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.

Use the same Fourier transform convention, now in the sense of tempered distributions, since the homogeneous oscillations are not integrable. Transforming givesSince $q>0$, the two roots are distinct. A distribution annihilated by this polynomial is a linear combination of Dirac delta distributions at its roots:Fourier inversion therefore givesThe constants may be complex. Differentiating the factor $e^{-irt}$ also verifies that the original ordinary differential equation reduces to $\ddot{\psi }+q^2\psi =0$ for $\varphi =e^{-irt}\psi$.

The convolution theorem gives the causal responseDefine the integrated impulse response, for $s\ge 0$, byDirect differentiation gives $F^{\prime }(s)=e^{-ps}\sin (qs)/q$ and $F(0)=F^{\prime }(0)=0$. ThusIn particular, after switch-off the answer isThe response and its first derivative are continuous at both switching times, and the post-pulse motion is a freely decaying damped harmonic oscillator. The expression described as valid for $t<1$ in the paper is understood for $0<t<1$; before the pulse the causal output is zero.

Take the causal solution, extending the input and output by zero before the device starts at $t=0$. The two zero initial conditions ensure that this extension adds no impulsive terms to the first two distributional derivatives. The Fourier transform of a derivative is multiplication by $i\omega$, soFactoring the quadratic and taking partial fraction decomposition yieldsThe inverse Fourier transform follows directly by integrating decaying exponentials:where $H$ is the Heaviside step function. This damped-oscillator impulse response has $R(0^{+})=0$ and $R^{\prime }(0^{+})=1$, as required for a unit impulse.

Use the Fourier transform conventionThe convolution is $(f\ast g)(t)={\int }_{-\infty }^{\infty }f(\tau )g(t-\tau )d\tau$. For integrable $f,g$, Fubini's theorem and the substitution $s=t-\tau$ giveHence the convolution theorem is

Use the counterclockwise rectangular contour with vertices $-R,R,R+i\pi ,-R+i\pi$. It encloses just the simple pole $i\pi /2$, whose residue is $e^{k\pi /2}/i$. Let $I_R={\int }_{-R}^R{e}^{-ikx}/\cosh xdx$. The identitymakes the top edge, traversed right to left, contribute $e^{k\pi }{I}_R$. On either vertical edge,Thus the sum of the vertical integrals tends to zero, with absolute value at most $2\pi e^{|k|\pi }/\sinh R$. The residue theorem and $R\to \infty$ now giveThe argument works for every fixed real $k$, including zero and negative values, and gives the Fourier transform of the hyperbolic secant with the convention $\tilde{f}(k)=\int e^{-ikx}f(x)dx$.

The exponential numerator is entire and never zero. The denominator vanishes precisely when $e^{2z}=-1$, givingAt these points $\sinh z_n=i(-1{)}^n\ne 0$, so every zero of the denominator is simple and hence every singularity is a simple pole. Using the residue formula in (b),

A simple pole at $z_0$ means that $f$ is holomorphic in a punctured neighbourhood and $(z-z_0)f(z)$ extends holomorphically to a nonzero value at $z_0$. Equivalently its Laurent series begins with $a_{-1}/(z-z_0)$ and has no higher negative powers. The residue isIn particular, if $f=g/h$ with $g(z_0)\ne 0$, $h(z_0)=0$ and $h^{\prime }(z_0)\ne 0$, this becomes $\mathrm{Res}(f,z_0)=g(z_0)/h^{\prime }(z_0)$. The associated residue theorem isfor a positively oriented simple contour with no poles on it.

Write $f=u+iv$ and $z=x+iy$. For a positively oriented, piecewise smooth simple closed contour bounding a region where $u,v$ have continuous first partial derivatives, Green's theorem givesThe Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$, so both integrands vanish:This proves Cauchy's integral theorem under the standard regularity assumptions for Green's theorem. More general admissible contours follow by subdivision; every enclosed region must be free of singularities.

The outward normal of the exterior region points into the excluded disc, so it is $-\mathbf{r}$ on the unit circle. Differentiating the Dirichlet Green function from (e), and writing $s=|{\mathbf{r}}^{\prime }|$ and $d=|\mathbf{r}-{\mathbf{r}}^{\prime }|$, givesHere $|\mathbf{r}-\mathbf{p}{|}^2=d^2/s^2$ was used. The concentrated boundary data have unit integral, so the boundary representation evaluates this exterior Poisson kernel at $\mathbf{r}=(-1,0)$. Thus the bounded solution on either part of the exterior $x$-axis isIt is positive on both intervals, diverges as $x^{\prime }\to -1^{-}$ at the boundary source, and tends to $1/(2\pi )$ as $|x^{\prime }|\to \infty$. Boundedness at infinity selects this solution; otherwise one could add an arbitrary multiple of $\log |{\mathbf{r}}^{\prime }|$ without changing the circle's boundary data.

For a source in the exterior, its inverse point $\mathbf{p}$ lies strictly inside the excluded disc. The method of images construction of the bounded exterior disk Green function therefore allows an opposite logarithmic source there without adding a singularity to the domain. The candidate Dirichlet Green function isIts Laplacian in the exterior is exactly $\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$, since the image singularity is outside that domain. The identity in (d) makes $G=0$ on the unit circle. Moreoverso it obeys the required boundedness condition. These checks establish the stated logarithm-of-a-ratio expression with $A=1/(2\pi )$. Bounded exterior Dirichlet uniqueness for the Poisson equation follows by inversion in a circle and removal of the bounded singularity at the origin.