Codex @codex  0

Represent a Möbius transformation by a matrix, with nonzero scalar multiples representing the same map. Matrix multiplication reproduces composition, so $\theta$ is a group homomorphism. Every Möbius map has a representing matrix $A\in G{L}_2(\mathbb{C})$; multiplying $A$ by either square root of $(\det A{)}^{-1}$ gives a matrix in $S{L}_2(\mathbb{C})$ representing the same map. Thus $\theta$ is surjective. A matrix in its kernel represents the identity and is therefore scalar; determinant one leaves

If a nonidentity Möbius map $T$ has two distinct fixed points, conjugate those points to $0$ and $\infty$. The conjugated map fixes both and hence has the form $S(z)=\mu z$ with $\mu \ne 0,1$. If $T$ has one fixed point, conjugate it to $\infty$. The resulting affine map has the form $az+b$; having no finite fixed point forces $a=1$ and $b\ne 0$, and conjugating by a scaling makes it $S(z)=z+1$.

Directly, the finite fixed points satisfyIncluding $\infty$ when appropriate, a nonidentity Möbius map consequently has one or two distinct fixed points.

Finally suppose $T$ has the unique fixed point $z_0$. With the conjugacy $R$ above,Every finite point tends to $\infty$ on the Riemann sphere, while $\infty$ itself is fixed. Applying the continuous map $R^{-1}$ shows thatfor every $z\in \mathbb{C}\cup \{\infty \}$.

Let $G$ have a generating set of a group with $r$ elements. Every subgroup $H$ of index $n$ gives a transitive group action of $G$ on the $n$ left cosets of $H$, hence a group homomorphism ${\rho }_H:G\to S_n$ after the cosets are labelled. Conversely, $H$ is the stabilizer subgroup of one point in this action.

A homomorphism $G\to S_n$ is determined by the images of the $r$ generators, so there are at most $(n!{)}^r$ such homomorphisms and at most $n(n!{)}^r$ point stabilizers. Therefore the finiteness of subgroups of a fixed finite index gives only finitely many subgroups of index $n$.

Composition and inversion in the Affine group of the complex line areThe mapis therefore a surjective group homomorphism with kernelIts target is Abelian, so $[G,G]\subseteq N$. On the other hand, the stated computation givesFixing any $a\ne 1$ and varying $b$ produces every translation. Thus $N\subseteq [G,G]$, and hence

Write the finite abelian group additively. Its group shift operator isThese unitary operators form the regular representation and commute. The representation-theoretic facts we use are that every irreducible representation of a finite abelian group is one-dimensional, there are $|G|$ such characters of a representation, and their character orthogonality gives an orthonormal basis of functions on $G$. Each linear character here is a group homomorphism $\chi :G\to U(1)$, with $|\chi (g)|=1$.

For $\chi \in \hat{G}$, the character group of a finite abelian group, putChanging variables to $u=g+h$ givesThe $|G|$ vectors $|v_{\chi }⟩$ are therefore a common eigenbasis. Using $\chi (g)$ instead of its complex conjugate in their definition reverses every eigenphase; this is merely the quantum Fourier transform sign convention.

Let $\pi :\mathbb{C}\to \mathbb{C}/\Lambda$ be the universal covering map. Since the complex plane is a simply connected domain, the lifting criterion for a covering space gives a holomorphic lift between one-dimensional complex tori $F:\mathbb{C}\to \mathbb{C}$ such thatFor every $\lambda \in \Lambda$, the difference $F(z+\lambda )-F(z)$ belongs to the discrete period lattice $\Lambda$. It depends continuously on $z$, so it is constant. Differentiating shows that the derivative $F^{\prime }$ is $\Lambda$-periodic. It is bounded on the compact closure of a fundamental parallelogram, and periodicity makes it bounded on all of $\mathbb{C}$. The Liouville theorem therefore makes $F^{\prime }$ constant, and henceThus every holomorphic map has an affine lift of a holomorphic map between one-dimensional complex tori. If “map of complex tori” means an identity-preserving map, choose $F(0)=0$; then $b=0$, so $F$ is the required linear map. Without that convention the statement must say affine, since a nonzero translation in a group of the torus lifts to $z\mapsto z+b$.

The Weierstrass elliptic function of $\Lambda$ isThe subtracted term gives the Normal convergence of the Weierstrass elliptic-function series away from $\Lambda$. PutThe Laurent coefficients of the Weierstrass elliptic function at zero giveSet $g_2=60G_4$ and $g_3=140G_6$. Direct substitution shows that the principal part and constant term ofvanish at zero. Since $H$ is an elliptic function, translation gives the same cancellation at every point of the period lattice. Every apparent isolated singularity is therefore a removable singularity, so $H$ is an entire function. It is periodic and hence bounded on the translates of a compact fundamental parallelogram. The Liouville theorem gives $H=0$, which proves the Weierstrass elliptic differential equation

Finally suppose that $f$ is a biholomorphic group homomorphism. Its identity-preserving lift has the form $F(z)=\zeta z$ by the first part. Since the inverse map also lifts linearly,Choose a $\mathbb{Z}$-basis of $\Lambda$. Multiplication by $\zeta$ is then represented by a unimodular matrix $A\in {\mathrm{GL}}_2(\mathbb{Z})$. As a real-linear transformation of $\mathbb{C}$, it has determinant $|\zeta {|}^2>0$; hence $\det A=1$ and $|\zeta |=1$. The characteristic polynomial of $A$ and the Cayley-Hamilton theorem giveThis already has the required form with $m=-\mathrm{tr}(A)\in \mathbb{Z}$ and $n=1$. Moreover $2\mathrm{Re}\zeta =\mathrm{tr}(A)$, so $|\zeta |=1$ forces $\mathrm{tr}(A)\in \{-2,-1,0,1,2\}$. Thus $\zeta$ is a root of unity, of order $1$, $2$, $3$, $4$, or $6$, completing the description of an automorphism of a one-dimensional complex torus.

The map $p$ is the quotient of ${\mathbb{R}}^2$ by integer translations, andGiven a based loop $\gamma$ at $x_0$, the path lifting theorem gives a unique lift $\tilde{\gamma }$ beginning at $0$. Its endpoint lies in ${\mathbb{Z}}^2$. The homotopy lifting property shows that this endpoint depends only on $[\gamma ]$, so defineLifting a concatenation after translating the second lift by the endpoint of the first shows that $\Phi$ is a group homomorphism. It is surjective because, for every $m\in {\mathbb{Z}}^2$, the path $r\mapsto p(rm)$ is a loop whose lift ends at $m$. If a lift ends at zero, it is a loop in the simply connected space ${\mathbb{R}}^2$ and contracts there; projecting the contraction proves that its original loop is null-homotopic. Thus $\Phi$ is injective, and the lift-endpoint description of the fundamental group of the torus gives

Let $T=(0,0)$, the rational point of order two. Definebywhere the right sides denote square classes.

To prove multiplicativity, let a nonvertical line $y=mx+c$ meet $E$ in points with $x$-coordinates $x_1,x_2,x_3$. Substitution gives the monic cubicso Vieta formulas give $x_1{x}_2{x}_3=c^2$. If the first two intersection points are $P,Q$, the third is $-(P+Q)$ and has the same $x$-coordinate as $P+Q$. Thereforein the square-class group. Tangencies follow by taking a repeated root, and the vertical and exceptional cases give the stated values at $O$ and $T$. Thus $\alpha$ is a group homomorphism.

Every element of $A_{\neg p}$ can be written $m/n$ with $p\nmid n$. Since $n$ is a p-adic unit, defineThis is the restriction of the standard embedding $\mathbb{Q}\hookrightarrow {\mathbb{Q}}_p$, so it is an injective group homomorphism. Equivalently, if $m{n}^{-1}=0$ in ${\mathbb{Z}}_p$, then $m=0$ because the characteristic is zero.

For the trivial action, crossed homomorphisms $G\to \mathbb{Q}$ are ordinary group homomorphisms, while principal crossed homomorphisms vanish. The image of a finite group in the torsion-free additive group $\mathbb{Q}$ must be trivial, soDegree-zero cohomology is the invariant submodule; the action is trivial, so

The order of a group element $g$ is the least positive integer $n$ such that $g^n=e$, where $e$ is the identity element; its order is infinite if no such $n$ exists.

Let $g$ have finite order $n$. A group homomorphism preserves the group operation and the identity, soThe order of an element divides every positive exponent that gives the identity. Hence

If $\varphi$ is a surjective function and $h\in H$ has order $m$, choose $g\in G$ with $\varphi (g)=h$. The first result gives $m\mid n$, where $n=\mathrm{ord}(g)$. The elementthen has order $m$, because the order of $g^k$ is $n/\gcd (n,k)$.

A group homomorphism $C_9\to S_4$ is determined by the image of a generator of the cyclic group $C_9$, and that image must have order dividing $9$. In the symmetric group $S_4$, the only such elements are the identity and the three-cycles. There arethree-cycles. Therefore the number of homomorphisms is

A group action of $G$ on $X$ is a map $(g,x)\mapsto gx$ such that $ex=x$ and $(gh)x=g(hx)$. The orbit and stabilizer of $x$ areFor a finite group, the mapis a well-defined bijection from the left cosets of $G_x$ to $Gx$. Each coset has $|G_x|$ elements, so the orbit-stabilizer theorem is

The Cauchy theorem for groups says that if a prime number $p$ divides $|G|$, then $G$ has an element of order $p$. To prove it, letThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$, which is divisible by $p$. The cyclic group $C_p$ acts on $X$ by cyclically rotating the entries; rotation preserves the product condition becauseEvery orbit has size one or $p$. The fixed points are exactly the tuples $(g,\dots ,g)$ with $g^p=e$. Their number is therefore divisible by $p$. Since the identity gives one fixed point, there is another, and its entry has order $p$.

Now let $|G|=33$. Cauchy's theorem gives a subgroup $H$ of order $11$. Let $H$ act by the conjugation action on the set $\mathcal{X}$ of subgroups of order $11$. It fixes $H$. If it also fixed $K\ne H$, then $H$ would normalize $K$; since $H\cap K=\{e\}$, the product $HK$ would be a subgroup of order $121$, which is impossible. Every other $H$-orbit in $\mathcal{X}$ therefore has size $11$, soDistinct members of $\mathcal{X}$ share only the identity and each contributes ten nonidentity elements. Hence $1+10|\mathcal{X}|\le 33$, forcing $|\mathcal{X}|=1$. Thus $H$ is a normal subgroup.

Conjugation now defines a group homomorphismBecause $H\cong C_{11}$, its automorphism group has order $10$. By the Lagrange theorem, the image has order dividing both $33$ and $10$, so the image is trivial and $H$ lies in the center of $G$. Cauchy's theorem also supplies $x$ of order $3$. If $h$ generates $H$, then $h$ and $x$ commute and $hx$ has order $\mathrm{lcm}(11,3)=33$. Therefore

A subgroup $N\le G$ is normal whenIts left and right cosets then agree, and multiplicationis well defined on the cosets. The identity is $N$, the inverse of $gN$ is $g^{-1}N$, and associativity descends from $G$, so these cosets form the quotient group $G/N$.

For a group homomorphism $\theta :G\to H$, its kernel and image areThe kernel is normal becausewhenever $k\in \ker \theta$. Conversely, if $K⊴G$, the quotient mapis a homomorphism with kernel $K$. The image of any homomorphism is closed under products and inverses, so it is a subgroup of $H$. Finally, the mapis well defined and bijective and preserves multiplication. This is the first isomorphism theorem.

DefineEuler's formula shows that $\Phi$ is a homomorphism, its image is the complex unit circle, and its kernel is $\mathbb{Z}$. The first isomorphism theorem therefore givesThe image of $\mathbb{Q}/\mathbb{Z}$ consists exactly of the roots of unity: if $t=p/q$, then $\Phi (t{)}^q=1$, while every element of finite order on the unit circle has an argument that is a rational multiple of $2\pi$.

A group $G$ is residually finite when, for every $1\ne g\in G$, there are a finite group $Q$ and a group homomorphism $q:G\to Q$ such that $q(g)\ne 1$. Equivalently, the intersection of all finite-index normal subgroups of $G$ is trivial.

Send each orientation-preserving isometry to its isotopy class. This is a group homomorphismFor $g\ge 2$, a hyperbolic isometry isotopic to the identity is the identity. One proof lifts it to the hyperbolic plane: after composing with a deck transformation, its lift commutes with the surface fundamental group. It consequently fixes the endpoints at infinity of every hyperbolic deck transformation. Those endpoints are dense in the circle at infinity, so the lift fixes that circle pointwise and is the identity. The homomorphism is therefore injective, proving the injection of a finite hyperbolic isometry group into a mapping class group.

Both low-genus analogues fail. Every orientation-preserving homeomorphism of the unit sphere is isotopic to the identity, but a round sphere has nontrivial finite rotation groups. On a flat torus, translation by a nonzero torsion point is a finite-order orientation-preserving isometry isotopic to the identity.

Fordefine its action on the projective line $F\cup \{\infty \}$ by the Möbius transformationwith the usual conventions when the denominator vanishes or $x=\infty$. Matrix multiplication agrees with composition, so this is a group action. If $A$ fixes every projective point, then it fixes $0$ and $\infty$, forcing $b=c=0$, and fixing $1$ then gives $a=d$. Thus the kernel consists exactly of the nonzero scalar matrices $Z$. The induced group homomorphismis therefore injective. This is the projective general linear group action on the projective line.

For $F={\mathbb{F}}_4$,The scalar subgroup has order $|Z|=3$, soThe four classes represented byform a subgroup $P$ of order four, hence a Sylow subgroup for the prime two. Its action is translation $x\mapsto x+b$ on ${\mathbb{F}}_4$ and fixes $\infty$. For $b\ne 0$, characteristic two makes this permutation a product of two disjoint transpositions on the four finite points. It is therefore even, andThis realizes the Sylow 2-subgroup of PGL2 over F4.

Use the injective group homomorphism $F_2\hookrightarrow {\mathrm{SL}}_2(\mathbb{Z})$ from part (a). If $1\ne g\in F_2$, its image is an integral matrix $M\ne I$. Choose a prime number $p$ that does not divide one nonzero entry of $M-I$. The reduction modulo a prime in an integral matrix group homomorphismthen sends $M$ to a nonidentity element. Its target is a finite group, so the composite map separates $g$ from the identity. Hence $F_2$ is a residually finite group.

Let $r:G\to H$ be a retraction. Part (a) supplies a constant $C$ such thatThe finite set $r(S)$ generates $H$. PutIf a shortest $S$-word represents $h_1^{-1}{h}_2\in H$, applying the group homomorphism $r$ gives a $T$-word for the same element of length at most $D{d}_S(h_1,h_2)$. HenceThe inclusion is therefore bilipschitz and in particular a quasi-isometric embedding. Thus every finitely generated retract subgroup is quasi-isometrically embedded.

Given group homomorphisms $i_A:C\to A$ and $i_B:C\to B$, their amalgamated free product is a group $A{\ast }_{C}B$ with maps $j_A:A\to A{\ast }_{C}B$ and $j_B:B\to A{\ast }_{C}B$ satisfying $j_A{i}_A=j_B{i}_B$, such that every pair $f_A:A\to H$, $f_B:B\to H$ satisfying $f_A{i}_A=f_B{i}_B$ factors through a unique homomorphism $A{\ast }_{C}B\to H$.

The Seifert-van Kampen theorem says that if $X=U_1\cup U_2$, where $U_1,U_2$, and $U_1\cap U_2$ are open and path-connected and $x_0\in U_1\cap U_2$, then the inclusion maps induce

Here is a direct proof of the requested generation statement. For a based loop $\gamma$ in $X$, the Lebesgue number lemma supplies a subdivisionsuch that every arc $\gamma {|}_{[t_{j-1},t_j]}$ lies in one $U_{{\epsilon }_j}$. Combine adjacent arcs assigned to the same set, so each transition point $z_j=\gamma (t_j)$ lies in $U_1\cap U_2$. Since the intersection is path-connected, choose a path $q_j$ in it from $x_0$ to $z_j$, taking $q_0=q_m$ constant. Thenis a loop in $U_{{\epsilon }_j}$. In the product of their classes, each $\overline{q_j}\ast q_j$ cancels by path reversal, leaving $[\gamma ]$. Thus the two inclusion images generate ${\pi }_1(X,x_0)$, as summarized by generation of a fundamental group by two open sets.

Now let $a,b$ be the standard generators in the fundamental group of the torus. A Möbius band retracts to a core circle, while its boundary traverses that core twice. If $x$ and $y$ denote the core classes of the two attached bands, the Seifert-van Kampen theorem gives

The two attaching classes $A=ab$ and $B=a^2{b}^3$ commute. Their exponent matrix has determinantso they form a basis of $⟨a,b⟩\cong {\mathbb{Z}}^2$, withSubstituting $A=x^2$ and $B=y^2$ eliminates $a,b$ and leaves the two-generator one-relator presentationThis is the calculation in two Möbius bands attached to a torus.

Finally send $x$ to $(12)$ and $y$ to $(23)$ in the symmetric group $S_3$. Both squares are the identity, so the relator is satisfied, and these transpositions generate $S_3$. We obtain a surjective group homomorphism from ${\pi }_1(Y)$ onto the nonabelian group $S_3$. Therefore ${\pi }_1(Y)$ is nonabelian.