= Paper 3 {scope} https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2022/paperia_3_2022.pdf = 1E {parent=Paper 3} {scope} {title2=Groups} = Solution {parent=1E} The $g$ is the least $n$ such that $g^n=e$, where $e$ is the ; its order is infinite if no such $n$ exists. Let $g$ have finite order $n$. A preserves the and the identity, so $$ \phi(g)^n=\phi(g^n)=\phi(e)=e. $$ The order of an element divides every positive exponent that gives the identity. Hence $$ \boxed{\operatorname{ord}(\phi(g))\mid\operatorname{ord}(g)}. $$ If $\phi$ is a and $h\in H$ has order $m$, choose $g\in G$ with $\phi(g)=h$. The first result gives $m\mid n$, where $n=\operatorname{ord}(g)$. The element $$ g^{n/m} $$ then has order $m$, because the order of $g^k$ is $n/\gcd(n,k)$. A $C_9\to S_4$ is determined by the image of a [generator] of the $C_9$, and that image must have order dividing $9$. In the $S_4$, the only such elements are the identity and the . There are $$ \binom43(3-1)!=4\cdot2=8 $$ three-cycles. Therefore the number of homomorphisms is $$ \boxed{1+8=9}. $$ Solved by gpt-5.6-sol high. = 2E {parent=Paper 3} {scope} {title2=Groups} = Solution {parent=2E} A $G$ is [abelian] when $xy=yx$ for every $x,y\in G$. It is [cyclic] when some $g$ satisfies $$ G=\langle g\rangle=\{g^n:n\in\mathbb Z\}. $$ If $x=g^r$ and $y=g^s$ belong to a cyclic group, then $$ xy=g^{r+s}=g^{s+r}=yx, $$ so every cyclic group is abelian. The $C_2\times C_2$ is abelian, but every nonidentity element has order two, so no element generates all four elements. Thus an abelian group need not be cyclic. The condition on proper does not force $G$ to be abelian. The $$ Q_8=\{\pm1,\pm i,\pm j,\pm k\} $$ is nonabelian because $ij=k$ whereas $ji=-k$. Its proper subgroups are the trivial subgroup, $\{\pm1\}$, and the three cyclic subgroups $$ \langle i\rangle,\qquad\langle j\rangle,\qquad\langle k\rangle, $$ each of order four. They are all cyclic, so $$ \boxed{\text{the answer is no}}. $$ Solved by gpt-5.6-sol high. = 3A {parent=Paper 3} {scope} {title2=Vector Calculus} = Solution {parent=3A} Set $$ u=\frac{x}{y},\qquad v=xy. $$ Because the region lies in the , its four inequalities become $$ 1\leq u\leq\alpha,\qquad 1\leq v\leq\alpha. $$ Thus this [change of variables] sends $D$ to a rectangle in the $uv$-plane. Its is $$ \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 1/y&-x/y^2\\ y&x \end{vmatrix} =\frac{2x}{y}=2u, $$ so $$ dx\,dy=\frac{du\,dv}{2u}. $$ Since $x^2=uv$, the is $$ \begin{aligned} \iint_Dx^2\,dx\,dy &=\int_1^\alpha\int_1^\alpha uv\,\frac{du\,dv}{2u}\\ &=\frac12\left(\int_1^\alpha du\right) \left(\int_1^\alpha v\,dv\right)\\ &=\boxed{\frac{(\alpha-1)(\alpha^2-1)}4}. \end{aligned} $$ Solved by gpt-5.6-sol high. = 4A {parent=Paper 3} {scope} {title2=Vector Calculus} = Solution {parent=4A} Use $x>0$ as the parameter of the : $$ \mathbf r(x)=(x,\log x,0). $$ Its and speed are $$ \mathbf r'(x)=\left(1,\frac1x,0\right), \qquad |\mathbf r'(x)|=\frac{\sqrt{x^2+1}}x. $$ The is therefore $$ \boxed{\mathbf t(x)=\frac{(x,1,0)}{\sqrt{x^2+1}}}. $$ The parameter-independent formula for the gives $$ \kappa(x) =\frac{|\mathbf r'(x)\times\mathbf r''(x)|} {|\mathbf r'(x)|^3} =\boxed{\frac{x}{(x^2+1)^{3/2}}}. $$ Differentiating, $$ \kappa'(x)=\frac{1-2x^2}{(x^2+1)^{5/2}}. $$ Hence $\kappa$ increases for $01/\sqrt2$. Its occurs at $x=1/\sqrt2$ and equals $$ \boxed{\kappa_{\max}=\frac{2}{3\sqrt3}}. $$ The corresponding point is $$ \boxed{\left(\frac1{\sqrt2},-\frac12\log2,0\right)}. $$ Solved by gpt-5.6-sol high. = 5E {parent=Paper 3} {scope} {title2=Groups} = Solution {parent=5E} A of $G$ on $X$ is a map $(g,x)\mapsto gx$ such that $ex=x$ and $(gh)x=g(hx)$. The [orbit] and [stabilizer] of $x$ are $$ Gx=\{gx:g\in G\}, \qquad G_x=\{g\in G:gx=x\}. $$ For a , the map $$ gG_x\longmapsto gx $$ is a well-defined from the left cosets of $G_x$ to $Gx$. Each coset has $|G_x|$ elements, so the is $$ \boxed{|G|=|Gx|\,|G_x|}. $$ The says that if a $p$ divides $|G|$, then $G$ has an element of order $p$. To prove it, let $$ X=\{(g_1,\ldots,g_p)\in G^p:g_1\cdots g_p=e\}. $$ The first $p-1$ entries determine the last, so $|X|=|G|^{p-1}$, which is divisible by $p$. The $C_p$ acts on $X$ by cyclically rotating the entries; rotation preserves the product condition because $$ g_2\cdots g_pg_1=g_1^{-1}(g_1\cdots g_p)g_1=e. $$ Every orbit has size one or $p$. The fixed points are exactly the tuples $(g,\ldots,g)$ with $g^p=e$. Their number is therefore divisible by $p$. Since the identity gives one fixed point, there is another, and its entry has order $p$. Now let $|G|=33$. Cauchy's theorem gives a subgroup $H$ of order $11$. Let $H$ act by the on the set $\mathcal X$ of subgroups of order $11$. It fixes $H$. If it also fixed $K\ne H$, then $H$ would normalize $K$; since $H\cap K=\{e\}$, the product $HK$ would be a subgroup of order $121$, which is impossible. Every other $H$-orbit in $\mathcal X$ therefore has size $11$, so $$ |\mathcal X|\equiv1\pmod{11}. $$ Distinct members of $\mathcal X$ share only the identity and each contributes ten nonidentity elements. Hence $1+10|\mathcal X|\leq33$, forcing $|\mathcal X|=1$. Thus $H$ is a . Conjugation now defines a $$ G\longrightarrow\operatorname{Aut}(H). $$ Because $H\cong C_{11}$, its has order $10$. By the , the image has order dividing both $33$ and $10$, so the image is trivial and $H$ lies in the
[center] of $G$. Cauchy's theorem also supplies $x$ of order $3$. If $h$ generates $H$, then $h$ and $x$ commute and $hx$ has order $\operatorname{lcm}(11,3)=33$. Therefore $$ \boxed{G=\langle hx\rangle\cong C_{33}}. $$ Solved by gpt-5.6-sol high. = 6E {parent=Paper 3} {scope} {title2=Groups} = Solution {parent=6E} A of the is a map $$ z\longmapsto\frac{az+b}{cz+d}, \qquad ad-bc\ne0, $$ with its natural values at the pole and at infinity. For three distinct points $z_1,z_2,z_3$, define $$ T_z(\zeta)= \frac{(\zeta-z_2)(z_1-z_3)} {(\zeta-z_3)(z_1-z_2)}. $$ The usual limiting conventions cover an infinite $z_i$. This Möbius transformation sends $(z_1,z_2,z_3)$ to $(1,0,\infty)$. Defining $T_w$ similarly, the map $$ \boxed{f=T_w^{-1}\circ T_z} $$ sends $z_i$ to $w_i$. If two Möbius transformations do so, their quotient fixes $0,1,\infty$. A Möbius transformation fixing infinity is affine, and fixing zero and one then makes it the identity. This proves uniqueness. With this convention, the is $$ [z_1,z_2,z_3,z_4] =T_z(z_4) =\frac{(z_4-z_2)(z_1-z_3)} {(z_4-z_3)(z_1-z_2)}. $$ Substitution shows that translations, nonzero scalings, and inversion preserve it; since these generate the Möbius group, every Möbius transformation preserves cross-ratios. Conversely, suppose a $f$ of the Riemann sphere preserves every cross-ratio. Let $m$ be the unique Möbius transformation agreeing with $f$ at three chosen points $z_1,z_2,z_3$. For any other $z$, preservation by $f$ and $m$ gives $$ [z_1,z_2,z_3,z] =[f(z_1),f(z_2),f(z_3),f(z)] =[f(z_1),f(z_2),f(z_3),m(z)]. $$ The last coordinate in a cross-ratio with three fixed distinct entries is injective, so $f(z)=m(z)$. The equality already holds at the three base points, hence $f=m$ everywhere and $f$ is Möbius. Finally, the map $z\mapsto a\overline z+b$ is constant when $a=0$ and therefore is not Möbius. If $a\ne0$, it is bijective and fixes infinity. Were it Möbius, it would have the affine form $\alpha z+\beta$. Equality on real $z$ forces $\alpha=a$ and $\beta=b$, whereas equality at $z=i$ would require $ai=-ai$, contradicting $a\ne0$. Thus $$ \boxed{\text{there are no such }a,b}. $$ Solved by gpt-5.6-sol high. = 7E {parent=Paper 3} {scope} {title2=Groups} = Solution {parent=7E} A $N\leq G$ is [normal] when $$ gNg^{-1}=N\qquad(g\in G). $$ Its left and right cosets then agree, and multiplication $$ (gN)(hN)=ghN $$ is well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the $G/N$. For a $\theta:G\to H$, its [kernel] and [image] are $$ \ker\theta=\{g:\theta(g)=e_H\}, \qquad \operatorname{im}\theta=\{\theta(g):g\in G\}. $$ The kernel is normal because $$ \theta(gkg^{-1}) =\theta(g)\theta(k)\theta(g)^{-1} =e_H $$ whenever $k\in\ker\theta$. Conversely, if $K\trianglelefteq G$, the quotient map $$ q:G\longrightarrow G/K,\qquad q(g)=gK, $$ is a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the map $$ G/\ker\theta\longrightarrow\operatorname{im}\theta, \qquad g\ker\theta\longmapsto\theta(g) $$ is well defined and bijective and preserves multiplication. This is the . Define $$ \Phi:(\mathbb R,+)\longrightarrow(\mathbb C\setminus\{0\},\cdot), \qquad \Phi(t)=e^{2\pi it}. $$ shows that $\Phi$ is a homomorphism, its image is the , and its kernel is $\mathbb Z$. The first isomorphism theorem therefore gives $$ \boxed{\mathbb R/\mathbb Z\cong S^1}. $$ The image of $\mathbb Q/\mathbb Z$ consists exactly of the : if $t=p/q$, then $\Phi(t)^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$. Solved by gpt-5.6-sol high. = 8E {parent=Paper 3} {scope} {title2=Groups} = Solution {parent=8E} The set $S(\mathbb N)$ consists of all $\mathbb N\to\mathbb N$. The composition of two bijections is a bijection, composition is associative, the identity map is an identity, and every bijection has an inverse, so $S(\mathbb N)$ is a . An element of $S_{\mathrm{fin}}(\mathbb N)$ has finite [support]. The support of a composition is contained in the union of the two supports, and a permutation and its inverse have the same support. The identity has empty support. Hence $S_{\mathrm{fin}}(\mathbb N)$ is a . Let $\sigma$ be a cycle and choose $n$ that it moves. Because $\sigma$ has finite support, the sequence $$ n,\sigma(n),\sigma^2(n),\ldots $$ must repeat. Since $\sigma$ is invertible, its first repetition returns to $n$; let the least positive return time be $l$. The cycle condition says that every moved point occurs in this orbit, so $\sigma^l$ fixes every point. No smaller positive power fixes $n$. Therefore $$ \boxed{\operatorname{ord}(\sigma)=l}, $$ which is finite. For $\tau\in S_{\mathrm{fin}}(\mathbb N)$, partition its finite support into the orbits of the $\langle\tau\rangle$. On each orbit, let $\sigma_i$ agree with $\tau$ and fix every point outside that orbit. Then each $\sigma_i$ is a , their supports are pairwise disjoint, and $$ \tau=\sigma_1\cdots\sigma_k. $$ Disjoint cycles commute because at every point at most one of them acts nontrivially. Writing $l_i=\operatorname{ord}(\sigma_i)$, a power $\tau^m$ is the identity exactly when every $\sigma_i^m$ is the identity, equivalently when every $l_i$ divides $m$. Thus $$ \boxed{\operatorname{ord}(\tau) =\operatorname{lcm}(l_1,\ldots,l_k)}. $$ Solved by gpt-5.6-sol high. = 9A {parent=Paper 3} {scope} {title2=Vector Calculus} = a {parent=9a} {scope} = Solution {parent=a} Since $r=(x_jx_j)^{1/2}$, the gives $$ \boxed{\frac{\partial r}{\partial x_i}=\frac{x_i}{r}}. $$ For the radial $g(r)\mathbf x$, the and give $$ \begin{aligned} \nabla\cdot\bigl(g(r)\mathbf x\bigr) &=\frac{\partial}{\partial x_i}\bigl(g(r)x_i\bigr)\\ &=g'(r)\frac{x_i}{r}x_i+3g(r)\\ &=\boxed{rg'(r)+3g(r)}. \end{aligned} $$ Solved by gpt-5.6-sol high. = b {parent=9a} {scope} = Solution {parent=b} Differentiating the $$ \mathbf x=(r\sin\theta\cos\phi,\, r\sin\theta\sin\phi,\, r\cos\theta) $$ gives the orthogonal decomposition $$ d\mathbf x =\mathbf e_r\,dr+r\mathbf e_\theta\,d\theta +r\sin\theta\,\mathbf e_\phi\,d\phi, $$ where $$ \begin{aligned} \mathbf e_r&=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta),\\ \mathbf e_\theta&=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta),\\ \mathbf e_\phi&=(-\sin\phi,\cos\phi,0). \end{aligned} $$ Thus the are $$ \boxed{h_r=1,\qquad h_\theta=r,\qquad h_\phi=r\sin\theta}. $$ Since the satisfies $$ df=f_r\,dr+f_\theta\,d\theta+f_\phi\,d\phi =d\mathbf x\cdot\nabla f, $$ comparison of coefficients gives the $$ \boxed{ \nabla f =\mathbf e_r\frac{\partial f}{\partial r} +\mathbf e_\theta\frac1r\frac{\partial f}{\partial\theta} +\mathbf e_\phi\frac1{r\sin\theta}\frac{\partial f}{\partial\phi}}. $$ Solved by gpt-5.6-sol high. = c {parent=9a} {scope} = Solution {parent=c} Each field has only an azimuthal component and is independent of $r$ and $\phi$. The therefore reduces to $$ (\nabla\times\mathbf A)_r =\frac1{r\sin\theta} \frac{\partial}{\partial\theta} \bigl(\sin\theta A_\phi\bigr), \qquad (\nabla\times\mathbf A)_\theta =-\frac1r\frac{\partial}{\partial r}(rA_\phi), $$ and its $\phi$ component vanishes. The $$ \sin\theta\tan\frac\theta2=1-\cos\theta, \qquad \sin\theta\left(-\cot\frac\theta2\right)=-(1+\cos\theta) $$ show in both cases that $$ \boxed{\nabla\times\mathbf A_+ =\nabla\times\mathbf A_- =\frac{\mathbf e_r}{r^2} =\frac{\mathbf x}{r^3}}. $$ Apply part (a) with $g(r)=r^{-3}$. Then $$ \nabla\cdot\frac{\mathbf x}{r^3} =r(-3r^{-4})+3r^{-3}=0 $$ away from the origin, explicitly confirming that each resulting has zero . Solved by gpt-5.6-sol high. = d {parent=9a} {scope} = Solution {parent=d} Part (c) gives $$ \nabla\times(\mathbf A_+-\mathbf A_-)=0. $$ The half-space $x_1>0$ is a , so the makes this curl-free field a : it is the gradient of a single-valued scalar potential. Using $\tan u+\cot u=2/\sin(2u)$, $$ \mathbf A_+-\mathbf A_- =\frac2{r\sin\theta}\mathbf e_\phi. $$ The spherical-coordinate formula from part (b) shows that a potential must satisfy $f_\phi=2$ and may be independent of $r,\theta$. On $x_1>0$ the azimuthal angle has the single-valued branch $$ \phi=\arctan\frac{x_2}{x_1}. $$ Hence one solution, up to an , is $$ \boxed{f(\mathbf x)=2\phi =2\arctan\frac{x_2}{x_1}}. $$ Solved by gpt-5.6-sol high. = 10A {parent=Paper 3} {scope} {title2=Vector Calculus} = Solution {parent=10A} On the part with $1\leq z\leq2$, write $$ \mathbf r(\rho,\phi) =\left(\rho\cos\phi,\rho\sin\phi, \sqrt{\rho^2-1}\right), \qquad \sqrt2\leq\rho\leq\sqrt5. $$ This is the upper portion of a [hyperboloid of one sheet]. Choosing the [outward orientation], whose normal points radially away from the $z$-axis, the is $$ \begin{aligned} d\mathbf S &=(\mathbf r_\phi\times\mathbf r_\rho)\,d\rho\,d\phi\\ &=\boxed{\left( \frac{\rho^2\cos\phi}{\sqrt{\rho^2-1}}, \frac{\rho^2\sin\phi}{\sqrt{\rho^2-1}}, -\rho\right)d\rho\,d\phi}. \end{aligned} $$ Its radial component is positive and its vertical component is negative. Reversing the normal reverses all the fluxes below. For $$ \mathbf A=(-yz^2,xz^2,0), $$ the is $$ \nabla\times\mathbf A=(-2xz,-2yz,2z^2). $$ On the surface, $z^2=\rho^2-1$, and therefore $$ (\nabla\times\mathbf A)\cdot d\mathbf S =-2\rho(2\rho^2-1)\,d\rho\,d\phi. $$ Direct integration gives $$ \begin{aligned} \int_S\nabla\times\mathbf A\cdot d\mathbf S &=-2\int_0^{2\pi}\int_{\sqrt2}^{\sqrt5} \rho(2\rho^2-1)\,d\rho\,d\phi\\ &=-2\pi[\rho^4-\rho^2]_{\sqrt2}^{\sqrt5}\\ &=\boxed{-36\pi}. \end{aligned} $$ To verify this with the , note that on a circle of fixed $\rho,z$, $$ \mathbf A=z^2\rho\,\mathbf e_\phi, \qquad d\mathbf r=\rho\mathbf e_\phi\,d\phi. $$ The induced runs in the negative $\phi$ direction on the top circle and the positive $\phi$ direction on the bottom circle. Hence $$ \oint_{\partial S}\mathbf A\cdot d\mathbf r =2\pi(1^2)(\sqrt2)^2 -2\pi(2^2)(\sqrt5)^2 =4\pi-40\pi =-36\pi. $$ For $S'$, Stokes' theorem again reduces the flux to its two boundary circles. The outward orientation gives positive $\phi$ direction at $z=-1$, where $\rho^2=2$, and negative $\phi$ direction at $z=\sqrt2$, where $\rho^2=3$. Thus $$ \boxed{ \int_{S'}\nabla\times\mathbf A\cdot d\mathbf S =2\pi(1)(2)-2\pi(2)(3) =-8\pi}. $$ Solved by gpt-5.6-sol high. = 11A {parent=Paper 3} {scope} {title2=Vector Calculus} = i {parent=11a} {scope} = Solution {parent=i} Let $\phi_1,\phi_2$ be two solutions with the same , and set $u=\phi_1-\phi_2$. Then $$ \nabla^2u-m^2u=0\quad\hbox{in }V, \qquad \frac{\partial u}{\partial n}=0\quad\hbox{on }S. $$ Multiplying by $u$ and applying gives $$ \int_V\left(|\nabla u|^2+m^2u^2\right)dV =\int_Su\frac{\partial u}{\partial n}\,dS =0. $$ The integrand is nonnegative. If $m>0$, both terms can vanish only when $u=0$, so the solution of the is unique. If $m=0$, the identity only forces $\nabla u=0$. On a connected region, $u$ may be any constant. Thus solutions of the with prescribed normal derivative are unique only up to an additive constant, so $$ \boxed{\text{uniqueness fails when }m=0}. $$ Solved by gpt-5.6-sol high. = ii {parent=11a} {scope} = Solution {parent=ii} Put $\eta=\psi-\phi$. The common gives $\eta=0$ on $S$. Expanding the , $$ \begin{aligned} E[\psi]-E[\phi] ={}&\int_V\left(|\nabla\eta|^2+m^2\eta^2\right)dV\\ &+2\int_V\left(\nabla\phi\cdot\nabla\eta +m^2\phi\eta\right)dV. \end{aligned} $$ and the field equation make the cross term zero: $$ \int_V\left(\nabla\phi\cdot\nabla\eta+m^2\phi\eta\right)dV =\int_S\eta\frac{\partial\phi}{\partial n}\,dS -\int_V\eta(\nabla^2\phi-m^2\phi)\,dV =0. $$ Consequently $$ \boxed{ E[\psi]-E[\phi] =\int_V\left(|\nabla\eta|^2+m^2\eta^2\right)dV\geq0}. $$ For $m>0$, equality holds exactly when $\eta=0$. For $m=0$, equality holds exactly when $\nabla\eta=0$, so $\eta$ is constant on the connected region; its zero boundary value then again forces $\eta=0$. The minimizing property therefore proves uniqueness for the Dirichlet problem in both cases: $$ \boxed{\psi=\phi\text{ is the equality condition for every }m\geq0}. $$ Solved by gpt-5.6-sol high. = 12A {parent=Paper 3} {scope} {title2=Vector Calculus} = a {parent=12a} {scope} = Solution {parent=a} Let a change of orthonormal coordinates be represented by a $R$. Since both and are vectors, $$ L'_i=R_{ip}L_p, \qquad \omega'_j=R_{jq}\omega_q. $$ Using $L_i=I_{ij}\omega_j$ in both frames gives $$ I'_{ij}R_{jq}\omega_q=R_{ip}I_{pq}\omega_q $$ for every vector $\boldsymbol\omega$. Multiplication by $R^{-1}=R^T$ yields $$ \boxed{I'_{ij}=R_{ip}R_{jq}I_{pq}}, $$ which is exactly the transformation law for a rank-two . Thus $I_{ij}$ is a rank-two tensor. Put $\mathbf r=\mathbf x-\mathbf a$. The gives $$ \mathbf r\times(\boldsymbol\omega\times\mathbf r) =r^2\boldsymbol\omega -\mathbf r(\mathbf r\cdot\boldsymbol\omega). $$ Therefore $$ \boxed{ I_{ij}(\mathbf a) =\rho\int_B \left[(x_k-a_k)(x_k-a_k)\delta_{ij} -(x_i-a_i)(x_j-a_j)\right]dV}. $$ At the centre of mass, $$ \boxed{ I_{ij}(\mathbf0) =\rho\int_B(x_kx_k\delta_{ij}-x_ix_j)\,dV}. $$ Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho\int_BdV=M$. Hence the in tensor form is $$ \boxed{ I_{ij}(\mathbf a) =I_{ij}(\mathbf0) +M(a_ka_k\delta_{ij}-a_ia_j)}. $$ Solved by gpt-5.6-sol high. = b {parent=12a} {scope} = Solution {parent=b} Reflection symmetry of the cube in each coordinate plane makes every off-diagonal integral $$ -\rho\int_Bx_ix_j\,dV\qquad(i\ne j) $$ vanish because its integrand is [odd] in one coordinate. Permuting the three coordinate axes leaves the cube unchanged, so its three diagonal moments are equal. Thus its at the centre has the isotropic form $$ I_{ij}(\mathbf0)=\lambda\delta_{ij}. $$ For $$ \mathbf a=\frac\ell2(1,1,0), \qquad |\mathbf a|^2=\frac{\ell^2}{2}, $$ part (a) and $\lambda=M\ell^2/6$ give $$ \boxed{ I(\mathbf a) =\frac{M\ell^2}{12} \begin{pmatrix} 5&-3&0\\ -3&5&0\\ 0&0&8 \end{pmatrix}}. $$ The displayed symmetric matrix has orthogonal $$ (1,1,0),\qquad(1,-1,0),\qquad(0,0,1) $$ with corresponding $$ \frac{M\ell^2}{6}, \qquad \frac{2M\ell^2}{3}, \qquad \frac{2M\ell^2}{3}. $$ For a unit angular velocity, $|\mathbf L|=|I\boldsymbol\omega|$ is minimized and maximized along eigenvectors for the smallest and largest eigenvalues. Their ratio is $$ \boxed{ \frac{|\mathbf L|_{\max}}{|\mathbf L|_{\min}} =\frac{2M\ell^2/3}{M\ell^2/6}=4}. $$ Solved by gpt-5.6-sol high.