past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-3.bigb

= Paper 3
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2025/paperia_3_2025.pdf

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

= Solution
{parent=1d}

The centre is
$$Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}.$$
Every <subgroup> $H\le Z(G)$ is normal, since $ghg^{-1}=h$ for all $g\in G$ and $h\in H$.

Write $D_{2n}=\langle r,s:r^n=s^2=1, srs=r^{-1}\rangle$. A central rotation must satisfy $r^k=r^{-k}$, and no reflection is central for $n\ge3$. Hence
$$Z(D_{2n})=\begin{cases}\{1\},&n\text{ odd},\\\{1,r^{n/2}\},&n\text{ even}.\end{cases}$$

Solved by gpt-5.6-sol high.

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

= i
{parent=2d}
{scope}

= Solution
{parent=i}

False. A finite cyclic <group> has nonidentity torsion, whereas $\mathbb Z$ is torsion-free, so an injective homomorphism cannot exist.

Solved by gpt-5.6-sol high.

= ii
{parent=2d}
{scope}

= Solution
{parent=ii}

False. A surjection $C_n\to C_m$ exists exactly when $m$ divides $n$; for example there is none from $C_3$ to $C_2$.

Solved by gpt-5.6-sol high.

= iii
{parent=2d}
{scope}

= Solution
{parent=iii}

False. The proper <subgroup> $2\mathbb Z\times2\mathbb Z$ of $\mathbb Z^2$ is isomorphic to $\mathbb Z^2$ and is not cyclic.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3a}

The parametrisation $r(x,y)=(x,y,F(x,y))$ has area factor
$$|r_x\times r_y|=\sqrt{1+F_x^2+F_y^2},$$
so $\operatorname{area}(S)=\iint_D\sqrt{1+F_x^2+F_y^2}\,dx\,dy$.

For the unbounded saddle $F=(x^2-y^2)/2$, this factor is $\sqrt{1+x^2+y^2}$. Thus the requested <integral> is
$$2\pi\int_0^\infty\frac{r\,dr}{(1+r^2)^{3/2}}=2\pi,$$
which converges because its radial tail is $O(r^{-2})$.

Solved by gpt-5.6-sol high.

= i
{parent=3a}
{scope}

= Solution
{parent=i}

Here the area factor is $\sqrt{1+2^2+3^2}=\sqrt{14}$ and the triangular base has area $3$. The surface area is therefore $3\sqrt{14}$.

Solved by gpt-5.6-sol high.

= ii
{parent=3a}
{scope}

= Solution
{parent=ii}

Write the upper cap as $z=\sqrt{1-x^2-y^2}$. Its area factor is $(1-r^2)^{-1/2}$, and the projected disc has radius $\sqrt a$. Therefore
$$\operatorname{area}(S)=2\pi\int_0^{\sqrt a}\frac r{\sqrt{1-r^2}}\,dr=2\pi(1-\sqrt{1-a}).$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=4a}

Green's theorem states, for a positively oriented simple boundary $C=\partial R$,
$$\oint_C(P\,dx+Q\,dy)=\iint_R(Q_x-P_y)\,dA.$$
The defining <polynomial> becomes
$$r^2(r^2-2r\cos\theta-\sin^2\theta),$$
so the region is the <cardioid>
$$0\le r\le1+\cos\theta,\qquad-\pi\le\theta\le\pi.$$

Solved by gpt-5.6-sol high.

= i
{parent=4a}
{scope}

= Solution
{parent=i}

The field is $e_\theta$, so on the polar boundary $F\cdot dr=r\,d\theta$. Hence
$$\oint_CF\cdot dr=\int_{-\pi}^{\pi}(1+\cos\theta)\,d\theta=2\pi.$$

Solved by gpt-5.6-sol high.

= ii
{parent=4a}
{scope}

= Solution
{parent=ii}

Here $Q_x-P_y=y-2y=-y$. Green's theorem gives the <integral> of $-y$ over a region symmetric about the $x$ axis, hence the answer is $0$.

Solved by gpt-5.6-sol high.

= iii
{parent=4a}
{scope}

= Solution
{parent=iii}

On the boundary the <integral> is $\int_{-\pi}^{\pi}\theta\,dy$. <Integration by parts> has zero endpoint term and leaves
$$-\int_{-\pi}^{\pi}y\,d\theta=-\int_{-\pi}^{\pi}(1+\cos\theta)\sin\theta\,d\theta=0.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=5d}
{scope}

= i
{parent=a}
{scope}

= Solution
{parent=i}

The orbit and stabiliser are
$$Gx=\{gx:g\in G\},\qquad \operatorname{Stab}_G(x)=\{g\in G:gx=x\}.$$
Such an action can be faithful: the natural action of $S_3$ on three points is faithful, although every point stabiliser has order two.

Solved by gpt-5.6-sol high.

= ii
{parent=a}
{scope}

= Solution
{parent=ii}

If $y=gx$, then
$$\operatorname{Stab}_G(y)=g\operatorname{Stab}_G(x)g^{-1}.$$
Indeed, $h$ fixes $x$ exactly when $ghg^{-1}$ fixes $gx$. Conjugation by $g$ is therefore the required isomorphism.

Solved by gpt-5.6-sol high.

= b
{parent=5d}
{scope}

= Solution
{parent=b}

For $1\le k<n$, any nonidentity permutation moves some $k$-subset: choose a moved point and complete a subset so that it contains that point but not its image. Thus the kernel is trivial. For $k=n$ the action is trivial, so it is faithful only in the degenerate case $n=1$.

Solved by gpt-5.6-sol high.

= c
{parent=5d}
{scope}

= Solution
{parent=c}

If $N\triangleleft S_n$, then $N\cap A_n\triangleleft A_n$, hence the intersection is either $1$ or $A_n$. In the latter case $N$ is $A_n$ or $S_n$. In the former, $N$ injects into $S_n/A_n\cong C_2$, so $|N|\le2$. A <normal subgroup> of order two would be central, but $Z(S_n)=1$ for $n\ge3$. Therefore the normal <subgroups> are
$$1,\quad A_n,\quad S_n.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=6d}
{scope}

= Solution
{parent=a}

With $D_8=\langle r,s:r^4=s^2=1, srs=r^{-1}\rangle$, its normal <subgroups> are
$$1,\qquad\langle r^2\rangle,\qquad\langle r\rangle,\qquad
\langle r^2,s\rangle,\qquad\langle r^2,rs\rangle,\qquad D_8.$$
The last three proper nontrivial examples have index two; the four individual reflection <subgroups> are not normal.

Solved by gpt-5.6-sol high.

= b
{parent=6d}
{scope}

= Solution
{parent=b}

If $K\triangleleft G$, the quotient map $G\to G/K$ is surjective with kernel $K$. Conversely, every kernel is normal because $\phi(gkg^{-1})=\phi(g)1\phi(g)^{-1}=1$.

Solved by gpt-5.6-sol high.

= c
{parent=6d}
{scope}

= Solution
{parent=c}

The statement is false. The <quaternion group> $Q_8$ is nonabelian, but each of its <subgroups> is normal: its nontrivial proper <subgroups> are its centre $\{\pm1\}$ and the three cyclic <subgroups> of order four, all of index two.

Solved by gpt-5.6-sol high.

= d
{parent=6d}
{scope}

= i
{parent=d}
{scope}

= Solution
{parent=i}

For $g\in G$ and $x\in\phi^{-1}(N)$,
$$\phi(gxg^{-1})=\phi(g)\phi(x)\phi(g)^{-1}\in N,$$
so the preimage is normal.

Solved by gpt-5.6-sol high.

= ii
{parent=d}
{scope}

= Solution
{parent=ii}

Given $h\in H$, choose $g\in G$ with $\phi(g)=h$. For $k\in K$,
$$h\phi(k)h^{-1}=\phi(gkg^{-1})\in\phi(K),$$
because $K$ is normal. Thus $\phi(K)\triangleleft H$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=7d}

For $f(z)=(az+b)/(cz+d)$, finite fixed points obey
$$cz^2+(d-a)z-b=0,$$
with the point at infinity included in the usual way when appropriate. The fundamental theorem of algebra on the Riemann sphere gives at least one fixed point. Unless $f$ is the identity, the equation is nonzero of degree at most two, so there are one or two distinct fixed points.

Let $\zeta$ be a primitive $m$th root of unity. For every $u\in\mathbb C$,
$$f_u(z)=u+\zeta(z-u)$$
has order $m$, and these transformations are distinct as $u$ varies.

Projectivising <matrices> is a homomorphism. Thus $B=CAC^{-1}$ in $SL_2(\mathbb C)$ implies $g=[C]f[C]^{-1}$ in the Möbius <group>.

The converse as stated is false because <matrix> representatives may be rescaled: $A=I$ and $B=2I$ define the same Möbius transformation, while neither $B$ nor $-B$ is conjugate to $A$ (their traces differ).

Solved by gpt-5.6-sol high.

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

= Solution
{parent=8d}

Invertible <matrices> are closed under multiplication, contain $I$, have associative multiplication, and have inverses by the adjugate formula over the field $\mathbb F_p$. The first column can be any nonzero <vector> and the second any <vector> outside its span, so
$$|GL_2(\mathbb F_p)|=(p^2-1)(p^2-p).$$

For $p=2$, the <matrix> $\begin{pmatrix}0&1\\1&1\end{pmatrix}$ has order three. There is no element of order six: $GL_2(\mathbb F_2)\cong S_3$, whose element orders are $1,2,3$.

For $p>2$, $SL_2(\mathbb F_p)=\ker\det$ is a proper normal <subgroup>. For $p=2$, the order-three <subgroup> in the copy of $S_3$ is proper and normal.

In $GL_2(\mathbb F_{11})$, take
$$H=\left\{\begin{pmatrix}a&b\\0&1\end{pmatrix}:a\in A, b\in\mathbb F_{11}\right\},$$
where $A\le\mathbb F_{11}^{\times}$ is the <subgroup> of order five. This <semidirect product> has order $55$ and is nonabelian because a nontrivial diagonal element does not commute with translations.

Solved by gpt-5.6-sol high.

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

= a
{parent=9a}
{scope}

= Solution
{parent=a}

The <divergence theorem> is $\iint_{\partial V}F\cdot n\,dS=\iiint_V\nabla\cdot F\,dV$. Here $\nabla\cdot F=3z+2yz$. The $y$ term integrates to zero by symmetry, while the ellipse has area $\pi/\sqrt{ab}$. Hence the flux is
$$\frac\pi{\sqrt{ab}}\int_0^3 3z\,dz=\frac{27\pi}{2\sqrt{ab}}.$$

Solved by gpt-5.6-sol high.

= b
{parent=9a}
{scope}

= Solution
{parent=b}

The top and bottom fluxes cancel because $F_z=x^2+y^2$ is independent of $z$. Parametrise the side by $(\cos\theta/\sqrt a,\sin\theta/\sqrt b,z)$; its outward <vector> area is
$$\left(\frac{\cos\theta}{\sqrt b},\frac{\sin\theta}{\sqrt a},0\right)d\theta\,dz.$$
The term involving $y^2z$ integrates to zero, and the remaining <integral> is
$$\int_0^3\int_0^{2\pi}\frac{3z\cos^2\theta}{\sqrt{ab}}\,d\theta\,dz
=\frac{27\pi}{2\sqrt{ab}},$$
confirming part (a).

Solved by gpt-5.6-sol high.

= c
{parent=9a}
{scope}

= Solution
{parent=c}

The curl is $(2y-y^2,x,0)$, so the field is not conservative. On $C$, $z=1$ and $dz=0$, giving
$$\oint_C(3x\,dx+y^2\,dy)=\oint_Cd\left(\frac32x^2+\frac13y^3\right)=0.$$

Solved by gpt-5.6-sol high.

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

= a
{parent=10a}
{scope}

= Solution
{parent=a}

Write $p=\sum_{j=0}^na_jx^{n-j}y^j$. Equating the coefficient of $x^{n-j-2}y^j$ in $\nabla^2p$ gives
$$(n-j)(n-j-1)a_j+(j+2)(j+1)a_{j+2}=0.$$
Thus all even coefficients are determined by $a_0$ and all odd coefficients by $a_1$, with no further constraints. These two choices give two independent harmonic homogeneous <polynomials>.

Solved by gpt-5.6-sol high.

= b
{parent=10a}
{scope}

= i
{parent=b}
{scope}

= Solution
{parent=i}

For radial <functions>, $\nabla^2h=h''+2h'/r$. Since $\nabla^2e^{-r}=e^{-r}-2e^{-r}/r$ and $\nabla^2r^{-4}=12r^{-6}$, decay and the boundary value give
$$u=e^{-r}+r^{-4}-\frac{e^{-1}}r.$$

Solved by gpt-5.6-sol high.

= ii
{parent=b}
{scope}

= Solution
{parent=ii}

The angular Laplacian sends $\sin\theta$ to $\cos(2\theta)/\sin\theta$, while $1/r$ is radially harmonic. Therefore
$$u=\frac{\sin\theta}{r}.$$

Solved by gpt-5.6-sol high.

= iii
{parent=b}
{scope}

= Solution
{parent=iii}

The $\phi$ part contributes $-\sin\phi/(r^2\sin^2\theta)$, and again $1/r$ is radially harmonic. Hence
$$u=\frac{\sin\phi}{r}.$$

Solved by gpt-5.6-sol high.

= c
{parent=10a}
{scope}

= i
{parent=c}
{scope}

= Solution
{parent=i}

Set $A=x(1-x)$, $B=y(1-y)$ and $C=z(1-z)$. The forcing is $AB+AC+BC$, while $\nabla^2(ABC)=-2(AB+AC+BC)$. Thus
$$u=-\frac12x(1-x)y(1-y)z(1-z),$$
which vanishes on every face.

Solved by gpt-5.6-sol high.

= ii
{parent=c}
{scope}

= Solution
{parent=ii}

Each sine product is a Dirichlet eigenfunction. The squared wave-number sums are $24\pi^2$ and $30\pi^2$, respectively, so
$$u=-\frac{\sin(2\pi x)\sin(2\pi y)\sin(4\pi z)}{24\pi^2}
+\frac{\sin(2\pi x)\sin(\pi y)\sin(5\pi z)}{30\pi^2}.$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=11a}

For coordinates $x=x(q)$, the Jacobian is $J_{ij}=\partial x_i/\partial q_j$. Expanding after separating $x_1=r\cos\theta_1$ from the remaining coordinates gives
$$|J_n|=r\sin^{n-2}\theta_1\,|J_{n-1}|.$$
Consequently
$$dV=r^{n-1}\prod_{j=1}^{n-2}\sin^{n-1-j}\theta_j\,dr\,d\theta_1\cdots d\theta_{n-2}\,d\phi,$$
and on $r=R$ the surface element is obtained by omitting $dr$ and replacing $r$ by $R$.

Solved by gpt-5.6-sol high.

= i
{parent=11a}
{scope}

= Solution
{parent=i}

The ball volume is
$$V_n(R)=\frac{\pi^{n/2}R^n}{\Gamma(n/2+1)}.$$
For $n=2m$ this is $\pi^mR^{2m}/m!$; for $n=2m+1$ it is $2^{2m+1}m!\pi^mR^{2m+1}/(2m+1)!$.

Solved by gpt-5.6-sol high.

= ii
{parent=11a}
{scope}

= Solution
{parent=ii}

Differentiating the ball volume with respect to $R$ gives
$$A_{n-1}(R)=\frac{2\pi^{n/2}R^{n-1}}{\Gamma(n/2)}=\frac nR V_n(R).$$

Solved by gpt-5.6-sol high.

= iii
{parent=11a}
{scope}

= Solution
{parent=iii}

Reflection symmetry makes the <integral> zero for $i\ne j$. Rotational symmetry makes all diagonal <integrals> equal, and their sum is $R^2A_{n-1}(R)$. Therefore
$$\int_{r=R}x_ix_j\,dS=\delta_{ij}\frac{R^2A_{n-1}(R)}n
=\delta_{ij}\frac{\pi^{n/2}R^{n+1}}{\Gamma(n/2+1)}.$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=12a}

The product rule gives
$$\partial_j(T_{ij}v_i)=(\partial_jT_{ij})v_i+T_{ij}\partial_jv_i.$$
Integrating and applying the divergence theorem proves the identity.

Here $\operatorname{div}T=4(x,y,z)$ and $v=x(x,y,z)$, so
$$\int_V\operatorname{div}T\cdot v\,dV
=4\int_Vx(x^2+y^2+z^2)\,dV=\frac73.$$
On the boundary, $(Tn)\cdot v=x(x\cdot n)(x^2+y^2+z^2-1)$. The nonzero contributions from the faces $x=1,y=1,z=1$ are respectively $2/3,5/12,5/12$, totaling $3/2$.

Finally, direct contraction gives
$$T_{ij}\partial_jv_i=2x(x^2+y^2+z^2)-4x,$$
whose <integral> is $-5/6$. Thus the right side is $3/2-(-5/6)=7/3$, equal to the left side.

Solved by gpt-5.6-sol high.