past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-3.bigb

= Paper 3
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/Paperia_3_2026.pdf

= 1D
{parent=Paper 3}
{scope}
{title2=Groups}

= a
{parent=1d}
{scope}

= Solution
{parent=a}

The second component must have order $4$ and the first is arbitrary, giving $2\cdot2=4$ elements.

Solved by gpt-5.6-sol high.

= b
{parent=1d}
{scope}

= Solution
{parent=b}

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

Solved by gpt-5.6-sol high.

= c
{parent=1d}
{scope}

= Solution
{parent=c}

The cycle types are $(4)(1)(1)$ and $(4)(2)$. Each contributes $\binom64(4-1)!=90$, for a total of $180$.

Solved by gpt-5.6-sol high.

= d
{parent=1d}
{scope}

= Solution
{parent=d}

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

Solved by gpt-5.6-sol high.

= 2D
{parent=Paper 3}
{scope}
{title2=Groups}

= a
{parent=2d}
{scope}

= Solution
{parent=a}

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

Solved by gpt-5.6-sol high.

= b
{parent=2d}
{scope}

= Solution
{parent=b}

True. If $G/Z(G)$ is generated by $gZ(G)$, write $x=g^az_1$, $y=g^bz_2$ with central $z_i$; then $xy=yx$.

Solved by gpt-5.6-sol high.

= c
{parent=2d}
{scope}

= Solution
{parent=c}

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

Solved by gpt-5.6-sol high.

= 3A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= Solution
{parent=3a}

$s(t)=\int^t|\dot x(u)|du$, $t=dx/ds$, $\kappa=|dt/ds|$, and $n=\kappa^{-1}dt/ds$. Differentiating $t\cdot t=1$ proves $t\perp n$. Here $|\dot x|=2e^t$ and
$$t=\tfrac12(\cos t-\sin t,\sin t+\cos t,\sqrt2),\quad n=\frac{(-2\sin t,2\cos t,\sqrt2)}{\sqrt{10}},\quad \kappa=\frac{\sqrt{10}}{4e^t}.$$

Solved by gpt-5.6-sol high.

= 4A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= Solution
{parent=4a}

<Divergence theorem> says $\int_V\nabla\cdot F\,dV=\int_{\partial V}F\cdot n\,dS$ for the outward normal. Here $\nabla\cdot(xe^{-r^2})=(3-2r^2)e^{-r^2}$, so radial integration gives $4\pi a^3e^{-a^2}$. On the sphere $F\cdot n=ae^{-a^2}$, whose <integral> is the same $4\pi a^3e^{-a^2}$.

Solved by gpt-5.6-sol high.

= 5D
{parent=Paper 3}
{scope}
{title2=Groups}

= a
{parent=5d}
{scope}

= Solution
{parent=a}

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 $\langle g\rangle$ shows $|g|\mid|G|$. Coset multiplication makes $G/N$ a <group> when $N\triangleleft G$. 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 $x\in N$.

Solved by gpt-5.6-sol high.

= b
{parent=5d}
{scope}

= Solution
{parent=b}

Every $b^m$ lies in $N$ because $(bN)^m=N$. Conversely choose $r$ with $mr\equiv1\pmod n$; for $a\in N$, $a=(a^r)^m$ because $a^n=e$.

Solved by gpt-5.6-sol high.

= 6D
{parent=Paper 3}
{scope}
{title2=Groups}

= i
{parent=6d}
{scope}

= Solution
{parent=i}

$GL_2(\mathbb C)$ consists of invertible $2\times2$ complex <matrices> and $SL_2(\mathbb C)=\ker\det$. It is normal, and the first isomorphism theorem gives $GL_2/SL_2\cong\mathbb C^*$.

Solved by gpt-5.6-sol high.

= ii
{parent=6d}
{scope}

= Solution
{parent=ii}

The Möbius <group> consists of maps $z\mapsto(az+b)/(cz+d)$ with $ad-bc\ne0$, including the usual action on $\infty$. <Scalar> <matrices> act trivially and are the whole kernel, so $M\cong GL_2(\mathbb C)/(\mathbb C^*I)$.

Solved by gpt-5.6-sol high.

= iii
{parent=6d}
{scope}

= Solution
{parent=iii}

The map $z\mapsto z-w$ sends $w$ to $0$, proving transitivity (with inversion handling $\infty$). Fixed points satisfy the quadratic $cz^2+(d-a)z-b=0$, with multiplicity and $\infty$ included, so a nonidentity has one or two. The stabiliser of $\infty$ is $S=\{z\mapsto az+b:a\ne0\}$.

Solved by gpt-5.6-sol high.

= iv
{parent=6d}
{scope}

= Solution
{parent=iv}

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.

Solved by gpt-5.6-sol high.

= 7D
{parent=Paper 3}
{scope}
{title2=Groups}

= Solution
{parent=7d}

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 $K\triangleleft G$ 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.

Solved by gpt-5.6-sol high.

= 8D
{parent=Paper 3}
{scope}
{title2=Groups}

= Solution
{parent=8d}

Disjoint-cycle decomposition follows by partitioning the set into permutation orbits. Every cycle $(a_1\ldots a_r)$ equals $(a_1a_r)\cdots(a_1a_2)$, 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 $S_n$.

Solved by gpt-5.6-sol high.

= 9A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=9a}
{scope}

= Solution
{parent=a}

Using $\varepsilon_{ijk}\varepsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$,
$[\nabla\times(F\times G)]_i=\partial_j(F_iG_j)-\partial_j(F_jG_i)$; expanding the <derivatives> gives the stated four terms.

Solved by gpt-5.6-sol high.

= b
{parent=9a}
{scope}

= Solution
{parent=b}

Irrotational means $\nabla\times E=0$. For $E=\nabla f$, its $i$th curl component is $\varepsilon_{ijk}\partial_j\partial_kf=0$ because mixed partials are symmetric.

Solved by gpt-5.6-sol high.

= c
{parent=9a}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

On $C$, $dx=a e_\phi d\phi$, so $I=\int_0^{2\pi}d\phi=2\pi$. Locally $f=\phi$, but no single-valued global potential exists on the punctured domain; this is why irrotationality does not force this closed <integral> to vanish.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

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

Solved by gpt-5.6-sol high.

= 10A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=10a}
{scope}

= Solution
{parent=a}

<Stokes theorem> is $\int_S(\nabla\times F)\cdot n\,dS=\oint_{\partial S}F\cdot dx$, with the boundary orientation given by the right-hand rule relative to $n$.

Solved by gpt-5.6-sol high.

= b
{parent=10a}
{scope}

= Solution
{parent=b}

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

Solved by gpt-5.6-sol high.

= c
{parent=10a}
{scope}

= Solution
{parent=c}

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.

Solved by gpt-5.6-sol high.

= d
{parent=10a}
{scope}

= Solution
{parent=d}

The clockwise boundary <integral> at $z=0$ is
$-\int_0^{2\pi}a^4(\sin^4\phi+\cos^4\phi)d\phi=-3\pi a^4/2$. Direct integration of the curl over the side and top gives the same value (the $h$-dependent terms cancel), verifying Stokes.

Solved by gpt-5.6-sol high.

= e
{parent=10a}
{scope}

= Solution
{parent=e}

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

Solved by gpt-5.6-sol high.

= 11A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=11a}
{scope}

= Solution
{parent=a}

For two solutions let $u$ be their difference. Green’s identity and $\partial_nu=0$ give $0=\int_Vu(\nabla^2u-m^2u)=-\int_V(|\nabla u|^2+m^2u^2)$, so $u=0$ when $m\ne0$. At $m=0$, solutions are unique only up to an additive constant.

Solved by gpt-5.6-sol high.

= b
{parent=11a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

On a simply connected region, $\nabla\times E=0$ gives $E=-\nabla\phi$. Then <Gauss's law> gives $\nabla^2\phi=-\rho/\varepsilon_0$.

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

Writing $r=|x|$ and imposing continuity gives
$$\phi(r)=\frac{\rho_0}{2\varepsilon_0}\begin{cases}2(b-a),&r\lt a,\\2b-r-a^2/r,&a\le r\le b,\\(b^2-a^2)/r,&r\gt b.\end{cases}$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

<Gauss's law> gives $E=Q\hat r/(4\pi\varepsilon_0r^2)$ 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.

Solved by gpt-5.6-sol high.

= 12A
{parent=Paper 3}
{scope}
{title2=Vector Calculus}

= a
{parent=12a}
{scope}

= Solution
{parent=a}

A rank-$k$ Cartesian <tensor> has components $T_{i_1\ldots i_k}$ transforming under an orthogonal change of <basis> $R$ as $T\prime_{i_1\ldots i_k}=R_{i_1j_1}\cdots R_{i_kj_k}T_{j_1\ldots j_k}$.

Solved by gpt-5.6-sol high.

= b
{parent=12a}
{scope}

= Solution
{parent=b}

Contracting one index from each tensor means summing them, for example $C_{i_1\ldots i_{p-1}j_1\ldots j_{q-1}}=A_{i_1\ldots i_{p-1}r}B_{rj_1\ldots j_{q-1}}$. 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.

Solved by gpt-5.6-sol high.

= c
{parent=12a}
{scope}

= Solution
{parent=c}

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

Solved by gpt-5.6-sol high.

= d
{parent=12a}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

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

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

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

Solved by gpt-5.6-sol high.

= iii
{parent=d}
{scope}

= Solution
{parent=iii}

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

Solved by gpt-5.6-sol high.

= e
{parent=12a}
{scope}

= Solution
{parent=e}

Expanding the integrand, the terms linear in $x$ integrate to zero. Therefore
$$T_{ij}=\frac{4\pi a^4e^{-a^2}}3\delta_{ij}+4\pi a^2e^{-a^2}u_iv_j,$$
so these are $\alpha$ and $\beta$.

Solved by gpt-5.6-sol high.