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In cylindrical coordinates $r=\sqrt{x^2+y^2}$, the bounded solid enclosed by the two conical sheets and the circular cylinder isIts meridional section consists of two triangular regions; revolving them about the $z$-axis gives the solid. The two conical sheets meet at the origin and meet the circular cylinder in its circles at $z=\pm 1$.

Since $\mathbf{F}=(x,y,z)$, its divergence is $3$. Integrate directly with the cylindrical coordinates volume element $rdzdrd\theta$:The same calculation gives $\mathrm{Vol}(V)=4\pi /3$.

At every regular point of the double circular cone, its normal is perpendicular to $\mathbf{F}$ by part (a). Therefore the integrand of the flux integral vanishes pointwise:The vertex, if included, is a single point of zero surface area and does not alter the surface integral. The answer is independent of both the subset and its orientation. Tangency, rather than a cancellation between different parts of the surface, explains the zero flux.

Away from the vertex, use the defining function $H=z^2-x^2-y^2$. Its gradient is normal to the smooth part of the double circular cone, andon the surface. Hence the field is tangent at every regular point of the cone. At the origin the cone has no unique tangent plane, but $\mathbf{F}(0)=0$, which belongs to its tangent cone. An equivalent argument, valid also at the vertex, uses the integral curves of a vector field: $\dot{\mathbf{x}}=\mathbf{x}$ has solution $\mathbf{x}(t)=e^t\mathbf{x}(0)$, and these dilations preserve $H=0$. This is an example of radial dilation tangency to homogeneous zero sets.

A subgroup $N$ is a normal subgroup of $G$ if $gN{g}^{-1}=N$ for every $g\in G$. Define multiplication on its left cosets byTo check that this is well-defined, replace $g,h$ by $g{n}_1,h{n}_2$ with $n_1,n_2\in N$. Thenbecause normality gives $h^{-1}{n}_{1}h\in N$. Thus the product depends only on the two cosets. Associativity follows from that of $G$, the identity is $N$, and the inverse of $gN$ is $g^{-1}N$. These are all the group axioms, so $G/N$ is a group under coset multiplication. The natural projection $g\mapsto gN$ is a surjective group homomorphism with kernel $N$.

The Möbius transformations are bijections, so they preserve which coordinates of a triple are equal. Conversely, part (i) implies transitivity within each equality pattern. For a pattern with fewer than three distinct entries, extend the chosen distinct points to a triple before applying part (i); this works also for $p=2$, when the projective line has exactly three points. There are therefore five orbits of a group action.

We first count $|G{L}_2({\mathbb{F}}_p)|$. Its first column can be any nonzero vector, giving $p^2-1$ choices. The second must be outside the first column's one-dimensional span, giving $p^2-p$ choices. HenceFor the all-equal pattern take $(\infty ,\infty ,\infty )$. Its stabilizer subgroup consists of upper triangular matricesof order $p(p-1{)}^2$. Its group orbit has order $p+1$.

The three exactly-two-equal patterns have representatives $(\infty ,\infty ,0)$, $(\infty ,0,\infty )$ and $(0,\infty ,\infty )$. Each stabilizer subgroup fixes $0$ and $\infty$ individually, so it is the diagonal subgroup of order $(p-1{)}^2$. Each group orbit has order $p(p+1)$; their repeated-coordinate positions keep these three orbits distinct.

For the all-distinct pattern use $(0,1,\infty )$. Its stabilizer subgroup is the scalar subgroup of order $p-1$, and the group orbit has order $(p+1)p(p-1)$. Stabilizers of arbitrary triples are conjugates of the displayed representative stabilizers. In summary the equality-pattern orbits of projective triples haveThe sizes add to $(p+1{)}^3$, and each orbit size times its stabilizer order is $|G|$, as in the orbit-stabilizer theorem.

Identify the projective line with one-dimensional subspaces of ${\mathbb{F}}_p^2$: use $v_t=(t,1{)}^T$ for finite $t$, and $v_{\infty }=(1,0{)}^T$. This convention turns the matrix action into the stipulated Möbius transformation.

Since $x$ and $z$ are distinct projective points, $v_z,v_x$ are a basis. Write $v_y=A{v}_z+B{v}_x$. Both $A$ and $B$ are nonzero, because $y$ is different from $x,z$. The matrix with columns $A{v}_z$ and $B{v}_x$ is invertible and sends the lines of $(1,0{)}^T,(0,1{)}^T,(1,1{)}^T$ to those of $z,x,y$, respectively. This constructs the requested map without separate exceptional formulas at infinity.

A matrix fixing $0$ and $\infty$ is diagonal, say $\mathrm{diag}(a,d)$ with $a,d\ne 0$. To fix $1$ it must also satisfy $a=d$, so the stabilizer subgroup of $(0,1,\infty )$ is precisely $\{\lambda I:\lambda \in {\mathbb{F}}_p^{\times }\}$. Any two solutions differ on the right by one of these matrices. ThusEquivalently, after dividing out scalar matrices the projective general linear group action on the projective line is sharply three-transitive on a projective line.

Write a matrix in the upper unitriangular group as $M=I+A$. Here $A^3=0$, and its only potentially nonzero entry in $A^2$ is $(A^2{)}_{13}=ab$. The binomial theorem therefore gives the unitriangular matrix power formulaFor an odd prime number $p$, both $p$ and $\left(\genfrac{}{}{0ex}{}{p}{2}\right)=p(p-1)/2$ vanish in the finite field ${\mathbb{F}}_p$. Thus $M^p=I$. A nonidentity $M$ has order dividing the prime $p$ (by the same exponent-division argument as in part (i)), and so has order exactly $p$.

For $p=2$, take $a=b=1$ and $x=0$. Then $M^2=I+A^2\ne I$, while $M^4=(I+A^2{)}^2=I$. This element has order $4$, not $2$. Hence every nonidentity element has order $p$ exactly for odd $p$. The counterexample at $p=2$ accounts for the quadratic term in the matrix power formula.

If a nonidentity power $x^j$ belongs to $H$, reduce $j$ modulo $p$ so that $1\le j<p$. There is an integer $a$ with $aj\equiv 1(\mathrm{mod}p)$, and therefore $(x^j{)}^a=x\in H$, a contradiction. ThusNow suppose that $G$ is abelian and finite. Start with $H_0=\{1\}$, and whenever $H_k\ne G$, choose $x_{k+1}\notin H_k$. The multiplication mapis a group homomorphism because all elements commute. It is surjective by construction and injective because the intersection of its two factors is trivial: $hx=h^{\prime }{x}^{\prime }$ implies $h^{\prime -1}h=x^{\prime }{x}^{-1}$ lies in that intersection. Hence $H_{k+1}\cong H_k\times C_p$, and its order is $p$ times the previous order. Strict growth and finiteness ensure that the construction ends at $G$. Thereforewhere $n$ is the number of factors, possibly zero. This is an elementary abelian group, obtained by an explicit direct product of groups construction.

We first prove the needed instance of Cauchy theorem for groups, rather than invoke it. Let a prime number $q$ divide $|G|$ and consider the setThere are $|G{|}^{q-1}$ such tuples: choose the first $q-1$ entries freely, and the last is forced. Cyclic rotation preserves this set, because moving the first entry to the end conjugates the product, which remains $1$. Each rotation orbit of a group action has either one or $q$ elements. Indeed a nontrivial shift fixing a tuple generates all shifts, since its step has an inverse modulo the prime $q$. The fixed tuples are precisely $(g,\dots ,g)$ with $g^q=1$.

Counting nonfixed orbits in multiples of $q$, and using $q\mid |G{|}^{q-1}$, shows that the number of solutions of $g^q=1$ is divisible by $q$. The identity supplies one solution, so there is a nonidentity solution. If its order of a group element is $d$, divide $q$ by $d$: the remainder would give a smaller positive exponent producing the identity, so $d\mid q$. Hence $d=q$. This proves the required prime-order existence result by the cyclic-tuple proof of Cauchy theorem.

Our hypothesis says that this nonidentity element has order $p$, so $q=p$. Every prime divisor of $|G|$ is therefore $p$, which givesThus a finite group of prime exponent has prime-power order. The case $n=0$ includes the trivial group. No group-theoretic counting theorem was assumed in the tuple argument.

Place the cube at the origin with face normals $\pm e_1,\pm e_2,\pm e_3$, and label opposite faces by $(1,2),(3,4),(5,6)$. Every cube symmetry is a signed permutation matrix: there are $2^{3}3!=48$ choices. Its group action on faces is faithful because fixing all faces fixes all their normal directions. Central inversion $-I$ exchanges every opposite pair and therefore acts as $h$. Since $-I$ commutes with every linear cube symmetry, the face-action image lies in $C_{S_6}(h)$. The preceding count gives $|C_{S_6}(h)|=48$, so the faithful face action is onto this centraliser:For the remaining isomorphism, let $R$ be the rotational symmetry group of a cube. Half the signed permutation matrices have determinant $1$, so $|R|=24$. Central inversion has determinant $-1$, and every orientation-reversing symmetry is $(-I)r$ for a unique $r\in R$. Since $-I$ is central and $R\cap \{I,-I\}=\{I\}$, multiplication gives a direct product of groups:The group action of $R$ on the four body diagonals is faithful. To prove this, choose direction vectors $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$. They span ${\mathbb{R}}^3$, and their only linear relation is that their sum is zero. A rotation fixing all four diagonal lines sends each listed vector to itself or its negative. Applying the relation forces all four signs to agree. The all-negative choice is $-I$, which is not in $R$, so the rotation is the identity. Thus $R$ embeds in $S_4$; both have order $24$, giving $R\cong S_4$. Finally $C_{S_6}(g)\cong C_2\times S_4$ from part (i). ConsequentlyThe direct-product decomposition of the cube symmetry group supplies the second isomorphism, while the first comes from the specified face action.

Conjugation in the symmetric group sends $(12)$ to $(\sigma (1)\sigma (2))$. Hence $\sigma$ commutes with $g$ exactly when it preserves the unordered pair $\{1,2\}$. It can independently swap that pair and arbitrarily permute the remaining points. Therefore the centraliser of a transposition isLet $m=n/2$. The orbits of a group action generated by $h$ are its $m$ two-point blocks. Any centraliser element must permute those blocks, and can choose independently whether to swap the two points in each image block. Conversely every such choice commutes with $h$, since it preserves the partner relation. This centraliser of a fixed-point-free involution is the permutation wreath product $C_2\wr S_m$, with

Use $\gamma (t)=(\cos t,\sin t,0)$ for $0\le t\le 2\pi$. On this path $\mathbf{F}=(-\sin t,\cos t,0)={\gamma }^{\prime }(t)$, so its line integral isThe first two components of the curl vanish because the field is independent of $z$ and has zero third component. Writing $q=x^2+y^2$, its third curl component isThus this azimuthal inverse-radius vector field is a curl-free vector field on $G={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, but its line integral is not path independent. This domain is not simply connected: the circle links the removed axis. A spanning disk through the axis is inadmissible for Stokes theorem, since the vector field is undefined there. Zero curl alone does not force global path independence.

A sufficient condition is that $G$ be an open simply connected domain, with $\mathbf{F}$ continuously differentiable. Under this condition a curl-free vector field has a global potential of a conservative vector field and hence path independence by the fundamental theorem for line integrals. One way to see the global step is to deform the closed loop formed by two paths into a point inside $G$: the integral is unchanged during the deformation because the curl is zero, as expressed by Stokes theorem. Therefore simple connectivity is a sufficient domain hypothesis. It is not a claim that every particular curl-free field needs such a domain; a given field can have a global potential on a domain with holes.

Parametrize the smooth path by $\gamma :[0,1]\to G$. The chain rule turns the line integral of the gradient into a one-variable derivative:This is the fundamental theorem for line integrals; in particular the integral has path independence.

For a twice continuously differentiable potential of a conservative vector field, the curl of its gradient vanishes because mixed partial derivatives commute:A smooth gradient field is necessarily curl-free.

The divergence formula reduces the problem to the ordinary differential equation $r{f}^{\prime }+nf=0$ on $r>0$. Multiplication by $r^{n-1}$ makes it an exact derivative:Since the interval $r>0$ is connected, $r^{n}f(r)$ is a single constant $C$. Conversely, substituting $f=C{r}^{-n}$ into the divergence formula gives zero. ThusThis proves both existence and uniqueness up to the constant within the specified radial class. For $n=1$, the two punctured half-lines still share this same constant because the stipulated coefficient is one function of $r=|x|$.