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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 $r=|x|$.

Away from the origin, differentiate $r^2=x_j{x}_j$ using Einstein notation:HenceThis gradient is the radial unit vector. Division by $r$ is legitimate on the punctured domain; the distance function is not differentiable at the origin.

DefineThey sum to $t_{ij}$, with $s_{ji}=s_{ij}$ and $a_{ji}=-a_{ij}$. Interchanging two slots of a tensor still produces a tensor, so these are a symmetric second-rank tensor and an antisymmetric second-rank tensor, not merely symmetric and antisymmetric arrays in a selected frame.

For uniqueness, if $t=s+a=s^{\prime }+a^{\prime }$, then $s-s^{\prime }=a^{\prime }-a$. The common difference is both symmetric and antisymmetric, so each of its components equals its own negative and vanishes over $\mathbb{R}$. Thus the decomposition is unique. These are the symmetric and antisymmetric parts of a matrix, with their tensor transformation law preserved.

Rotational symmetry of the vector field means $\mathbf{F}(r\mathbf{n})=h(r)\mathbf{n}$: rotations fixing $\mathbf{n}$ force the field to be parallel to $\mathbf{n}$, and rotations between directions make its magnitude depend only on $r$. Its flux integral over the radius-$r$ sphere is $4\pi r^{2}h(r)$. Applying the flux identity from part (i) to the ball, with the spherically symmetric source, givesTherefore the origin-regular spherical Poisson flux law isOnly source values at $s\le |\mathbf{x}|$ occur. Changing the exterior source can change the scalar potential by an interior constant, but cannot change this radial gradient. Regularity on all of ${\mathbb{R}}^3$, inherited from the Poisson equation setting, matters: on a punctured domain alone an additional $C\mathbf{x}/r^3$ would have zero divergence away from the origin and would not be fixed by the local source. It is excluded here by the origin regularity needed for the ball flux identity.

The electric field of this point charge is a translated and scaled version of the radial field in part (ii). If $\mathbf{a}\notin V$ and $\mathbf{a}\notin \partial V$, its divergence vanishes throughout $V$, so the divergence theorem gives zero flux integral.

If $\mathbf{a}$ lies inside $V$, remove a small ball centered at $\mathbf{a}$. On the remaining volume the divergence is zero. Its boundary is $\partial V$ together with the small sphere, whose outward normal for the punctured volume points towards $\mathbf{a}$. That inner sphere contributes $-q/{ϵ}_0$, by the $4\pi$ sphere flux computed in part (ii). Thus the exterior flux must be $q/{ϵ}_0$:This is Gauss's law for a point charge. The assumption that the charge is not on the boundary is essential to these alternatives.

On the outward-oriented sphere of radius $R$, $\mathbf{n}=\mathbf{x}/R$, so $\mathbf{x}\cdot d\mathbf{S}=RdS$. HenceAway from the origin, the divergence and curl of a radial vector field with $f(r)=r^{-3}$ giveIf the enclosed volume excludes the origin and its boundary avoids it, this field is smooth throughout the volume. The divergence theorem then gives $I=0$. The nonzero sphere flux is compatible with zero divergence away from the origin because the enclosed singularity prevents applying that smooth-volume theorem directly to the full ball.

For $\mathbf{F}=\nabla \varphi$, the definition of the Laplacian gives $\nabla \cdot \mathbf{F}={\nabla }^2\varphi =f$. Integrating over $V$ and applying the divergence theorem yieldsThis is the flux form of Poisson equation, for an outward-oriented piecewise smooth boundary and the usual smoothness of $\varphi$ on a neighborhood of the volume. In particular, a singularity cannot be ignored when applying this identity.