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Take nonzero real trial functions $y$ with $y,y^{\prime }$ square-integrable and sufficient decay for integration by parts. For a compactly supported variation $y\mapsto y+\epsilon \eta$, the first variations areThe quotient rule for the functional then yieldsBy the fundamental lemma of the calculus of variations, stationarity for all such $\eta$ is equivalent toConversely, multiplying a decaying eigenfunction equation by $y$ and using integration by parts gives $F[y]=EG[y]$, so its Rayleigh quotient equals $E$ and its first variation vanishes. Thus the stationary values of the functional are the energy eigenvalues. The word "extrema" in the question includes stationary points here: higher eigenfunctions can be saddle points of the unrestricted Rayleigh quotient.

The Rayleigh-Ritz variational principle givesChoose an admissible trial family, calculate its Rayleigh quotient, and minimize over its parameters to obtain the best upper bound within that family. Equality holds for the exact ground state; a restricted family generally gives only an upper bound.

Close the hyperboloid segment with the two disks at $z=\pm 1$, and let $V$ be its interior. The flux through either disk is zero because $A_z=0$. Thus the desired surface integral is the total outward flux through the closed boundary. Sincethe divergence theorem and cylindrical coordinates giveThe extra factor $r$ is the volume integral measure in cylindrical coordinates; omitting it would give the wrong flux.

The divergence theorem states that, for a continuously differentiable vector field $\mathbf{H}$ on a bounded region $V$ with piecewise smooth boundary $S$ and outward unit normal $\hat{\mathbf{n}}$,To apply it, first derive the divergence of a cross product using the Levi-Civita symbol and the product rule:In the second term the minus sign comes from interchanging the first two indices of the Levi-Civita symbol. The divergence theorem with $\mathbf{H}=\mathbf{F}\times \mathbf{G}$ now yields

Use Einstein summation convention, writing ${\partial }_j=\partial /\partial x_j$. The contraction of two Levi-Civita symbols givesThe last equality is the product rule. Recognizing the divergences and directional derivatives proves the curl of a cross product identity:

The center of a group contains $e$. If $z,w$ lie in the center, then for every $g\in G$,It is therefore closed under the group operation and inverse elements, and inherits associativity, proving it is a subgroup. Each of its elements commutes with every element of $G$, in particular with every other central element. HenceThus the center is an abelian subgroup.

In additive notation every element $(a,b)\in {\mathbb{Z}}_2\times {\mathbb{Z}}_4$ obeys $4(a,b)=(0,0)$. Its order of a group element therefore divides four, whereas $1\in {\mathbb{Z}}_8$ has order eight. A group isomorphism preserves element orders, soFor the second product, define the group homomorphismIt is well-defined because multiples of six are divisible by both two and three. Its kernel contains only zero: simultaneous divisibility by two and three implies divisibility by six. Thus $\Psi$ is injective, and equal finite group orders make it surjective. Explicitly the six images areThereforeFor general positive integers $n,m$, the order of a direct-product element is the least common multiple of its component orders. Hence every element order divides $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, while $(1,1)$ attains this order. It generates the whole product precisely when $n,m$ are coprime:This is the cyclicity of a product of two finite cyclic groups, with the explicit isomorphism furnished by the Chinese remainder theorem.

Both subgroups contain the identity element, so $H\cap K$ is nonempty. If $x,y\in H\cap K$, then $xy\in H$ and $xy\in K$, so $xy\in H\cap K$. Similarly $x^{-1}$ belongs to both subgroups and hence to their intersection. Associativity is inherited from $G$. ThusThis verifies the group axioms; finiteness is unnecessary.

Closure of the componentwise operation follows from closure in each factor. For three pairs, associativity holds coordinatewise:The identity element is $(e_1,e_2)$ and the inverse element of $(g_1,g_2)$ is $(g_1^{-1},g_2^{-1})$. These verify all group axioms for the direct product of groups. Counting independent choices givesThe subset $N=G_1\times \{e_2\}$ contains the identity and is closed under products and inverses, so it is a subgroup. For any $x=(a,b)$ in the product and $(g,e_2)\in N$,Conjugation by $x^{-1}$ gives the reverse inclusion, and henceThis is the definition of a normal subgroup.

By (d), every group of group order three is a cyclic group. Writing its elements as $e,a,a^2$, every product is determined by $a^3=e$; the multiplication table isThe map $a^j\mapsto j(\mathrm{mod}3)$ is a bijective group homomorphism to addition modulo three. Thereforeso there is exactly one such group up to isomorphism.

Choose $g\ne e$ in a group of prime group order $p$. Its order of a group element divides $p$ by (c) and is not $1$, hence equals $p$. Consequently its cyclic subgroup already contains all elements of $G$:Thus every nonidentity element is a group generator.

Let $d=\mathrm{ord}(g)$. The powersare distinct: equality of two with exponents between $0$ and $d-1$ would give a smaller positive power equal to $e$. Reducing exponents modulo $d$ proves closure and supplies inverses, so this is a cyclic subgroup of group order $d$. Lagrange's theorem now gives

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$,The left cosets of $H$ partition $G$, and multiplication by a coset representative is a bijection from $H$ onto that coset, explaining the formula. If $|G|=p$ is prime, then $|H|$ is either $1$ or $p$, so $H=\{e\}$ or $H=G$. Thus the correct conclusion is no proper nontrivial subgroup. The trivial subgroup $\{e\}$ is proper for every group of prime order, so the wording in the question needs this qualification.

The group order is its cardinality, $\boxed{|G|=\#\{g:g\in G\}}$. The order of a group element $g$ iswhere $e$ is the identity element. In a finite group such an $n$ always exists, since two nonnegative powers of $g$ must coincide.

A common shift of all three angles leaves every spring length unchanged. This continuous rotational symmetry gives the zero mode: there is no restoring force against rigid rotation.

The equilateral equilibrium also has the dihedral group symmetry $D_3$. Its action on the three angular displacements preserves the uniform one-dimensional subspace and the two-dimensional sum-zero subspace. The latter is an irreducible representation over the real numbers; a reflection may reverse all angular signs as well as exchange bead labels, which does not change these invariant subspaces. The symmetric stiffness matrix must act as a scalar on the sum-zero subspace, explaining the equal vibrational angular frequencies. Directly, each sum-zero vector satisfies $(3I-11^T)\mathbf{v}=3\mathbf{v}$. This is an example of degenerate normal modes from permutation symmetry, and of normal modes of three equal masses with a symmetric gap potential.

Put ${\theta }_i={\theta }_i^0+{\eta }_i$. Differentiating the spring length givesThe springs are unextended at equilibrium, so the quadratic potential energy contains only the squares of these first-order extensions:The linearized Euler-Lagrange equations are thereforeThe stiffness matrix is $(k/4)(3I-11^T)$. Its eigenvector $1=(1,1,1{)}^T$ has zero eigenvalue, and every vector whose components sum to zero has eigenvalue $3k/4$. One convenient orthonormal set of normal modes isAny orthonormal basis of the sum-zero plane is equally valid for the two degenerate normal modes. The general small motion isThe linear drift in the zero mode describes a slow uniform rotation. These are angular frequencies; the ordinary frequencies of the two vibrational modes are ${\omega }_1/(2\pi )$.

Use the bead angles ${\theta }_1,{\theta }_2,{\theta }_3$ as generalized coordinates. The straight-line spring length between two beads follows from the law of cosines:The hoop is horizontal, so the gravitational potential energy is constant and can be omitted. The kinetic energy is $(m/2){\sum }_i{\dot{\theta }}_i^2$, giving the LagrangianAt ${\theta }_i^0=\Theta +2\pi (i-1)/3$, every spring has its natural length, so the potential energy is zero and its first derivatives vanish. This is an equilibrium for every constant $\Theta$. It is stable against relative displacement because each spring energy is nonnegative; the overall rotation is a neutral direction.

An axial electric field produces $\mathbf{J}=d\mathbf{E}$, so the observation sets $d=0$ and imposes no restriction on the transverse or antisymmetric parts. ThusA symmetric conductivity tensor would further require $h=0$, but axial rotational invariance alone does not require this.

Choose an orthonormal basis with $\mathbf{n}={\mathbf{e}}_3$. A half-turn about $\mathbf{n}$ kills all mixed axial-transverse components. Commutation with a quarter-turn makes the transverse block a scalar identity matrix plus an antisymmetric matrix. The resulting oriented-axis invariant Cartesian tensor, in coordinate-free form, iswith real constants $a,d,h$. The Kronecker delta, the outer product $n_i{n}_j$, and the displayed Levi-Civita symbol contraction are invariant under every proper rotation about $\mathbf{n}$, so this form is also sufficient. In the chosen basis,The antisymmetric part must be retained unless further symmetry is imposed.

Decompose the electric field as $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=(\mathbf{n}\cdot \mathbf{E})\mathbf{n}$. The electric current density isFor a nonzero real electric field, the alignment of current and field in an axisymmetric conductor gives the complete classification of collinearity:

For completeness, if ${\mathbf{E}}_{\perp }\ne 0$ and $\mathbf{J}=\lambda \mathbf{E}$, taking the dot product with $\mathbf{n}\times {\mathbf{E}}_{\perp }$ forces $h=0$. The transverse equation then gives $\lambda =a$, while a nonzero axial component gives $\lambda =d$. This proves there are no missing directions. Zero current is included as the zero scalar multiple; if “parallel” requires a direction for $\mathbf{J}$, exclude cases where that scalar is zero. For a strictly anisotropic crystal the last possibility is absent.

Under a proper rotation matrix $R$, the product of coordinates transforms as a rank-$n$ tensor. The unit ball is unchanged by the substitution $\mathbf{y}=R\mathbf{x}$, whose Jacobian determinant is one. HenceThus $S$ is an isotropic tensor. An isotropic second-rank tensor has the form $S_{ij}=A{\delta }_{ij}$. Taking the trace and using spherical coordinates,Therefore the isotropic second moment of a ball isAn isotropic third-rank tensor has the form $B{ϵ}_{ijk}$, but $S_{ijk}$ is symmetric in every pair of indices, forcing $B=0$. Alternatively, $\mathbf{x}\mapsto -\mathbf{x}$ makes its integrand odd on the ball. Thus

For a Cartesian coordinate change $x_i^{\prime }=R_{ij}{x}_j$ with $R$ an orthogonal matrix, a rank-$n$ tensor obeyswith repeated indices summed according to Einstein notation. A tensor is isotropic if $T_{i_1\cdots i_n}^{\prime }=T_{i_1\cdots i_n}$ for every proper rotation matrix $R$. Under proper rotations the Levi-Civita symbol is an isotropic third-rank tensor; if reflections are included, it is a pseudotensor. The convention using proper rotations is the one relevant to the rank-three classification here.