The kernel is a normal subgroup, so it contains the identity class and is a union of the classes listed in part (c). Its candidate orders are therefore $1$, $1+3=4$, $1+2=3$ and $1+3+2=6$. Lagrange's theorem excludes four, since it does not divide six. The remaining possibilities areAll three are normal: the first and last trivially, while $A_3$ is the kernel of the map giving the sign of a permutation, or the index-two subgroup generated by a 3-cycle.

Because $\Phi$ is onto, the first isomorphism theorem gives $G^{\prime }\cong S_3/K$. Explicitly, the map $gK\mapsto \Phi (g)$ is well defined, multiplicative and surjective; equality of its values means $h^{-1}g\in K$, so it is also injective. ThusEach is realized by the identity map, the sign map and the trivial map, respectively. This completes the classification of normal subgroups and quotients of the symmetric group on three letters.

The cycle type records the lengths of the disjoint cycles in a permutation, including fixed points as length-one cycles. The possibilities for three objects areThese exhaust the six elements of the symmetric group $S_3$, denoted ${\Sigma }_3$ in the source. They are also the conjugacy classes of the symmetric group on three letters.

The conjugacy class of $g$ isIf $N$ is a normal subgroup and $g\in N$, then normality gives $hg{h}^{-1}\in N$ for every $h$. Thus the whole conjugacy class of each member of $N$ is contained in $N$, andConjugacy classes partition the group, so this is a union of complete disjoint classes, never selected fractions of classes.

The kernel of a group homomorphism isThe homomorphism sends identity to identity and inverses to inverses. Thus $e_G\in K$, and if $k_1,k_2\in K$, then $\Phi (k_1{k}_2^{-1})=e_{G^{\prime }}$, so $K$ is a subgroup by the subgroup test.

For any $g\in G$ and $k\in K$,Hence $gK{g}^{-1}\subseteq K$. Applying the same inclusion with $g^{-1}$ gives equality, provingThis normality argument does not require either group to be finite.

Perform the product and inverse formulas from part (b) in the commutative ring $\mathbb{Z}/4\mathbb{Z}$. They involve only addition, multiplication and negatives, so closure and inverses remain valid; no division by a nonunit is needed. Associativity again comes from matrix multiplication, and $A(0,0,0)$ remains the identity. The Heisenberg group over a commutative ring therefore hasThese are residues modulo four, not arithmetic in the finite field of four elements.

For the additional subgroup task, write an element of $H$ as $A(t,y,t)$. Its product and inverse areBoth remain in $H$, and the product is unchanged on exchanging the factors because $ts=st$. The identity is included, so $H$ is an Abelian subgroup. There are four independent choices each for $t,y$, givingThus there are four distinct left cosets. Explicitly the surjective group homomorphism $A(x,y,z)\mapsto x-z(\mathrm{mod}4)$ has kernel $H$; the four cosets are the level sets of that difference, with representatives $A(d,0,0)$, $d=0,1,2,3$. This is the Abelian plane subgroup of the Heisenberg group modulo four.

Denote a matrix by $A(x,y,z)$, with $y$ the upper-right entry. Direct matrix multiplication givesThis proves closure. Associativity is inherited from matrix multiplication. The identity is $A(0,0,0)$, and the inverse isBoth products with this matrix are the identity. Thus these real matrices form the Heisenberg group, in the coordinate convention with central coordinate $y$.

Let $N=\{A(0,b,0):b\in \mathbb{R}\}$. Its multiplication adds $b$, its identity is $b=0$ and its inverse negates $b$, so it is a subgroup. MoreoverThus every element of $N$ is central, and $gN{g}^{-1}=N$. Therefore $N$ is a normal subgroup. In fact it is the full center of a group, as described by the center quotient of the real Heisenberg group.

The left coset isEvery $g\in G$ belongs to $gH$, because $e\in H$, so the cosets cover $G$. If $gH$ and $kH$ intersect, choose $g{h}_1=k{h}_2$ in the intersection. Then $k^{-1}g=h_2{h}_1^{-1}\in H$, so $g=kh$ for some $h\in H$. Thus $gH=khH=kH$.

Hence two left cosets are either identical or disjoint, and the distinct left cosets form a partition of a set whose union is $G$. Finiteness is unnecessary for this partition argument. In the finite case each coset has $|H|$ members, since multiplication by $g$ is a bijection, giving Lagrange's theorem.

For equal masses, write $M=mB$ with $B=\left(\begin{array}{ccc}3& 1& 1\\ 1& 1& 0\\ 1& 0& 1\end{array}\right)$. The generalized eigenvalue problem for small oscillations is $kv={\omega }^{2}mBv$.

Exchange symmetry of climbers 2 and 3 gives $v_a=(0,1,-1{)}^T$, for which $B{v}_a=v_a$. HenceThe physical displacements are $(y_1,y_1+y_2,y_1+y_3)$, so climber 1 is stationary and the other two move vertically with equal amplitudes in opposite directions. Their changes in force on climber 1 cancel.

For the remaining symmetric modes, put $v=(a,b,b{)}^T$ and let $\mu =k/(m{\omega }^2)$. The eigenproblem becomes $3a+2b=\mu a$, $a+b=\mu b$. Eliminating $a,b$ gives $(3-\mu )(1-\mu )-2={\mu }^2-4\mu +1=0$. Therefore $\mu =2\pm \sqrt{3}$. Since $(2+\sqrt{3})(2-\sqrt{3})=1$,For clarity, their relative physical climber displacements, rather than rope-length amplitudes, areThe lower mode moves all climbers in phase; the higher mode moves the lower pair against the top climber. These are the normal modes of three vertically suspended masses with a branched spring connection.

Set $L_i={\hat{L}}_i+y_i$. The equilibrium equations cancel the constant gravitational and elastic terms, leaving the small-oscillation mass and stiffness matricesThe off-diagonal mass terms arise because rope-length coordinates are not absolute climber displacements. For any nonzero velocity vector, ${\dot{y}}^{T}M\dot{y}=m_1{\dot{y}}_1^2+m_2({\dot{y}}_1+{\dot{y}}_2{)}^2+m_3({\dot{y}}_1+{\dot{y}}_3{)}^2>0$, so $M$ is a positive-definite matrix.

Within the vertical Hookean model, the displacement equations are exactly linear while all ropes stay taut. Small amplitudes ensure that none reaches the slack regime $L_i\le L_0$.

The Euler-Lagrange equations giveFor example, differentiating $T$ with respect to ${\dot{L}}_1$ collects contributions from all three climbers, because changing $L_1$ moves all of them. At equilibrium, all accelerations vanish, soThe top rope supports the total weight, while each lower rope supports only its own climber. These extensions are positive, consistent with the taut-rope tension law.

Take downward vertical distances from the support. The climber positions are $q_1=L_1$, $q_2=L_1+L_2$ and $q_3=L_1+L_3$. Their kinetic energy is thereforeIntegrating the tension-extension relation gives the elastic potential energy $k(L_i-L_0{)}^2/2$ for each taut rope. Because downward displacement lowers gravitational potential,An arbitrary additive constant in $U$ is immaterial. This is the vertical-motion model implicit in using only rope lengths as generalized coordinates; swinging angles are not additional degrees of freedom in this model.

The scalar, symmetric-traceless and axial tensor decomposition fixes the scalar part by the trace:Subtracting this isotropic part from the symmetric tensor in part (c) givesIts trace is zero. The antisymmetric tensor part isAdding $PI+\hat{S}+\hat{A}$ reconstructs every entry of $F$.

Subtracting a scalar multiple of the identity preserves every eigenvector: if $Sv=\lambda v$, then $(S-PI)v=(\lambda -P)v$. Thus the principal axes are the same as in part (c), with shifted principal valuesThe two tensors are coaxial symmetric tensors. At $r=0$ both vanish, so neither specifies unique axes.

Let $r=\sqrt{x_1^2+x_2^2}$. Averaging each pair of transposed entries gives the symmetric tensorFor $r>0$, the upper block is the outer product of $(x_1,x_2)$ with itself. It sends that vector to $r^2$ times itself and annihilates its perpendicular. Thus its orthonormal principal axes and corresponding eigenvalues areTaking these vectors as the rows of the change-of-components matrix givesThis matrix is orthogonal with determinant one. These are the principal axes of a planar rank-one tensor block.

The matrix is a pointwise frame transformation: because its angle varies with position, it does not represent one fixed Cartesian rotation diagonalizing the entire field everywhere. On the axis $r=0$, the symmetric tensor is zero; all principal values coincide and any orthonormal axes may be chosen. The displayed formula for $M$ is therefore used only off that axis.

Define the symmetric tensor and antisymmetric tensor parts by $S=(T+T^T)/2$ and $A=(T-T^T)/2$. The transformed transpose satisfies $(T^{\prime }{)}^T=(MT{M}^T{)}^T=M{T}^T{M}^T$. Thereforeand similarlyBoth obey the tensor component transformation law. Thus the symmetric and antisymmetric parts are themselves second-order tensors. Taking a transpose and forming linear combinations commute with this orthogonal change of frame.

Preserving Cartesian lengths and Euclidean inner products requires an orthogonal matrix:Its determinant is $\pm 1$. Relative to an originally right-handed frame, $\det M=+1$ gives a right-handed frame and $\det M=-1$ a left-handed frame. More generally, this sign determines whether the two frames have the same or opposite orientation.

The tensor component transformation law isFor a tensor field, the components are compared at the same physical point, whose coordinates satisfy $x=M^T{x}^{\prime }$.

Use the Fourier transform convention $\tilde{y}(\omega )={\int }_{\mathbb{R}}y(t)e^{-i\omega t}dt$. For a transformable response with the necessary boundary terms vanishing, integration by parts givesTransforming the differential equation therefore gives $(5-{\omega }^2+2i\omega )\tilde{y}=\tilde{f}$, soThe identities also hold for tempered distributions. The denominator factors as $(\omega -i-2)(\omega -i+2)$; its two poles are in the upper half-plane, corresponding under inverse factor $e^{i\omega t}$ to the causal Green function of a damped oscillator.

The frequency-domain formula selects the Fourier-transformable response. Decay of the force alone would not remove arbitrary homogeneous solutions, which grow towards negative time and do not have an ordinary bilateral Fourier transform.

Choose a branch of the complex logarithm with $-\pi /2<\arg z<3\pi /2$, placing its branch cut down the negative imaginary axis. Thus $\mathrm{Log}x=\ln x$ for $x>0$, while $\mathrm{Log}(-x)=\ln x+i\pi$ on the negative real axis for $x>0$.

Integrate $\mathrm{Log}z/(z^2+a^2)$ around an upper semicircle, with a small clockwise semicircular indentation above zero. The large arc contributes $O(\ln R/R)\to 0$, and the small arc contributes $O(\epsilon |\ln \epsilon |)\to 0$. The only enclosed pole is $ia$, withWriting $I(a)$ for the required integral, the negative real segment contributes $I(a)+i\pi {\int }_0^{\infty }dx/(x^2+a^2)$. Hence the residue theorem givesThe imaginary terms cancel, leavingThe integral converges at both endpoints because $|\ln x|$ is integrable near zero and $\ln x/x^2$ is integrable at infinity. As a real-variable check, scaling $x=au$ reduces the result to the vanishing first logarithmic moment at $a=1$, one of the logarithmic moments of the Cauchy kernel.