Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 3

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.

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 $\mathrm{Aut}(C_2)$ is trivial, so $G=N\times P\cong C_2\times C_3\cong C_6$.

Solved by gpt-5.6-sol high.

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.

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. Ifthe finite fixed points satisfy $c{z}^2+(d-a)z-b=0$. 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.

A field is irrotational when $\nabla \times F=0$. In ${\mathbb{R}}^3$ definewhere $\gamma$ 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 isfor 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.

Differentiation of $x=\rho \cos \varphi$, $y=\rho \sin \varphi$ givesThe two radial face areas are ${\rho }_0\Delta \varphi \Delta z$ and $({\rho }_0+\Delta \rho )\Delta \varphi \Delta z$. Each constant-$\varphi$ face has area $\Delta \rho \Delta z$, and each constant-$z$ face has areaThe 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 $\Delta \rho \to 0$ givesat $\rho ={\rho }_0$.

Solved by gpt-5.6-sol high.

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 $x{y}^{-1}$, 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 $k=qm+r$, $0\le r<m$, 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 $⟨x⟩\cap ⟨y⟩=\{e\}$, 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 $G=C_{13}⋊C_2$. The homomorphism $C_2\to \mathrm{Aut}(C_{13})\cong C_{12}$ 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.

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.

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.

Define $(xN)(yN)=xyN$. If representatives are changed to $x{n}_1,y{n}_2$, normality moves $n_1$ past $y$ and leaves the product in $xyN$, so multiplication is well-defined.

Let $G^{\prime }$ be the set of finite products of commutators. It is closed under products, and $[x,y{]}^{-1}=[y,x]$, so it is a subgroup. Moreoverso conjugation preserves products of commutators and $G^{\prime }◃G$. In $G/G^{\prime }$, one has $xy{G}^{\prime }=yx{G}^{\prime }$ because $(yx{)}^{-1}xy=[x,y]$, hence the quotient is abelian. Conversely, if $G/N$ is abelian, every commutator maps to the identity, so $G^{\prime }\le N$. This is the universal property of the commutator subgroup.

Solved by gpt-5.6-sol high.

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

Solved by gpt-5.6-sol high.

The subgroup $A_5^{\prime }$ is normal in $A_5$. It is nontrivial because $A_5$ is nonabelian; simplicity of $A_5$ therefore gives

Solved by gpt-5.6-sol high.

The sign map has abelian quotient, so $S_5^{\prime }\le A_5$. On the other hand $A_5^{\prime }=A_5$ and $A_5\le S_5$ imply $A_5\le S_5^{\prime }$. Hence

Solved by gpt-5.6-sol high.

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

The standard intersection-index inequality gives $[G:H\cap K]\le [G:H][G:K]=ab$. 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 areand orbit--stabilizer gives $|{\mathrm{Conj}}_G(x)|=[G:C_G(x)]$. 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 $h{g}^{-1}=ab$ with $a\in C_G(y)$ and $b\in C_G(x)$. The hint then rewrites the product aswhich is conjugate to $xy$. Therefore

Solved by gpt-5.6-sol high.

The first isomorphism theorem says that for a homomorphism $\varphi :G\to H$,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 $G{L}_n(\mathbb{R})$ consists of invertible real $n\times n$ matrices and $S{L}_n(\mathbb{R})$ consists of those with determinant one. The latter is the kernel of the surjective determinant map to ${\mathbb{R}}^{\ast }$, so it is normal and

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 $\det A=\pm 1$, and the adjugate formula proves the converse. The matrices $\left(\begin{array}{cc}1& m\\ 0& 1\end{array}\right)$ show infinitude.

Reduction modulo two is a homomorphism $G\to G{L}_2({\mathbb{F}}_2)$. 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.

Using the product rule for directional derivatives,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 $u\cdot w=-2a{\rho }^4{\sin }^{2}z$. Therefore

Solved by gpt-5.6-sol high.

Expanding with the Levi-Civita symbol and the product rule givesThe Stokes theorem stateswhere 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 $n$ leavesThe 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 isOn the positively oriented equator, $dx=R{e}_{\varphi }d\varphi$, hence $dx\times n=R{e}_{\theta }d\varphi$ and the boundary integral is also $2\pi R^2$.

Solved by gpt-5.6-sol high.

The identity follows immediately from the product rule:If ${\varphi }_1,{\varphi }_2$ have the same boundary data, put $\psi ={\varphi }_1-{\varphi }_2$. Integrating the identity with $\varphi =\psi$ givesbecause the volume equation and boundary term vanish. Positivity of $\kappa$ makes $\psi$ constant, and its boundary value makes it zero, proving uniqueness.

For any $w=\varphi +\psi$ with $\psi =0$ on the boundary, expansion givesbecause 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.

Since $q=-\kappa \nabla T$ and $\nabla \cdot q=H$, the divergence theorem givesSpherical symmetry gives $(r^2\kappa T^{\prime }{)}^{\prime }=-r^{2}H$. Regularity at zero and decay at infinity yieldFor $\alpha \le -1$, 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 $\varphi =r^{-\alpha -1}$. Applying the Dirichlet principle to any admissible $w$ gives

Solved by gpt-5.6-sol high.

Let the new orthonormal basis satisfy $e_i^{\prime }=R_{ij}{e}_j$. Then an $n$th-rank tensor transforms bySince $\partial /\partial x_i^{\prime }=R_{ij}\partial /\partial x_j$, 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

Solved by gpt-5.6-sol high.

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 givesNext contract $i=k$ and $j=l$. Sinceone obtainsSolving yields

Solved by gpt-5.6-sol high.