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.

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

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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 $180$.

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

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$Z(G)=\{z:zg=gz\text{ for every }g\in G\}$; 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.

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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 $xy=yx$.

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

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

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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 $4\pi a^3{e}^{-a^2}$.

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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 $x\in N$.

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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 $a^n=e$.

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$G{L}_2(\mathbb{C})$ 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 $G{L}_2/S{L}_2\cong {\mathbb{C}}^{\ast }$.

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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 $M\cong G{L}_2(\mathbb{C})/({\mathbb{C}}^{\ast }I)$.

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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 $S=\{z\mapsto az+b:a\ne 0\}$.

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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.

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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.

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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 $S_n$.

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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.

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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.

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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.

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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.

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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 $n$.

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

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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.

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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.

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

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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.

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

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

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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.

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

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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.

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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).

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

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

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

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

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