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The Möbius group consists of all Möbius transformations of the Riemann sphere ${\mathbb{C}}_{\infty }=\mathbb{C}\cup \{\infty \}$, with operation function composition. A nonzero-determinant matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ represents the map $z\mapsto (az+b)/(cz+d)$. If $c\ne 0$, its pole $-d/c$ maps to $\infty$ and $\infty$ maps to $a/c$; if $c=0$, the map is affine and fixes $\infty$. Nonzero scalar multiples of $M$ represent the same transformation.

Define $\varphi (M)$ by this action for $M\in S{L}_2(\mathbb{C})$. Substitution of fractional-linear expressions shows $\varphi (MN)=\varphi (M)\circ \varphi (N)$, so $\varphi$ is a group homomorphism. Any invertible matrix can be rescaled into the special linear group: choose $\lambda \in {\mathbb{C}}^{\times }$ with ${\lambda }^2\det M=1$, which is possible because every nonzero complex number has a square root. This does not change the transformation, proving surjectivity. A transformation represented by $M$ is the identity only if fixing $0$, $\infty$ and $1$ forces $b=c=0$ and $a=d$. The determinant-one condition then gives $a^2=1$. ThusFor the fixed-point claim, a nonidentity Möbius transformation has at least one fixed point of a Möbius transformation: when $c\ne 0$ the finite fixed points solve a quadratic, and when $c=0$ infinity is fixed. A quadratic shows there can be no more than two distinct fixed points unless the transformation is the identity. Suppose there is exactly one. Conjugate it to infinity by a Möbius transformation. The resulting map has form $z\mapsto az+b$, since it fixes infinity. If $a\ne 1$ it also has the finite fixed point $b/(1-a)$, a contradiction. Hence it is $z\mapsto z+b$ with $b\ne 0$. Its $m$th power is $z\mapsto z+mb$, never the identity for $m>0$. Conjugation preserves order, so a nonidentity finite-order Möbius transformation has exactly two fixed points.

Now let $K$ be a finite abelian subgroup of the Möbius group. First suppose it contains an element $g$ of order greater than two. Conjugate the two fixed points of $g$ to $0$ and $\infty$, so $g(z)=\zeta z$, with $\zeta$ a root of unity of order greater than two. Every element $h$ commuting with $g$ permutes its fixed-point set, because $g(h(p))=h(g(p))=h(p)$. A Möbius transformation preserving each of $0,\infty$ is $h(z)=az$; one exchanging them is $h(z)=a/z$. In the second case $hg(z)=a/(\zeta z)$ and $gh(z)=\zeta a/z$, so commuting would force ${\zeta }^2=1$, impossible. Thus all elements of $K$ are scalings. A finite subgroup of a field multiplicative group is cyclic; here one can see this directly by choosing the least common multiple $m$ of all element orders and observing that the scalars lie among the $m$th roots of unity, a cyclic group. Hence $K$ is cyclic.

Otherwise every nonidentity element of $K$ has order two. If $K$ is nontrivial, choose $g$ and conjugate it to $z\mapsto -z$. The action of $K$ on $\{0,\infty \}$ has image of order at most two. Its kernel consists of scalings $z\mapsto az$, and the order-two condition forces $a^2=1$, so the kernel has at most two elements. Hence $|K|\le 4$. Orders one and two give cyclic groups; order four with all nonidentity elements involutions gives $C_2\times C_2$, generated by any two distinct nonidentity elements. Both possibilities occur: scalings by roots of unity give every finite cyclic group, and $\{z,-z,1/z,-1/z\}$ gives the Klein four group. The classification is

For a group action of a finite group $G$ on a set and a point $x$, the orbit-stabilizer theorem iswhere $G_x=\{g:gx=x\}$ is the stabilizer subgroup. The map $g{G}_x\mapsto gx$ is well defined because elements of the same coset act identically on $x$. If $gx=hx$, then $h^{-1}g\in G_x$, so their cosets agree; every orbit point has the form $gx$, giving surjectivity. This is a bijection. Lagrange's theorem, $|G|=[G:H]|H|$ for any subgroup $H$, yields the counting formula.

Apply the conjugation action of the finite p-group $G$ to its nontrivial normal subgroup $N$. Normality ensures the action stays inside $N$. The singleton orbits of a group action are exactly the elements of $N\cap Z(G)$. Every larger orbit of a group action has size a positive power of $p$, by the orbit-stabilizer theorem and Lagrange's theorem. ThereforeSince $N$ is nontrivial, its order is divisible by $p$. The intersection contains the identity and has cardinality a positive multiple of $p$, so it contains a nonidentity element. This proves the central intersection property of normal subgroups of finite p-groups.

For the proper subgroup $H$, let $H$ act by left multiplication on the left cosets $G/H$. A coset $gH$ is fixed by every $h\in H$ exactly when $g^{-1}Hg\subseteq H$. Both groups have the same finite size, so this inclusion is equality; the fixed cosets are precisely those with $g\in N_G(H)$. Hence their number is $[N_G(H):H]$. All non-singleton orbits of a group action have size divisible by $p$, soSince $H$ is proper, $[G:H]$ is a positive power of $p$ greater than one. The fixed-coset count is nonzero, because $H$ itself is fixed, and is divisible by $p$. Therefore the normalizer strictly contains $H$:This is the normalizer condition for finite p-groups. The proof also covers $H=\{1\}$, whose action fixes all cosets.

Every permutation is a product of transpositions, since a cycle can be writtenwith rightmost factors acting first. Define the sign of a permutation to be $(-1{)}^r$ for a product of $r$ transpositions. To prove that it is well defined without assuming parity, let $P_{\sigma }$ be the permutation matrix specified by $P_{\sigma }{e}_i=e_{\sigma (i)}$. Then $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, and a transposition exchanges two columns of the identity, so its determinant is $-1$. Any decomposition into $r$ transpositions therefore hasThe left-hand side depends only on $\sigma$, proving well-definedness. Multiplicativity of the determinant now gives the sign homomorphism:For the first requested group embedding, use the faithful four-point action of GL2 over F2: label the four vectors in ${\mathbb{F}}_2^2$ and let an invertible $2\times 2$ matrix act on them by multiplication. Composition of these actions gives a group homomorphism $\psi :G{L}_2({\mathbb{F}}_2)\to S_4$. If a matrix fixes every vector, it fixes both standard basis vectors and is the identity, so the action is faithful. The matrixfixes $(0,0)$ and $(1,1)$ and exchanges $(1,0)$ with $(0,1)$. Its image is a single transposition, hence $\mathrm{sgn}(\psi (B))=-1$. The composite sign map is nontrivial.

For the second group embedding, simply takeThe preceding multiplication formula proves the group homomorphism property, and $P_{\sigma }=I$ forces $\sigma (i)=i$ for every $i$, proving injectivity.

Let $R$ be an orthogonal matrix describing a Cartesian change of coordinates, with transformed vector components $v_i^{\prime }=R_{ia}{v}_a$ and $w_j^{\prime }=R_{jb}{w}_b$. Their outer product transforms aswhich is the transformation law of a second-order Cartesian tensor. Here “rank two” counts tensor indices, not the rank of a matrix; a nonzero outer product has matrix rank one.

For an isotropic second-rank tensor, invariance means $RT{R}^T=T$ for every proper rotation matrix. The half-turn matrices $\mathrm{diag}(1,-1,-1)$, $\mathrm{diag}(-1,1,-1)$ and $\mathrm{diag}(-1,-1,1)$ force all off-diagonal entries to vanish. Quarter-turns around coordinate axes then force the three diagonal entries to be equal. Conversely $RI{R}^T=I$, so the most general second-rank isotropic tensor isSince $v{w}^T$ has matrix rank at most one, it cannot equal a nonzero scalar multiple of the three-dimensional identity. Thus its isotropy forces $v{w}^T=0$. If $v\ne 0$, a nonzero component $v_i$ makes the entire $i$th row vanish only when $w=0$; the converse is immediate. The required choices are exactlyFor the Levi-Civita symbol, multilinearity and antisymmetry of the determinant givefor every $R\in SO(3)$. This proves the rotation invariance of the Levi-Civita symbol, hence its third-rank isotropy. The convention is invariance under proper rotations: under an orthogonal reflection $\det R=-1$, the sign reverses. Thus it is a pseudotensor if transformations of both orientations are included.

The chain rule along a smooth curve isDifferentiating the circular parametrization gives the tangent vectorThus the chain rule turns the angular derivative of $u$ into $-y{u}_x+x{u}_y$. The given partial differential equation makes this equal to $u$, so along every portion of the circle inside the upper half-plane, $U(t)=u(\mathbf{x}(t))$ satisfies the ordinary differential equation $U^{\prime }=U$. Its solution is $U(t)=U(t_0)e^{t-t_0}$.

For the two specified points take $a=\sqrt{2}$. The initial point corresponds to $t_0=\pi /4$ and the terminal point to $t_1=3\pi /4$. The entire intervening arc lies in $y>0$, so no continuation outside the domain is used. The requested value isThese circular characteristic curves also show locally that the general solution has the polar-coordinate form $u(r,\theta )=C(r)e^{\theta }$, with $0<\theta <\pi$; there is no requirement of angular periodicity on the upper half-plane.

Lagrange's theorem states that, for a finite group $G$ and a subgroup $H$,To prove it, define an equivalence relation by $x\sim y$ when $x^{-1}y\in H$. Its classes are the left cosets $xH$, which therefore partition $G$. Left multiplication by $x$ gives a bijection $H\to xH$, so every coset has $|H|$ elements. Counting this partition gives the formula; in particular $|H|$ divides $|G|$. Applying the theorem to the cyclic subgroup generated by an element shows that its order of a group element divides $|G|$.

Write the given order-two normal subgroup as $N=\{1,z\}$. Conjugation must preserve $N$ and cannot send its nonidentity element to the identity, so $gz{g}^{-1}=z$ for all $g\in G$. Thus $z$ is in the center of a group: a normal subgroup of order two is central.

Choose $g\notin N$. By Lagrange's theorem, its order of a group element is $2$, $p$ or $2p$. It cannot have order two: since $z$ is central, the four distinct elements $1,z,g,zg$ would form a subgroup of order four, whereas $4$ does not divide $2p$ for odd $p$. If $g$ has order $2p$, it already generates $G$. Otherwise $g$ has order $p$, commutes with $z$, and $(gz{)}^k=1$ implies $g^k=z^{-k}\in ⟨g⟩\cap N=\{1\}$. Hence both $p$ and $2$ divide $k$, so $gz$ has order $2p$. In either case