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A group homomorphism satisfies $\phi (g_1{g}_2)=\phi (g_1)\phi (g_2)$; a bijective one is a group isomorphism. Its kernel of a group homomorphism is $K=\{g:\phi (g)=1_H\}$. It is a subgroup, and for $k\in K$, $g\in G$,Thus $gK{g}^{-1}\subseteq K$, and applying the same argument to $g^{-1}$ gives equality. Therefore $K$ is a normal subgroup.

Write $B_n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$. Direct matrix multiplication gives $B_n{B}_m=B_{n+m}$, $B_0=I$ and $B_n^{-1}=B_{-n}$. Consequently these matrices form the integer shear group, a subgroup of $GL(2,\mathbb{R})$. It is an infinite cyclic group, generated by $B_1$, since $B_1^n=B_n$ and no nonzero integer power is the identity.

The map $n\mapsto B_n$ is a bijective group homomorphism from the additive group of integers, proving the required group isomorphism. There are proper subgroups: for example $\{B_{2n}:n\in \mathbb{Z}\}$ omits $B_1$. In fact every subgroup is $\{B_{mn}:n\in \mathbb{Z}\}$ for an integer $m\ge 0$, by the classification of subgroups of an infinite cyclic group; $m=0$ gives the trivial subgroup and $m=1$ the whole group.

Lagrange's theorem states that the group order of a finite group divides the group order; more precisely $|G|=[G:H]|H|$.

If every element squares to the identity element, every element is its own inverse element, and $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$. Thus a group of exponent two is abelian. For distinct nonidentity $a,b$, the four elements $1,a,b,ab$ are distinct: for example $ab=1$ would imply $a=b$, and $ab=a$ would imply $b=1$. Products remain in this set because $a,b$ commute and square to one. Thus they form a Klein four-group subgroup of order four.

Now let $|G|=2p$ with odd prime number $p$. By Lagrange's theorem, every element order is one of $1,2,p,2p$. If an element has order $2p$, its square has order $p$. If no element had order $p$, all nonidentity elements would therefore have order two. There are at least two distinct such elements, producing a subgroup of order four. This contradicts $4\nmid 2p$. Hence there is an element of order $p$.

For a conservative system near a stable equilibrium, the quadratic Lagrangian has the form $L=\frac{1}{2}{\dot{q}}^{T}M\dot{q}-\frac{1}{2}{q}^{T}Kq$, with symmetric mass matrix $M$ and stiffness matrix $K$. A normal mode $q=v{e}^{i\omega t}$ satisfies the generalized eigenvalue problem for small oscillations $(K-{\omega }^{2}M)v=0$, so the frequency equation is $\det (K-{\omega }^{2}M)=0$.

To quadratic order, vertical velocities contribute only fourth-order corrections to the kinetic energy, leaving $T=({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)/2$. A string of length $\ell$ with small horizontal difference $d$ shortens its vertical projection by $d^2/(2\ell )$. The upper segment carries all three masses, the middle segment two, and the lower segment one. Hence their increases in gravitational potential energy sum toSubtracting $V$ from $T$ gives the stated approximate Lagrangian. Here $M=I$ andFor a mode with middle displacement zero, write $v=(X,0,Z{)}^T$. The middle row requires $X/b+Z/c=0$, so $Z=-cX/b$. A nonzero mode then has $X,Z\ne 0$, and the outer rows require ${\omega }^2/g=a^{-1}+b^{-1}=c^{-1}$. Therefore the necessary and sufficient condition isWhen it holds, the mode shape is $(1,0,-c/b)$ with $\omega =\sqrt{g/c}$. This is the force-cancellation mechanism of the stationary middle mass in a three-mass spring chain, applied to the equivalent quadratic stiffnesses here.

A two-index Cartesian tensor has components transforming as $T_{ij}^{\prime }=R_{ik}{R}_{jl}{T}_{kl}$ under an orthogonal change of orthonormal basis. The tensor order here counts indices, rather than the matrix rank. Its scalar, symmetric-traceless and axial tensor decomposition follows by settingThen $T_{ij}=Y{\delta }_{ij}+{\Omega }_{ij}+S_{ij}$, with $\Omega$ antisymmetric and $S$ symmetric and traceless. There are three independent antisymmetric components and five independent symmetric traceless components: three off-diagonal entries and two independent diagonal ones in the latter case.

For the symmetric tensor relative to a nonzero vector $A$, put $q=A_i{A}_i>0$, $s=A_i{T}_{ij}{A}_j$, and $t=T_{ii}$. The symmetric tensor decomposition relative to an axis givesIn particular $A_i{B}_i=s/q-s/q=0$. Define the orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/q$ onto the plane perpendicular to $A$, and setThis tensor is symmetric and kills $A$. Its trace is $t-s/q-2\alpha =0$. To verify the full decomposition, set $u=A/\sqrt{q}$ and split $T$ into its longitudinal component $(u^{T}Tu)u{u}^T$, two mixed components, and $PTP$. The mixed terms are exactly $B{A}^T+A{B}^T$, and $\alpha +\beta q=s/q$ gives the longitudinal coefficient. Thus all requested constraints and the reconstruction hold.

The residue theorem states that the contour integral of a meromorphic function around a positively oriented simple closed contour equals $2\pi i$ times the sum of its enclosed residues, provided there are no singularities on the contour. The roots of $1+z^n$ are $z_k=e^{(2k+1)i\pi /n}$, $k=0,\dots ,n-1$. They are simple poles, with residues $1/(n{z}_k^{n-1})=-z_k/n$.

Use the sector contour of angle $2\pi /n$, running outward along the positive axis and inward along the rotated ray. It encloses only $z_0=e^{i\pi /n}$. The large-arc contribution is $O(R^{1-n})$, hence vanishes. With $J={\int }_0^{\infty }dx/(1+x^n)$, the two ray contributions giveSince $1-e^{2i\alpha }=-2i{e}^{i\alpha }\sin \alpha$,For the second integral, use $z^{n/2}/(1+z^n)$ on a branch continuous across the sector away from zero, with a small inner arc of radius $ϵ$. Its large arc is $O(R^{1-n/2})$ and its small arc is $O({ϵ}^{1+n/2})$, both vanishing for $n\ge 3$. On the rotated ray, $z^{n/2}$ changes by $e^{i\pi }=-1$, so the two radial contributions add rather than subtract. The enclosed residue is $z_0^{1-n/2}/n=-i{e}^{i\pi /n}/n$. ConsequentlyBoth answers are instances of the Mellin integral of a reciprocal binomial. The direct contour argument also tracks the branch needed when $n$ is odd.

A real two-dimensional group representation is a group homomorphism $D:G\to GL(2,\mathbb{R})$, so $D(AB)=D(A)D(B)$ and $D(I)=I$.

Substitute $\tilde{x}=\alpha x+\beta y$ and $\tilde{y}=\gamma x+\delta y$ into the quadratic expression. Its coefficients of $x^2$ and $y^2$ are ${\alpha }^2\hat{a}+{\gamma }^2\hat{b}$ and ${\beta }^2\hat{a}+{\delta }^2\hat{b}$. The cross term is zero for every matrix in this particular cyclic group, because each has one nonzero entry per row. ThusLet $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ generate the group. For $A=\pm I$, $M=I$; for $A=\pm J$, $M=S:=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Since $S^2=I$, each $M$ is its own inverse and $D(J^k)=S^k$. The group law is respected: $S^{k+l}=S^k{S}^l$, and $S^4=I$ makes it independent of the representative modulo four. This proves that $D=M^{-1}$ is a group representation.

This quarter-turn action on diagonal quadratic forms hasso it is not faithful. Quadratic forms cannot distinguish a simultaneous sign reversal of both coordinates. The invariant coefficient directions $(1,1)$ and $(1,-1)$ correspond to the fixed form $x^2+y^2$ and the form $x^2-y^2$ whose sign changes under a quarter-turn.

Put $\Phi (r,\theta )=R(r)Y(\theta )$ in Laplace's equation, multiply by $r^2/(RY)$, and separate the two variables:With $s=\cos \theta$, the angular equation becomes $(1-s^2)Y_{ss}-2s{Y}_s+\lambda Y=0$. Requiring a smooth axisymmetric harmonic function at both poles selects $\lambda =n(n+1)$ and the Legendre polynomial $Y=P_n(s)$, for nonnegative integers $n$. The radial Cauchy-Euler equation has solutions $r^n$ and $r^{-n-1}$. Superposition over the angular eigenfunctions therefore givesThis expansion assumes regularity on each whole angular sphere; without it, singular angular solutions would also be possible. The boundary data in the shell involve only the $n=0,1$ modes. Define the spherical-shell harmonic interpolation weightsEach $p_n$ equals one at the inner boundary and zero at the outer one, and $q_n$ does the reverse. Hence the steady temperature isEach term is a harmonic function, and substitution verifies both prescribed boundary values. Uniqueness follows from the maximum principle for harmonic functions.

Take a variation $y_{\epsilon }=y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. Differentiating the functional at zero and using integration by parts givesThe boundary term vanishes. Since the remaining integral must vanish for every smooth endpoint-zero variation, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation $d(f_{y^{\prime }})/dx=f_y$.

For the mechanical problem put $A(r)=1-2m/r>0$. The Lagrangian is independent of $\theta$, a cyclic coordinate, so its conjugate momentum $\partial L/\partial \dot{\theta }=2r^2\dot{\theta }=2J$ is conserved. Therefore $J$ is constant, including the factor of two between $J$ and the canonical momentum for this normalization of $L$.

For the radial coordinate, $\partial L/\partial \dot{r}=2\dot{r}/A$ and $\partial L/\partial r=-{\dot{r}}^2{A}^{\prime }/A^2+2r{\dot{\theta }}^2-\mu A^{\prime }/A^2$. Its Euler-Lagrange equation isFor a circular orbit $r=R>2m$, the radial velocity and acceleration vanish. Using $A^{\prime }(R)=2m/R^2$ and $\dot{\theta }=J/R^2$ leaves $2J^2/R^3=2m\mu /[R^{2}A(R{)}^2]$, henceThis is the circular orbit condition for a radial kinetic Lagrangian applied to the coefficients here.

Choose the left homogeneous solution $v(x)=\sinh x$ and the right one $w(x)=\sinh (1-x)$. Their Wronskian is $v{w}^{\prime }-v^{\prime }w=-\sinh 1$. The Dirichlet modified Helmholtz Green function on an interval, with operator sign $D_x^2-1$, is thereforeIt vanishes at both endpoints, is continuous at $x=\xi$, and its first derivative jumps by $+1$, giving $(D_x^2-1)G=\delta (x-\xi )$. The Green-function representation then givesThe two integrals are $(e^{2x}-1)/4-x/2$ and $e(1-x)/2-(e-e^{2x-1})/4$, respectively. Substituting them and simplifying givesDirect differentiation verifies $u^{\prime \prime }-u=e^x$ and both zero boundary values. The negative kernel also explains why the solution is negative inside the interval despite the positive right-hand side.