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Without dust collection, $dM=-dm$. The rocket equation reduces toIntegrating from $v=0$, $M=M_0$ givesThis is the usual rocket equation with constant relative exhaust speed, valid while the mass remains positive and the prescribed ejection continues.

Neglect the centrifugal acceleration, which is of order ${\omega }^2$, and use the local flat-surface approximation. To zeroth order in $\omega$, the height above $P$ and flight time areInserting this vertical velocity into the Coriolis acceleration gives the eastward componentWith $x(0)=\dot{x}(0)=0$, two integrations yieldto first order. Horizontal velocity is already of order $\omega$, so its effect on vertical acceleration is of order ${\omega }^2$. Thus the flight time remains $T_0$ at this order. The Coriolis deflection of a vertical projectile at landing isThe displacement is westward, and its magnitude is $4\omega v^3\sin \theta /(3g^2)$. The horizontal speed remains westward during the flight: $\dot{x}=-2\omega \sin \theta z_0(t)$. This explains why the eastward acceleration on descent does not cancel the earlier displacement.

At $P$, use the local right-handed basis ${\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U$ pointing east, north and vertically outward. The Earth's rotation vector has componentsInitially $\dot{\mathbf{r}}=0$ in the rotating reference frame, so the Coriolis acceleration is zero. Evaluating the centrifugal acceleration givesThe gravitational acceleration is $-g{\mathbf{e}}_U$. HenceThe small angle $\delta$ from the downward vertical therefore satisfiesFor the northern hemisphere, $0\le \theta \le \pi /2$, this is the printed expression without absolute values. More generally, the printed signed meridional expression specifies the direction of the tilt; an unsigned angle uses its absolute value. The horizontal component is toward the equator and vanishes at both poles and at the equator.

A simple group has no normal subgroup except the identity subgroup and the whole group. To prove the Simplicity of the alternating group A5, we compute its conjugacy classes. Since the sign homomorphism is onto, $|A_5|=5!/2=60$. Its possible nonidentity cycle types are a three-cycle, a product of two disjoint transpositions, and a five-cycle.

There are $\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2=20$ three-cycles. The centralizer in $S_5$ of $(123)$ consists of its three powers and an independent interchange of the two unused letters. Only the three powers are even. Its centralizer in $A_5$ therefore has order three, giving one conjugacy class of size $60/3=20$ by the orbit-stabilizer theorem.

There are $5\cdot 3=15$ products of two disjoint transpositions: choose the fixed letter and then a pairing of the other four letters. For $(12)(34)$, a commuting permutation fixes $5$ and may interchange the two pairs and swap within them, giving a centralizer of order eight in $S_5$. It contains an odd element, $(12)$, so exactly half its elements are even. The centralizer in $A_5$ has order four, and these fifteen elements form one conjugacy class of size $60/4=15$.

Finally, there are $4!=24$ five-cycles. A permutation commuting with a five-cycle is determined by the image of one letter and must be a power of that cycle. All five powers are even, so the centralizer in $A_5$ has order five. Each such conjugacy class has size $60/5=12$, giving two classes of size twelve.

A normal subgroup must be a union of conjugacy classes and must include the identity. Its order therefore has the formThe resulting distinct possibilities areOf these, only $1$ and $60$ divide $60$. Lagrange's theorem rules out every other possible order. Hence

When $[G:H]=2$, there are exactly two left cosets and two right cosets. For $g\notin H$, both $gH$ and $Hg$ are the complement $G\setminus H$; for $g\in H$, both equal $H$. This is the index-two subgroup is normal argument: $gH=Hg$ for every $g$, which is equivalent to $gH{g}^{-1}=H$. Therefore every subgroup of index two is normal.

In the dihedral group $D_{2n}=⟨r,s⟩$, the rotation cyclic subgroup $⟨r⟩$ has order $n$ and index two, so it is a normal subgroup. It is nontrivial and proper for the usual dihedral convention $n\ge 2$ (polygon symmetries have $n\ge 3$). If the degenerate convention $D_2=C_2$ were allowed, the claimed nontrivial proper subgroup would not exist; the rotation argument uses $n>1$.

For any $k\ge 3$, let $G=S_k$ and take the stabilizer subgroup $H$ of the letter $1$. Its elements freely permute the other $k-1$ letters, soConjugation by $(12)$ sends $H$ to the stabilizer of $2$, which is different: for instance $(23)$ belongs to $H$ but does not fix $2$. Thusgives a nonnormal subgroup of every prescribed index $k\ge 3$.

By Lagrange's theorem, a stabilizer subgroup of $G$ has order $p^j$ for some $0\le j\le n$. The orbit-stabilizer theorem therefore givesThus an orbit of a group action either has one element or has size divisible by $p$, even if the acted-on set is infinite.

Apply this to the conjugation action. Its singleton orbits of a group action are the elements of the centre of a group, while every other orbit has size divisible by $p$. Partitioning the finite group into these orbits gives the class equation, and henceSince the identity belongs to $Z(G)$, its order is positive and divisible by $p$. ConsequentlyNow suppose that $|Z(G)|=p^{n-1}$. It is then proper, so choose $x\notin Z(G)$. The centralizeris a subgroup: commuting with $x$ is preserved by multiplication and inverses. It contains $Z(G)$ and also $x$, so $|C_G(x)|>|Z(G)|=p^{n-1}$. By Lagrange's theorem, its order is a power of $p$ dividing $p^n$, forcing $C_G(x)=G$. But that would make $x$ central, a contradiction. Therefore

For a group action, the orbit-stabilizer theorem identifies the orbit of a group action of $x$ with the left cosets of its stabilizer subgroup:This is well defined and bijective because $gx=hx$ exactly when $h^{-1}g\in G_x$. In particular, for finite $G$,Under the conjugation action $g\cdot x=gx{g}^{-1}$, a singleton orbit of a group action means that $gx{g}^{-1}=x$ for every $g$. This is equivalent to $gx=xg$ for every $g$, so precisely these elements constitute the centre of a groupThe identity is central, and if $x,y$ commute with every element then so do $xy$ and $x^{-1}$. Thus $Z(G)$ is a subgroup. Each central element is fixed under every conjugation, giving $gZ(G)g^{-1}=Z(G)$: the centre is a normal subgroup.

The general linear group $G{L}_2({\mathbb{F}}_5)$ consists of invertible two-by-two matrices over the finite field ${\mathbb{F}}_5$, under matrix multiplication. Such a matrix has two linearly independent columns. There are $5^2-1=24$ choices for a nonzero first column, and $5^2-5=20$ choices for a second column outside its one-dimensional span. ThusThe determinant is a group homomorphismwhere ${\mathbb{F}}_5^{\times }=\{1,2,3,4\}$ is the multiplicative group of nonzero field elements. It is surjective because $\mathrm{diag}(t,1)$ has determinant $t$. Its kernel of a group homomorphism is $S{L}_2({\mathbb{F}}_5)$. The first isomorphism theorem gives a quotient of order four, whenceThe nonzero squares form the subgroup $\{1,4\}$ of ${\mathbb{F}}_5^{\times }$. Its inverse imageis a subgroup with index two. Indeed, composing the determinant with the quotient map to ${\mathbb{F}}_5^{\times }/\{1,4\}$ makes $H$ the kernel of a surjective group homomorphism to a group of order two. Thus $|H|=240$.

For either coefficient system $\mathbb{Z}$ or ${\mathbb{F}}_p$, multiplying two matrices keeps all entries in that system, and multiplicativity of the determinant preserves determinant one. The identity matrix has determinant one, and for a determinant-one matrix the matrix inverse isIts entries again belong to the same coefficient system and its determinant is one. Together with associativity of matrix multiplication, these facts establish all the group axioms. Thus both sets are special linear groups.

Reduce each entry modulo five to defineReduction respects sums and products, so $\overline{MN}=\overline{M}\overline{N}$. It also respects the determinant, ensuring that the image has determinant one. Hence $\pi$ is a group homomorphism. Its kernel of a group homomorphism consists exactly of determinant-one integer matrices congruent to the identity modulo five:The determinant-one condition remains part of this set; the four integers cannot be chosen independently without it. A kernel of a group homomorphism is a normal subgroup, since $\pi (gk{g}^{-1})=\pi (g)e\pi (g{)}^{-1}=e$ for $k\in \ker \pi$. This is the principal congruence subgroup of level five.

Extend $\sigma \in S_n$ to a permutation $\hat{\sigma }$ of $n+2$ letters that fixes the last two letters. Let $t=(n+1n+2)$, and let $ϵ(\sigma )$ be $0$ or $1$ according as $\sigma$ is even or odd. DefineThe two factors have disjoint supports and commute. Their signs cancel, so $\iota (\sigma )\in A_{n+2}$. Additivity of parity of a permutation modulo two givesThus $\iota$ is a group homomorphism. Its restriction to the first $n$ letters recovers $\sigma$, making it injective. Its image is therefore a subgroup isomorphic to $S_n$:This group embedding works for every positive $n$, including the trivial case $n=1$.

The alternating group is the subgroup of the symmetric group consisting of even permutations:For a permutation cycle of length $l$,using rightmost-first composition. Its sign of a permutation is therefore $(-1{)}^{l-1}$. Multiplicativity of the sign homomorphism givesThusEquivalently, $n-t$ is even when $t$ counts all cycles, including fixed points. Odd-length cycles contribute positive sign and impose no further restriction.

For a permutation $g$, start with a letter $i$ and follow $i,g(i),g^2(i),\dots$. Finiteness gives a repetition; since $g$ is invertible, the first return is to $i$. The visited letters form a permutation cycle on which $g$ acts by moving to the next letter. If there are letters left, repeat the construction starting with one of them. Its trajectory cannot meet an earlier one, because applying a suitable inverse power of $g$ would put its starting letter in that earlier trajectory. Continuing partitions the letters into disjoint cycles, including fixed points as cycles of length one.

The resulting disjoint permutation cycles have disjoint supports, and their product agrees with $g$ on every letter. This proves the cycle decomposition of a permutation. A power $g^m$ fixes a cycle of length $l$ exactly when $l\mid m$, so if the cycle lengths are $l_1,\dots ,l_t$,The least common multiple is essential: multiplying the lengths would overcount whenever cycles share a factor.

The order of a group element $g$ is the least positive integer $m$ for which $g^m=e$, if one exists. In a finite group, two terms of $e,g,g^2,\dots ,g^{|G|}$ must agree. If $g^i=g^j$ with $i<j$, cancellation gives $g^{j-i}=e$. Thus such positive integers exist and have a least member: every element of a finite group has finite order.

If $h=ag{a}^{-1}$, repeated cancellation shows $h^m=a{g}^m{a}^{-1}$ for every positive integer $m$. Consequently $h^m=e$ exactly when $g^m=e$. The least positive exponents are therefore equal:This proof also shows that conjugation preserves infinite order when the ambient group is not finite.

A permutation preserves whether the two coordinates are equal. Consequently the diagonal and its complement are invariant under this group action. Any $(i,i)$ can be carried to any $(j,j)$ by a permutation sending $i$ to $j$. Likewise, for $i\ne j$ and $k\ne l$, the assignment $i\mapsto k$, $j\mapsto l$ extends to a permutation of the remaining letters.

For $n\ge 2$ the equality-pattern orbits of ordered pairs are the two orbits of a group action are thereforewith sizes $n$ and $n(n-1)$. For $n=1$ only the diagonal orbit of a group action exists.

Take the cube with vertices $(\pm 1,\pm 1,\pm 1)$, and choose the edge $E=\{(1,1,z):-1\le z\le 1\}$. Here the symmetry group of a cube consists of all rigid symmetries, including reflections. A symmetry fixes the centre and permutes the three pairs of opposite faces, so its matrix is a signed permutation matrix.

A symmetry in the stabilizer subgroup of $E$ must preserve its midpoint $(1,1,0)$ and send its direction $(0,0,1)$ to itself or its negative. In this cube symmetry action on edges, its only choices are to interchange the first two coordinates and to reverse the third coordinate. Thus the stabilizer subgroup consists ofThe two generating reflections commute and have order two, so this subgroup is a Klein four-group, isomorphic to $C_2\times C_2$ and has order four. Permuting coordinates and changing signs carries $E$ to any edge. Its orbit of a group action is therefore the set of all twelve edges. The orbit-stabilizer theorem givesIf one restricts to orientation-preserving symmetries, only two of the four stabilizer elements remain, giving the rotational symmetry group of a cube of order $24$; the full symmetry group asked for here has order $48$.

Use the auxiliary point charge of strength $q\ne 0$ at the origin, with potential ${\varphi }_2=q/(4\pi {\epsilon }_{0}r)$ from part (a). Apply the electrostatic reciprocity identity of part (b) to the exterior region $a<r<R$, then let $R\to \infty$. Its inner boundary has outward normal $-{\mathbf{e}}_r$, and ${\rho }_2=0$ everywhere in this exterior region. The outer boundary terms vanish under the stipulated decay assumption.

On $r=a$, ${\varphi }_1=\Phi$ andThus the surviving boundary term on the left is $\Phi q/{\epsilon }_0$. On the right, ${\varphi }_2$ is constant on the inner sphere, so its boundary term isIndeed, $\nabla {\varphi }_1=-{\mathbf{E}}_1$, and this integral is the outward electric flux from the interior ball, whose enclosed charge is zero by the hypothesis and Gauss's law. No symmetry of the exterior charge distribution is needed.

The electrostatic reciprocity identity now reduces toCanceling $q/{\epsilon }_0$ givesThe inward normal on the inner boundary is essential to the positive sign in this result.

Combining the electric-field equations gives the Poisson equation ${\nabla }^2{\varphi }_i=-{\rho }_i/{\epsilon }_0$. Apply the product rule for divergence toThe two cross terms in its divergence cancel, leavingThis electrostatic reciprocity identity follows from the divergence theorem, and is exactly Green's second identity for these potentials. Integrating and rearranging givesThe boundary orientation is outward from the chosen region. For smooth charge densities this follows directly from classical differentiation. For isolated point charges, one may excise small spheres and take their radii to zero, provided the other potential is regular at each charge.

Integrating the local electric-field divergence equation and applying the divergence theorem gives Gauss's law:For a point charge at the origin, spherical symmetry requires $\mathbf{E}=E_r(r){\mathbf{e}}_r$. A sphere of radius $r>0$ encloses charge $q$, so its electric flux is $4\pi r^2{E}_r(r)$. ThusThe electrostatic potential satisfies $E_r=-d\varphi /dr$. Integrating and choosing the reference value $\varphi (\infty )=0$ yieldsAn arbitrary additive constant gives any other choice of potential reference. The singularity at the origin is represented by the Dirac delta distribution; it is not a point where the ordinary derivatives of this field are finite.

Let $N$ denote the mattress count and $U$ the event of undisturbed sleep. Use the intended model in which the presence of the pea is independent of $N$, with probability $1/2$, and without a pea sleep is undisturbed. Conditioning on $N$ givesThe law of total probability and the equal prior probabilities giveBy Bayes' theorem, the posterior probabilities of $N=6,7,8$ given $U$ are respectively $5/18,6/18,7/18$. Therefore the conditional expectation isUndisturbed sleep shifts the posterior toward larger mattress counts because those counts make the observation more likely.

By the definition of conditional probability,Since the $B_j$ form a disjoint partition of the sample space, the law of total probability gives $\mathbb{P}(A)={\sum }_j\mathbb{P}(A\mid B_j)\mathbb{P}(B_j)$. Substitution proves Bayes' theorem in the required form:The hypotheses make the denominator positive and each conditioning probability well defined.