past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-3.bigb

= Paper 3
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2023/paperia_3_2023.pdf

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

= i
{parent=1d}
{scope}

= Solution
{parent=i}

False. Take $G=C_2\times C_2$ and let $N$ be either factor. Then $N\cong C_2$ and $G/N\cong C_2$, but $G$ is not cyclic and hence is not $C_4$.

Solved by gpt-5.6-sol high.

= ii
{parent=1d}
{scope}

= Solution
{parent=ii}

True. The <group> has order six. A Sylow $3$-subgroup $P$ has order three, while the given normal <subgroup> $N$ has order two. Their intersection is trivial and $NP=G$. Conjugation by $P$ on $N$ is trivial because $\operatorname{Aut}(C_2)$ is trivial, so $G=N\times P\cong C_2\times C_3\cong C_6$.

Solved by gpt-5.6-sol high.

= iii
{parent=1d}
{scope}

= Solution
{parent=iii}

False. Let $G=C_4$ and $H=C_2\times C_2$, and in each choose a <subgroup> of order two. Then the chosen normal <subgroups> are isomorphic and both quotients are $C_2$, while $G$ and $H$ are not isomorphic.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=2D}

Existence follows from the three-transitivity of <Möbius transformations>: one can send three distinct source points to $0,1,\infty$ and then send those to the three distinct target points.

For uniqueness, suppose $S$ and $T$ have the same values at three distinct points. Then $M=T^{-1}S$ fixes those points. If
$$
M(z)=\frac{az+b}{cz+d},
$$
the finite fixed points satisfy $cz^2+(d-a)z-b=0$. A nonidentity Möbius map therefore has at most two fixed points on the Riemann sphere. Since $M$ has three, it is the identity, and $S=T$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=3B}

A field is irrotational when $\nabla\times F=0$. In $\mathbb R^3$ define
$$
V(x)=-\int_\gamma F\cdot d\ell,
$$
where $\gamma$ is any path from $x_0$ to $x$. The difference between two choices is the <circulation> around a closed loop. By the <Stokes theorem> it is the flux of $\nabla\times F$ through a spanning surface and hence zero. Thus $V$ is well-defined, $V(x_0)=0$, and <differentiation> gives $F=-\nabla V$.

If $\nabla\times A_0=B$, every solution is
$$
A=A_0+\nabla\chi
$$
for an arbitrary <differentiable> <scalar> $\chi$, since the difference is curl-free. A necessary condition is $\nabla\cdot B=0$, because the divergence of a curl vanishes.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=4B}

<Differentiation> of $x=\rho\cos\phi$, $y=\rho\sin\phi$ gives
$$
h_\rho=1,\quad e_\rho=(\cos\phi,\sin\phi,0),\qquad
h_\phi=\rho,\quad e_\phi=(-\sin\phi,\cos\phi,0).
$$
The two radial face areas are $\rho_0\Delta\phi\Delta z$ and $(\rho_0+\Delta\rho)\Delta\phi\Delta z$. Each constant-$\phi$ face has area $\Delta\rho\Delta z$, and each constant-$z$ face has area
$$
\frac12\{(\rho_0+\Delta\rho)^2-\rho_0^2\}\Delta\phi.
$$
The volume is this last expression times $\Delta z$.

Only the radial faces contribute to the flux of $F(\rho)e_\rho$. Dividing their difference by the volume and taking $\Delta\rho\to0$ gives
$$
\frac1\rho\frac{d}{d\rho}\{\rho F(\rho)\}
$$
at $\rho=\rho_0$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=5D}

The <Lagrange theorem> states that $|H|$ divides $|G|$ for every <subgroup> $H$ of a finite <group> $G$, with $|G|=|G:H||H|$. The left cosets partition $G$, and multiplication by a representative bijects $H$ with each coset, proving the formula.

The intersection $H\cap K$ contains the identity and is closed under $xy^{-1}$, so it is a <subgroup>. Its order divides both $|H|$ and $|K|$; if these are coprime, $H\cap K=\{e\}$.

The order $o(x)$ is the least positive $m$ with $x^m=e$. Division $k=qm+r$, $0\leq r<m$, shows that $x^k=e$ exactly when $r=0$, so exactly when $o(x)\mid k$.

If $x,y$ commute and have coprime orders $m,n$, then $(xy)^{mn}=e$. Conversely $(xy)^r=e$ implies $x^r=y^{-r}$ lies in $\langle x\rangle\cap\langle y\rangle=\{e\}$, so $m,n\mid r$ and $mn\mid r$. Hence $o(xy)=mn$.

<Cauchy theorem for groups> says that every prime divisor $p$ of $|G|$ occurs as the order of an element of $G$. For $|G|=26$, the Sylow $13$-subgroup $P\cong C_{13}$ is unique and normal, and Cauchy's theorem supplies a complement $C_2$. Thus $G=C_{13}\rtimes C_2$. The homomorphism $C_2\to\operatorname{Aut}(C_{13})\cong C_{12}$ is either trivial or has the unique image of order two, inversion. These give exactly $C_{26}$ and the dihedral <group> of order $26$.

Solved by gpt-5.6-sol high.

= i
{parent=5d}
{scope}

= Solution
{parent=i}

No. In $S_3$, take $x=(12)$ and $y=(123)$. Their orders are two and three, but $xy$ is a transposition and has order two.

Solved by gpt-5.6-sol high.

= ii
{parent=5d}
{scope}

= Solution
{parent=ii}

No. In any <group> choose a nonidentity $x$ and put $y=x^{-1}$. They commute, but $xy=e$ has order one rather than $o(x)o(y)$.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=6D}

Define $(xN)(yN)=xyN$. If representatives are changed to $xn_1,yn_2$, normality moves $n_1$ past $y$ and leaves the product in $xyN$, so multiplication is well-defined.

Let $G'$ be the set of finite products of commutators. It is closed under products, and $[x,y]^{-1}=[y,x]$, so it is a <subgroup>. Moreover
$$
g^{-1}[x,y]g=[g^{-1}xg,g^{-1}yg],
$$
so conjugation preserves products of commutators and $G'\triangleleft G$. In $G/G'$, one has $xyG'=yxG'$ because $(yx)^{-1}xy=[x,y]$, hence the quotient is abelian. Conversely, if $G/N$ is abelian, every commutator maps to the identity, so $G'\leq N$. This is the universal property of the <commutator subgroup>.

Solved by gpt-5.6-sol high.

= i
{parent=6d}
{scope}

= Solution
{parent=i}

Writing $D_8=\langle r,s:r^4=s^2=e, srs=r^{-1}\rangle$, one has $[r,s]=r^{-2}=r^2$. The quotient by $\langle r^2\rangle$ is $C_2\times C_2$, so
$$
D_8'=\{e,r^2\}.
$$

Solved by gpt-5.6-sol high.

= ii
{parent=6d}
{scope}

= Solution
{parent=ii}

The <subgroup> $A_5'$ is normal in $A_5$. It is nontrivial because $A_5$ is nonabelian; simplicity of $A_5$ therefore gives
$$
A_5'=A_5.
$$

Solved by gpt-5.6-sol high.

= iii
{parent=6d}
{scope}

= Solution
{parent=iii}

The sign map has abelian quotient, so $S_5'\leq A_5$. On the other hand $A_5'=A_5$ and $A_5\leq S_5$ imply $A_5\leq S_5'$. Hence
$$
S_5'=A_5.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=7D}

The map $H\times K\to HK$, $(h,k)\mapsto hk$, has fibers of size $|H\cap K|$: two pairs have the same product exactly when their quotient is represented by an element of the intersection. Hence
$$
|HK|=\frac{|H||K|}{|H\cap K|}.
$$

The standard intersection-index inequality gives $[G:H\cap K]\leq [G:H][G:K]=ab$. This index is divisible by both coprime numbers $a$ and $b$, so it equals $ab$. The displayed formula then gives $|HK|=|G|$, hence $HK=G$.

The conjugacy class and centralizer are
$$
\operatorname{Conj}_G(x)=\{g^{-1}xg:g\in G\},
\qquad C_G(x)=\{g:gx=xg\},
$$
and orbit--stabilizer gives $|\operatorname{Conj}_G(x)|=[G:C_G(x)]$. Thus the coprimality hypothesis gives $C_G(x)C_G(y)=G$.

One inclusion in the desired equality follows by conjugating $xy$. Conversely take $g^{-1}xg\,h^{-1}yh$. Write $hg^{-1}=ab$ with $a\in C_G(y)$ and $b\in C_G(x)$. The hint then rewrites the product as
$$
h^{-1}a(xy)a^{-1}h,
$$
which is conjugate to $xy$. Therefore
$$
\operatorname{Conj}_G(xy)=
\operatorname{Conj}_G(x)\operatorname{Conj}_G(y).
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=8D}

The <first isomorphism theorem> says that for a homomorphism $\phi:G\to H$,
$$
G/\ker\phi\cong\operatorname{im}\phi,
\qquad g\ker\phi\mapsto\phi(g).
$$
The map is well-defined because equal cosets differ by a kernel element; it is a surjective homomorphism onto the image, and injectivity follows because its kernel is the identity coset.

Here $GL_n(\mathbb R)$ consists of invertible real $n\times n$ <matrices> and $SL_n(\mathbb R)$ consists of those with <determinant> one. The latter is the kernel of the surjective <determinant> map to $\mathbb R^*$, so it is normal and
$$
GL_n(\mathbb R)/SL_n(\mathbb R)\cong\mathbb R^*.
$$

The given <integral> <matrices> form a <group> because products and inverses remain <integral>. The inverse formula shows that an <integral> inverse exists exactly when the <determinant> divides every cofactor; taking <determinants> shows more directly that $\det A\det A^{-1}=1$, hence $\det A=\pm1$, and the adjugate formula proves the converse. The <matrices> $\left(\begin{smallmatrix}1&m\\0&1\end{smallmatrix}\right)$ show infinitude.

Reduction modulo two is a homomorphism $G\to GL_2(\mathbb F_2)$. Its kernel is exactly $H$, so $H$ is normal; its index is finite because the target has only six elements.

Solved by gpt-5.6-sol high.

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

= Solution
{parent=9B}

Using the product rule for directional <derivatives>,
$$
\partial_t(u\cdot w)
=(w\cdot\nabla)\left(\frac12|u|^2-P\right)
+(u\cdot\nabla)(-u\cdot w).
$$
Thus $f=|u|^2/2-P$ and $g=-h$. If $\nabla\cdot u=0$ and $w=\nabla\times u$, then $\nabla\cdot w=0$. Integrating the two directional <derivatives> as divergences gives only boundary fluxes, which vanish because $u\cdot n=w\cdot n=0$. This is <helicity conservation by tangent boundary conditions>, so $dH/dt=0$.

For the stated field, direct use of the cylindrical curl gives $u\cdot w=-2a\rho^4\sin^2z$. Therefore
$$
H=-2a\int_0^{2\pi}d\phi\int_0^\pi\sin^2z\,dz
\int_0^a\rho^5d\rho
=-\frac{\pi^2a^7}{3}.
$$

Solved by gpt-5.6-sol high.

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

= Solution
{parent=10B}

Expanding with the Levi-Civita symbol and the product rule gives
$$
\nabla\times(a\times b)=a\,\nabla\cdot b-b\,\nabla\cdot a
+(b\cdot\nabla)a-(a\cdot\nabla)b.
$$
The <Stokes theorem> states
$$
\int_S(\nabla\times F)\cdot n\,dS=\oint_C F\cdot dx,
$$
where the boundary orientation follows the right-hand rule about $n$.

Apply it to $F=m\times v$. On $S$, $m=n$ and $v\cdot n=0$. Substitution of the <vector> identity and contraction with $n$ leaves
$$
(\delta_{ij}-n_in_j)\partial_jv_i.
$$
The boundary integrand satisfies $(n\times v)\cdot dx=v\cdot(dx\times n)$, proving the formula. The <vector> $dx\times n$ is the outward co-normal in the tangent plane of $S$.

For the hemisphere, $n=e_r$ and $v=r\sin\theta e_\theta$. The projected divergence is $2\cos\theta$, so the left side is
$$
\int_0^{2\pi}\int_0^{\pi/2}2\cos\theta R^2\sin\theta\,d\theta d\phi=2\pi R^2.
$$
On the positively oriented equator, $dx=Re_\phi d\phi$, hence $dx\times n=Re_\theta d\phi$ and the boundary <integral> is also $2\pi R^2$.

Solved by gpt-5.6-sol high.

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

= a
{parent=11b}
{scope}

= Solution
{parent=a}

The identity follows immediately from the product rule:
$$
\nabla\cdot(\kappa\psi\nabla\phi)
=\psi\nabla\cdot(\kappa\nabla\phi)+\kappa\nabla\psi\cdot\nabla\phi.
$$
If $\phi_1,\phi_2$ have the same boundary data, put $\psi=\phi_1-\phi_2$. Integrating the identity with $\phi=\psi$ gives
$$
\int_V\kappa|\nabla\psi|^2dV=0
$$
because the volume equation and boundary term vanish. Positivity of $\kappa$ makes $\psi$ constant, and its boundary value makes it zero, proving uniqueness.

For any $w=\phi+\psi$ with $\psi=0$ on the boundary, expansion gives
$$
\int\kappa|\nabla w|^2
=\int\kappa|\nabla\phi|^2+int\kappa|\nabla\psi|^2,
$$
because the cross term vanishes by the same integration by parts. This proves the inequality and the <Dirichlet principle>.

Solved by gpt-5.6-sol high.

= b
{parent=11b}
{scope}

= Solution
{parent=b}

Since $q=-\kappa\nabla T$ and $\nabla\cdot q=H$, the divergence theorem gives
$$
\oint_Sq\cdot dS=\int_VH\,dV.
$$
Spherical symmetry gives $(r^2\kappa T')'=-r^2H$. Regularity at zero and decay at infinity yield
$$
T(r)=\begin{cases}
\displaystyle \frac{H_0}{3(\alpha+1)}+\frac{H_0}{3(2-\alpha)}(1-r^{2-\alpha}),&r\leq1,\\[6pt]
\displaystyle \frac{H_0}{3(\alpha+1)}r^{-\alpha-1},&r>1.
\end{cases}
$$
For $\alpha\leq-1$, the exterior solution cannot tend to zero: it is logarithmic at $-1$ and grows below it.

On the exterior domain, the energy-minimizing solution with boundary value one is $\phi=r^{-\alpha-1}$. Applying the <Dirichlet principle> to any admissible $w$ gives
$$
\int_1^\infty r^{\alpha+2}(w')^2dr
\geq\int_1^\infty r^{\alpha+2}(\phi')^2dr
=\alpha+1.
$$

Solved by gpt-5.6-sol high.

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

= a
{parent=12b}
{scope}

= Solution
{parent=a}

Let the new orthonormal <basis> satisfy $e'_i=R_{ij}e_j$. Then an $n$th-rank tensor transforms by
$$
T'_{i_1\ldots i_n}=R_{i_1j_1}\cdots R_{i_nj_n}T_{j_1\ldots j_n}.
$$
Since $\partial/\partial x'_i=R_{ij}\partial/\partial x_j$, the Hessian of a <scalar> obeys this law with two factors of $R$.

An isotropic tensor is unchanged under every rotation. In three dimensions the most general isotropic tensors of ranks zero through three are respectively
$$
a,\qquad0,\qquad b\delta_{ij},\qquad c\epsilon_{ijk}.
$$

Solved by gpt-5.6-sol high.

= b
{parent=12b}
{scope}

= Solution
{parent=b}

The measure, Gaussian factor, $r^{-1}$, coordinates, and <derivatives> all transform covariantly under rotations, so $T_{ijkl}$ is an <isotropic tensor integral>. Symmetry gives $\beta=\gamma$.

Contracting $i=j$ and using $\nabla^2(1/r)=0$ away from the origin gives
$$
3\alpha+2\beta=0.
$$
Next contract $i=k$ and $j=l$. Since
$$
\sum_i x_i\partial_i\partial_j(r^{-1})=\frac{2x_j}{r^3},
$$
one obtains
$$
3\alpha+12\beta
=2\int_{\mathbb R^3}\frac{e^{-r^2}}r,dV
=4\pi.
$$
Solving yields
$$
\boxed{\alpha=-\frac{4\pi}{15},\qquad
\beta=\gamma=\frac{2\pi}{5}}.
$$

Solved by gpt-5.6-sol high.