Use the angular-frequency Fourier transform conventionFor the piecewise smooth functions here, the inverse gives the function at continuity points and the mean of one-sided limits at a jump. Values assigned at individual points do not affect the transform.

The zero initial value and slope select a forward Green's function: set it to zero for $x<\xi$, and choose the right piece to start at zero with slope one. This givesFor $f=1$ on $(0,1)$ and zero elsewhere,Indeed $y_2={\int }_0^{\min (x,1)}(x-{\xi }^2/x)d\xi /2$. It and its first derivative match at one, while $y_2,y_2^{\prime }\to 0$ at zero. The only homogeneous solution with these two initial limits is zero, so this solution too is unique. Its growth at infinity is allowed by these boundary conditions.

The denominator factors as $(z+2-i)(z+2+i)$. Neither root cancels against the numerator, and both are simple, so the finite singularities areThis is a rational function, so it has no branch points. At infinity, putting $w=1/z$ gives $w/(1+4w+5w^2)$, analytic at zero; infinity is therefore a removable singularity with value zero.

Reflect the source across the equatorial plane and subtract the two ball kernels. This gives the Dirichlet Green function inside a hemisphere $G_{\mathrm{hemi}}(\mathbf{x};{\mathbf{x}}_0)=G_B(\mathbf{x};{\mathbf{x}}_0)-G_B(\mathbf{x};{\mathbf{x}}_0^{-})$. It is zero on the spherical boundary because each ball kernel is zero there, and on the plane because reflection makes their values equal.

For a point charge $e$, Poisson's equation in SI units is $\Delta V=-e\delta (\mathbf{x}-{\mathbf{x}}_0)/{\epsilon }_0$, so $V=-e{G}_{\mathrm{hemi}}/{\epsilon }_0$. Writing ${\mathbf{x}}_0^{\ast -}$ for the reflected inverse point givesThe construction uses the real charge, its negative reflected image, a negative inversion image of magnitude $ea/r_0$, and its positive reflected image. Reflection and inversion commute. All three images lie outside the open upper half-ball, and the prescribed zero boundary values determine the electrostatic potential uniquely.

If an element has order six, it generates $G$, so $G\cong C_6$. Otherwise choose an element $x$ of order three. The preceding result excludes any further order-three elements outside $\{e,x,x^2\}$, and there are no elements of order six by assumption. Each of the other three elements must therefore have order two. ThusThe noncyclic case is the symmetric group on three points, as the classification of a group of order six further identifies.

For a complex derivative to exist, the difference quotient must agree along real and imaginary increments. These give $u_x+i{v}_x$ and $v_y-i{u}_y$, respectively. Equating their real and imaginary parts gives the Cauchy-Riemann equationsFor $g(z)=z\mathrm{Re}z$, the components are $u=x^2$ and $v=xy$. The first equation requires $2x=x$, so $x=0$; the second requires $0=-y$, so $y=0$. Thus zero is the only possible differentiability point. Directly,so $\boxed{g^{\prime }(0)=0}$. The function is differentiable at that isolated point but is not holomorphic on any neighborhood of it.

For large $|z|$, expand in $a^2/z^2$ to obtainThe axisymmetric exterior multipole expansion replaces each $|z{|}^{-n-1}$ on the positive axis by $r^{-n-1}{P}_n(\cos \theta )$. Reflection symmetry eliminates the odd multipoles, soThe leading term is the potential of total charge $Q$ concentrated at the center; the next is its electric quadrupole correction. On either axis $P_2(\pm 1)=1$, recovering the same even axis potential.

Write $A=Q/(2\pi {ϵ}_0{a}^2)$. The Taylor expansion $\sqrt{a^2+z^2}=a+z^2/(2a)+O(z^4/a^3)$ gives the near-axis resultTo extend off the axis, work separately above and below the charged plane. In either source-free half-space the regular axisymmetric solid harmonics are $r^n{P}_n(\cos \theta )$. Matching the positive-axis coefficients, using $P_n(1)=1$, and then reflecting in the disk givesHere $P_2(t)=(3t^2-1)/2$ is a Legendre polynomial. The quadratic term is $A(z^2-{\rho }^2/2)/(2a)$. The linear term is $-Az$ above and $+Az$ below; it represents the discontinuity of the normal electric field. A single regular spherical expansion across the entire charged sheet would incorrectly remove that jump.

If $h=ag{a}^{-1}$, the representation law gives $D(h)=D(a)D(g)D(a{)}^{-1}$. Trace invariance under this similarity transformation givesThus the character of a representation is constant on every conjugacy class.