Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 3

The centre isEvery subgroup $H\le Z(G)$ is normal, since $gh{g}^{-1}=h$ for all $g\in G$ and $h\in H$.

Write $D_{2n}=⟨r,s:r^n=s^2=1,srs=r^{-1}⟩$. A central rotation must satisfy $r^k=r^{-k}$, and no reflection is central for $n\ge 3$. Hence

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

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

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False. The proper subgroup $2\mathbb{Z}\times 2\mathbb{Z}$ of ${\mathbb{Z}}^2$ is isomorphic to ${\mathbb{Z}}^2$ and is not cyclic.

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The parametrisation $r(x,y)=(x,y,F(x,y))$ has area factorso $\mathrm{area}(S)={\iint }_D\sqrt{1+F_x^2+F_y^2}dxdy$.

For the unbounded saddle $F=(x^2-y^2)/2$, this factor is $\sqrt{1+x^2+y^2}$. Thus the requested integral iswhich converges because its radial tail is $O(r^{-2})$.

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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 $3\sqrt{14}$.

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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

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Green's theorem states, for a positively oriented simple boundary $C=\partial R$,The defining polynomial becomesso the region is the cardioid

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The field is $e_{\theta }$, so on the polar boundary $F\cdot dr=rd\theta$. Hence

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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 $0$.

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On the boundary the integral is ${\int }_{-\pi }^{\pi }\theta dy$. Integration by parts has zero endpoint term and leaves

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The orbit and stabiliser areSuch an action can be faithful: the natural action of $S_3$ on three points is faithful, although every point stabiliser has order two.

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If $y=gx$, thenIndeed, $h$ fixes $x$ exactly when $gh{g}^{-1}$ fixes $gx$. Conjugation by $g$ is therefore the required isomorphism.

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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 $n=1$.

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If $N◃S_n$, then $N\cap A_n◃A_n$, 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 $|N|\le 2$. A normal subgroup of order two would be central, but $Z(S_n)=1$ for $n\ge 3$. Therefore the normal subgroups are

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With $D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$, its normal subgroups areThe last three proper nontrivial examples have index two; the four individual reflection subgroups are not normal.

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

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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 $\{\pm 1\}$ and the three cyclic subgroups of order four, all of index two.

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

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Given $h\in H$, choose $g\in G$ with $\varphi (g)=h$. For $k\in K$,because $K$ is normal. Thus $\varphi (K)◃H$.

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For $f(z)=(az+b)/(cz+d)$, finite fixed points obeywith 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 $u\in \mathbb{C}$,has order $m$, and these transformations are distinct as $u$ varies.

Projectivising matrices is a homomorphism. Thus $B=CA{C}^{-1}$ in $S{L}_2(\mathbb{C})$ 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).

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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

For $p=2$, the matrix $\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$ has order three. There is no element of order six: $G{L}_2({\mathbb{F}}_2)\cong S_3$, whose element orders are $1,2,3$.

For $p>2$, $S{L}_2({\mathbb{F}}_p)=\ker \det$ is a proper normal subgroup. For $p=2$, the order-three subgroup in the copy of $S_3$ is proper and normal.

In $G{L}_2({\mathbb{F}}_{11})$, takewhere $A\le {\mathbb{F}}_{11}^{\times }$ 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.

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The divergence theorem is ${\iint }_{\partial V}F\cdot ndS={\iiint }_V\nabla \cdot FdV$. 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

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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 isThe term involving $y^{2}z$ integrates to zero, and the remaining integral isconfirming part (a).

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The curl is $(2y-y^2,x,0)$, so the field is not conservative. On $C$, $z=1$ and $dz=0$, giving

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Write $p={\sum }_{j=0}^n{a}_j{x}^{n-j}{y}^j$. Equating the coefficient of $x^{n-j-2}{y}^j$ in ${\nabla }^{2}p$ givesThus 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.

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

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

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

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

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Each sine product is a Dirichlet eigenfunction. The squared wave-number sums are $24{\pi }^2$ and $30{\pi }^2$, respectively, so

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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 givesConsequentlyand on $r=R$ the surface element is obtained by omitting $dr$ and replacing $r$ by $R$.

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

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Differentiating the ball volume with respect to $R$ gives

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Reflection symmetry makes the integral zero for $i\ne j$. Rotational symmetry makes all diagonal integrals equal, and their sum is $R^2{A}_{n-1}(R)$. Therefore

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The product rule givesIntegrating and applying the divergence theorem proves the identity.

Here $\mathrm{div}T=4(x,y,z)$ and $v=x(x,y,z)$, soOn 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 giveswhose integral is $-5/6$. Thus the right side is $3/2-(-5/6)=7/3$, equal to the left side.

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