Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 3

The second component must have order $4$ and the first is arbitrary, giving $2\cdot 2=4$ elements.

Exactly the six $4$-cycles have order $4$: $(4-1)!=6$.

The cycle types are $(4)(1)(1)$ and $(4)(2)$. Each contributes $\left(\genfrac{}{}{0ex}{}{6}{4}\right)(4-1)!=90$, for a total of $180$.

Only type $(4)(2)$ is even, so the answer is $90$.

$Z(G)=\{z:zg=gz\text{ for every }g\in G\}$; it is closed under products and inverses, abelian, and invariant under conjugation. If $K=\{e,k\}$ is normal, conjugation fixes its unique nonidentity element $k$, so $k\in Z(G)$. Thus the statement is true.

True. If $G/Z(G)$ is generated by $gZ(G)$, write $x=g^a{z}_1$, $y=g^b{z}_2$ with central $z_i$; then $xy=yx$.

True. $S_3$ has trivial centre. A central permutation would commute with both $(12)$ and $(123)$, which only the identity does.

$s(t)={\int }^t|\dot{x}(u)|du$, $t=dx/ds$, $\kappa =|dt/ds|$, and $n={\kappa }^{-1}dt/ds$. Differentiating $t\cdot t=1$ proves $t\perp n$. Here $|\dot{x}|=2e^t$ and

Divergence theorem says ${\int }_V\nabla \cdot FdV={\int }_{\partial V}F\cdot ndS$ for the outward normal. Here $\nabla \cdot (x{e}^{-r^2})=(3-2r^2)e^{-r^2}$, so radial integration gives $4\pi a^3{e}^{-a^2}$. On the sphere $F\cdot n=a{e}^{-a^2}$, whose integral is the same $4\pi a^3{e}^{-a^2}$.

The order of $g$ is the least positive $r$ with $g^r=e$. Division $n=qr+s$ shows $g^n=e$ forces $s=0$. Lagrange theorem follows because left cosets of a subgroup partition $G$ into equal blocks; applying it to $⟨g⟩$ shows $|g|\mid |G|$. Coset multiplication makes $G/N$ a group when $N◃G$. If $x\in N$, then $x^n=e$. Conversely, if $x^n=e$, the order of $xN$ divides both $m$ and $n$, hence is one, so $x\in N$.

Every $b^m$ lies in $N$ because $(bN{)}^m=N$. Conversely choose $r$ with $mr\equiv 1(\mathrm{mod}n)$; for $a\in N$, $a=(a^r{)}^m$ because $a^n=e$.

$G{L}_2(\mathbb{C})$ consists of invertible $2\times 2$ complex matrices and $S{L}_2(\mathbb{C})=\ker \det$. It is normal, and the first isomorphism theorem gives $G{L}_2/S{L}_2\cong {\mathbb{C}}^{\ast }$.

The Möbius group consists of maps $z\mapsto (az+b)/(cz+d)$ with $ad-bc\ne 0$, including the usual action on $\infty$. Scalar matrices act trivially and are the whole kernel, so $M\cong G{L}_2(\mathbb{C})/({\mathbb{C}}^{\ast }I)$.

The map $z\mapsto z-w$ sends $w$ to $0$, proving transitivity (with inversion handling $\infty$). Fixed points satisfy the quadratic $c{z}^2+(d-a)z-b=0$, with multiplicity and $\infty$ included, so a nonidentity has one or two. The stabiliser of $\infty$ is $S=\{z\mapsto az+b:a\ne 0\}$.

Choose a fixed point $p$ of a nonidentity transformation and conjugate by a Möbius map carrying $p$ to $\infty$; the conjugate lies in $S$. The identity already lies there.

An action is a map $G\times X\to X$ satisfying $ex=x$ and $(gh)x=g(hx)$; it induces $G\to S_X$. Acting on the $n$ left cosets of $H$ gives a homomorphism with kernel $K◃G$ contained in $H$. If $K$ were trivial, $|G|$ would divide $n!$, so under the stated hypothesis $K$ is nontrivial. For $[G:L]=p$ with $p$ the least prime divisor of $|G|$, the transitive image in $S_p$ has order divisible by $p$; its other possible prime divisors are smaller, hence absent, so its order is $p$. Thus the kernel has index $p$ and, being contained in $L$, equals $L$. Prime index alone is insufficient: a subgroup generated by a transposition in $S_3$ has index $3$ and is not normal.

Disjoint-cycle decomposition follows by partitioning the set into permutation orbits. Every cycle $(a_1\dots a_r)$ equals $(a_1{a}_r)\cdots (a_1{a}_2)$, so transpositions generate $S_n$; $A_n$ is the kernel of the sign map. For even $m$, an $m$-cycle is odd. Products of pairs of $m$-cycles generate $A_n$ (express the standard generators $(ab)(ac)$ using the hinted overlapping cycles), while adjoining any one $m$-cycle yields all of $S_n$. Thus the set of all $m$-cycles generates $S_n$. For odd $m$, every $m$-cycle is even, so they cannot generate $S_n$.

Using ${\epsilon }_{ijk}{\epsilon }_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$,
$[\nabla \times (F\times G){]}_i={\partial }_j(F_i{G}_j)-{\partial }_j(F_j{G}_i)$; expanding the derivatives gives the stated four terms.

Irrotational means $\nabla \times E=0$. For $E=\nabla f$, its $i$th curl component is ${\epsilon }_{ijk}{\partial }_j{\partial }_{k}f=0$ because mixed partials are symmetric.

On $C$, $dx=a{e}_{\varphi }d\varphi$, so $I={\int }_0^{2\pi }d\varphi =2\pi$. Locally $f=\varphi$, but no single-valued global potential exists on the punctured domain; this is why irrotationality does not force this closed integral to vanish.

The field is $\nabla f$ for $f=\rho \sin \varphi e^{-{\rho }^2-z^2}$. Hence $I=0$, either directly or by the fundamental theorem for line integrals.

Stokes theorem is ${\int }_S(\nabla \times F)\cdot ndS={\oint }_{\partial S}F\cdot dx$, with the boundary orientation given by the right-hand rule relative to $n$.

$\nabla \times F=(3y^2,3x^2,3x^2+3y^2)$, whose divergence is $0$.

Take the outward normal on the cylindrical side and the upward normal on the top disc. The induced boundary direction on the bottom circle is clockwise as viewed from above.

The clockwise boundary integral at $z=0$ is
$-{\int }_0^{2\pi }{a}^4({\sin }^4\varphi +{\cos }^4\varphi )d\varphi =-3\pi a^4/2$. Direct integration of the curl over the side and top gives the same value (the $h$-dependent terms cancel), verifying Stokes.

$S\prime$ has the same bottom boundary and induced orientation, so Stokes’ theorem makes the answer independent of its shape: $-3\pi a^4/2$.

For two solutions let $u$ be their difference. Green’s identity and ${\partial }_{n}u=0$ give $0={\int }_{V}u({\nabla }^{2}u-m^{2}u)=-{\int }_V(|\nabla u{|}^2+m^2{u}^2)$, so $u=0$ when $m\ne 0$. At $m=0$, solutions are unique only up to an additive constant.

On a simply connected region, $\nabla \times E=0$ gives $E=-\nabla \varphi$. Then Gauss's law gives ${\nabla }^2\varphi =-\rho /{\epsilon }_0$.

Writing $r=|x|$ and imposing continuity gives

Gauss's law gives $E=Q\hat{r}/(4\pi {\epsilon }_0{r}^2)$ outside any spherically symmetric charge distribution. The shell in (ii) has $Q=2\pi {\rho }_0(b^2-a^2)$, and differentiating its exterior potential gives exactly this field.

A rank-$k$ Cartesian tensor has components $T_{i_1\dots i_k}$ transforming under an orthogonal change of basis $R$ as $T{\prime }_{i_1\dots i_k}=R_{i_1{j}_1}\cdots R_{i_k{j}_k}{T}_{j_1\dots j_k}$.

Contracting one index from each tensor means summing them, for example $C_{i_1\dots i_{p-1}{j}_1\dots j_{q-1}}=A_{i_1\dots i_{p-1}r}{B}_{r{j}_1\dots j_{q-1}}$. Its rank is $p+q-2$. Substitution of the transformation laws for $p=q=3$ and $R_{rs}{R}_{rt}={\delta }_{st}$ proves the tensor law.

Isotropic means invariant under every proper rotation. The general ranks $0,1,2,3$ are respectively a scalar $a$, zero, $b{\delta }_{ij}$, and $c{\epsilon }_{ijk}$ (for full orthogonal invariance the last must also vanish).

Rotational invariance makes this an isotropic vector, hence it is $0$; oddness gives the same result.

It equals $C{\delta }_{ij}$. Taking the trace gives $3C=4\pi a^4{e}^{-a^2}$, hence $C=4\pi a^4{e}^{-a^2}/3$.

This isotropic rank-three integral is $0$ by oddness under $x\mapsto -x$.

Expanding the integrand, the terms linear in $x$ integrate to zero. Thereforeso these are $\alpha$ and $\beta$.