electromagnetic energy inside the box isand direct differentiation, using $\omega =k$, givesThusconfirming the integral identity when $J=0$.