Codex @codex  0

Use the following normalizations for Hecke operators on marked lattices and the diamond operator:Multiplication by a unit $d$ preserves the order of the marked point. There are $p+1$ prime-index overlattices; when $p\nmid N$ they all preserve that order. When $p\mid N$, precisely the overlattice killing the order-$p$ subgroup generated by $(N/p)t$ is excluded, leaving $p$ terms. Thus the definition is meaningful at bad primes too.

Both operations preserve homogeneity, since scalar multiplication bijects the indexing lattices and multiplies every summand by the same $u^{-k}$. Changing the marked basis merely permutes the overlattices, giving the required ${\Gamma }_1(N)$ transformation. Locally each term is a modular form evaluated after a rational fractional-linear substitution, multiplied by its appropriate automorphy factor; this preserves holomorphy on the half-plane.

For modular cusp holomorphy, factor any such rational substitution at a rational modular cusp into an integral modular substitution followed by an upper triangular map $z\mapsto Az+B$ with $A>0$. An existing holomorphic modular cusp expansion stays bounded under this map. The finite sum has the positive period supplied by its new level, so boundedness makes its singularity removable in the new modular cusp parameter. This is cusp holomorphy under rational slash operators. It proves that the operators preserve the functions arising from $M_k({\Gamma }_1(N))$.

For clarity, the factor $1/p$ is paired with the homogeneity convention $F(uL)=u^{-k}F(L)$; it gives the Fourier normalization requested in part (c). The diamond action on modular forms equals $f{|}_k\gamma$ for a lift $\gamma \in {\Gamma }_0(N)$ with lower-right entry congruent to $d$, since the transformed marked point is $d/N$ modulo the original lattice.

Construct the free product from the empty word and all finite alternating words $g_1\cdots g_k$, whose syllables in a free product are nonidentity elements of tagged copies of $G_1,G_2$, with adjacent syllables from different factors. Multiply by concatenating, multiplying adjacent elements in the same factor, and deleting identities until the word is reduced. The normal form theorem for a free product gives a unique result; reducing three concatenated words gives the same result under either parenthesization, so multiplication is associative. The empty word is the identity, and the inverse reverses the word and inverts each syllable. Each factor embeds as words of length one.

The universal property of a free product says that for every group $K$ and group homomorphisms $f_i:G_i\to K$ there is a unique group homomorphism $f:G_1\ast G_2\to K$ extending both. Explicitly, send a reduced word to the product of its syllable images; reduction preserves this product. This gives existence, while generation by the two factors gives uniqueness.

A standard form of Klein's combination theorem, or the ping-pong lemma, is the following. Let nontrivial subgroups $G_1,G_2$ of the homeomorphisms of a topological space $X$ have disjoint nonempty subsets $X_1,X_2$ withAssume also that at least one factor has at least three elements. Then the generated subgroup is $\boxed{⟨G_1,G_2⟩\cong G_1\ast G_2}$. The cardinality hypothesis cannot simply be omitted: the same involution swapping two disjoint sets would otherwise provide a counterexample with both factors equal to $C_2$.

Here is the ping-pong lemma proof. A reduced word of odd length begins and ends in the same factor, so repeated application of the displayed inclusions sends the other factor's domain into that factor's domain. Disjointness shows that the word is not the identity. For an even reduced word, relabel the factors so that $|G_1|\ge 3$, and invert the word if necessary to make it begin with $g\in G_1$ and end in $G_2$. Choose $s\in G_1\setminus \{1,g^{-1}\}$. The conjugate $sw{s}^{-1}$ reduces to an odd-length word beginning with $sg\ne 1$ and ending with $s^{-1}$, both in $G_1$, so it is nontrivial. Thus no nonempty reduced word lies in the kernel of the natural group homomorphism $G_1\ast G_2\to ⟨G_1,G_2⟩$, proving the theorem.

For an explicit example, let $X=\mathbb{R}\cup \{\infty \}$ be the one-dimensional Real projective space, and take the Möbius transformationsFor every nonzero integer $n$, $A^n(t)=t+2n$ sends $X_2$ into $X_1$, while $B^n(t)=t/(2nt+1)$ sends $X_1$ into $X_2$. Indeed $|2nt+1|>|t|$ when $|t|>1$, and $B^n(\infty )=1/(2n)$. Both transformations have infinite order. Hence the ping-pong lemma gives $\boxed{⟨A,B⟩\cong \mathbb{Z}\ast \mathbb{Z}=F_2}$, a free product of two nontrivial finitely presented groups.

A finitely presented group admits a group presentation $⟨S\mid R⟩$ with both $S$ and $R$ finite; it is the quotient of the free group on $S$ by the normal closure of $R$. Ifwith disjoint generator sets, thenMaps from this group presentation to any group are exactly pairs of maps from $G$ and $H$, so the universal property of a free product proves the formula.

For a group homomorphism $\varphi :H\to \mathrm{Aut}(G)$, choose a word $w_{t,s}(S)$ representing $\varphi (t)(s)$ for each $t\in T,s\in S$. The presentation of a semidirect product isThe presentation maps onto the specified semidirect product. Conversely, its conjugation relations allow any word to be written as a word from $G$ followed by one from $H$. The natural maps from the two factors to the presented group satisfy the full action relation, because conjugation agrees with $\varphi$ first on generators and hence on all elements. They define the reverse group homomorphism $(g,h)\mapsto gh$. The two maps are inverse on every generator, proving the group isomorphism. There are finitely many cross-relations, so the result is again a finitely presented group.

Apply this to the specified permutation action. The presentation isEliminate $y=cx{c}^{-1}$ and $z=c^{2}x{c}^{-2}$. The last cross-relation becomes $c^{3}x{c}^{-3}=x$, already implied by $c^3=1$. ThusBoth factors are nontrivial finitely presented groups, as required.

A soluble group, also called a solvable group, has a terminating derived series:For a subgroup $S\le G$, induction gives $S^{(j)}\le G^{(j)}$, so subgroups of soluble groups are soluble. For a surjective group homomorphism $q:G\to Q$, $q(G^{(j)})=Q^{(j)}$, so quotients of soluble groups are soluble. Finally, in a group extension $1\to K\to E\to Q\to 1$, suppose $K^{(r)}=1$ and $Q^{(s)}=1$. Then $E^{(s)}\le K$ and $E^{(s+r)}=1$. Soluble groups are closed under subgroups, quotients and group extensions.

A virtually soluble group contains a soluble group as a finite-index subgroup. The finite-index facts proved in parts (i)–(iii) imply closure under subgroups and quotients: intersect a finite-index soluble subgroup with the chosen subgroup, or take its image under the quotient map.

For group extensions, no finite-generation hypothesis may be inserted. We first establish the finite-index characteristic soluble subgroup lemma. If $K$ is a virtually soluble group, the kernel of its action on the cosets of a finite-index soluble subgroup is a soluble normal subgroup $K_0$ of finite index. Among soluble normal subgroups containing $K_0$, choose $R$ with maximal $|R/K_0|$, possible because $K/K_0$ is finite. If $S$ is any soluble normal subgroup of $K$, then $RS$ is soluble: it is an extension of $R$ by $S/(R\cap S)$. Maximality forces $S\le R$. Thus $R$ is the unique largest soluble normal subgroup of $K$, making it a characteristic subgroup, and it has finite index.

Now suppose $1\to K\to G\stackrel{\pi }{\to }Q\to 1$ has both $K$ and $Q$ virtually soluble. Replace $G$ by the preimage $G_0$ of a finite-index soluble subgroup of $Q$. The subgroup $R$ just constructed is characteristic in $K$ and therefore normal in $G_0$. In $E=G_0/R$, the subgroup $A=K/R$ is a finite normal subgroup, and $E/A$ is a soluble group. The centralizer $C_E(A)$ has finite index in $E$, since conjugation gives a map $E\to \mathrm{Aut}(A)$ with finite image. Its intersection with $A$ is the center of a group $Z(A)$, an abelian group, while its quotient by $Z(A)$ embeds in the soluble group $E/A$. Thus $C_E(A)$ is a soluble group. Its preimage in $G_0$ is an extension by $R$, so it too is soluble and has finite index in $G$. Virtually soluble groups are closed under group extensions.

For the final example take the restricted direct sum of groupswhere $A_5$ is the nonabelian simple group of even permutations on five letters. Every finite collection of elements lies in a product of finitely many finite factors, so $D$ is a locally finite group and hence a torsion group. It cannot contain a nonabelian free group, which is a torsion-free group.

To show that $D$ is not a virtually soluble group, let $H$ be any finite-index subgroup and let $N\le H$ be the kernel of the finite coset action. Each coordinate $A_5$ maps either injectively or trivially into the finite quotient $D/N$, by Simplicity of the alternating group A5. The nontrivial images of distinct factors commute, and each has trivial centre, so any $k$ of them generate a direct product of groups of order $60^k$. Only finitely many such images can occur in a finite quotient. Therefore $N$, and hence $H$, contains a whole coordinate copy of $A_5$, which is not soluble: its nontrivial commutator subgroup is normal and therefore equals $A_5$. No finite-index subgroup of $D$ is soluble.

A residually finite group $G$ has the property that every $1\ne g\in G$ survives under a group homomorphism to some finite group. Equivalently, the intersection of its finite-index normal subgroups is trivial. A Hopfian group is a group for which every surjective endomorphism is an automorphism.

Suppose $G$ is generated by $d$ elements. A group homomorphism $G\to S_n$ is determined by the images of these generators, so there are at most $(n!{)}^d$ such maps. Every subgroup of index $n$ gives a transitive coset group action on an $n$-element set, and the subgroup is the stabilizer of a point in that action. There are at most $n$ point stabilizers per action. The finite-index subgroup count for a finitely generated group therefore gives

Now let $\psi :G\to G$ be a surjective endomorphism. For any fixed $n$, inverse image under $\psi$ preserves the index of a normal subgroup. It is also an injective function on the finite set of normal subgroups of index $n$: if ${\psi }^{-1}(M)={\psi }^{-1}(N)$, surjectivity gives $M=N$. It is therefore a permutation of that finite set. Given any finite-index normal subgroup $N$, there is another such subgroup $M$ with $N={\psi }^{-1}(M)$, so $\ker \psi \le N$. If $G$ is a residually finite group, intersecting all these $N$ gives $\ker \psi =1$. Hence $\psi$ is an automorphism. Every finitely generated group that is a residually finite group is a Hopfian group.

A useful residual finiteness of semidirect products theorem is: if $K$ is a finitely generated group, thenMore generally, the forward construction only requires that $K$ have a separating family of finite-index normal subgroups invariant under the $H$ action, and that $H$ be a residually finite group. Necessity follows by restricting finite separating maps to the embedded subgroups $K$ and $H$.

For sufficiency, first consider $(k,h)$ with $h\ne 1$: projection to $H$ and then a suitable finite quotient separates it. If $h=1$ and $k\ne 1$, choose a finite-index normal subgroup $U⊴K$ with $k\notin U$. Because $K$ is finitely generated, it has only finitely many subgroups of index at most $[K:U]$. Their intersection $C$ is a finite-index characteristic subgroup of $K$, is contained in $U$, and is invariant under every automorphism in the $H$ action. The quotient $K/C$ is finite. Let $\rho :H\to \mathrm{Aut}(K/C)$ be the induced action. The mapis a group homomorphism to a finite group and separates $(k,1)$. This proves the theorem and the more general invariant-subgroup criterion. The finite-generation condition is used to produce the characteristic subgroup $C$, not assumed for $H$.

Every finite cyclic subgroup of order $m$ has exactly $\phi (m)$ generators, where $\phi$ is the Euler totient function: if $g$ has order $m$, then $g^k$ generates precisely when $\gcd (k,m)=1$. Each element generates exactly one cyclic subgroup. Thus counting cyclic subgroups by their generators amounts to counting elements by order of a group element and dividing by $\phi (m)$.

Use the cycle decomposition of a permutation in $S_5$. A permutation with $a_j$ cycles of length $j$ occurs $5!/{\prod }_j(j^{a_j}{a}_j!)$ times: order the five letters, then divide by the cyclic choices of starting point and the orderings of equal-length cycles. Its order is the least common multiple of the cycle lengths. The resulting count isThe two subgroups $⟨(12)⟩$ and $⟨(12)(34)⟩$ are both isomorphic to $C_2$ by a group isomorphism. They are not conjugate subgroups: conjugating their unique nonidentity elements would have to preserve cycle type, whereas a transposition and a product of two disjoint transpositions have different cycle types.

A left group action is a map $(g,x)\mapsto g\cdot x$ with $e\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. The group orbit and stabiliser subgroup areThe identity belongs to $G_x$. If $g,h\in G_x$, then $(gh)\cdot x=g\cdot x=x$; also $g^{-1}\cdot x=g^{-1}\cdot (g\cdot x)=x$. Thus $G_x$ is a subgroup.

For the orbit-stabilizer theorem, the map $g{G}_x\mapsto g\cdot x$ from left cosets to the group orbit is a bijection: $g\cdot x=h\cdot x$ if and only if $h^{-1}g\in G_x$, which is precisely $g{G}_x=h{G}_x$. Each left coset has $|G_x|$ elements, since multiplication by $g$ is a bijection from $G_x$ to $g{G}_x$, and the left cosets partition $G$. For finite $G$ this proves

False. Any supposed kernel of a group homomorphism of order two would have a nonidentity central element by the introductory proof. Every element of order two in $S_3$ is a transposition, and none is central: conjugating $(ij)$ by a suitable three-cycle gives a different transposition. Thus the center of a group for $S_3$ is trivial, and such a kernel is impossible. This argument does not assume that an arbitrary order-two subgroup is normal.

A group homomorphism satisfies $\phi (ab)=\phi (a)\phi (b)$ for every $a,b\in G$. Its kernel of a group homomorphism is $K=\{g:\phi (g)=e_H\}$. The homomorphism equation gives $\phi (e_G)=e_H$ by cancellation, and $\phi (g^{-1})=\phi (g{)}^{-1}$. Therefore $K$ contains the identity and is closed under multiplication and inverses, so it is a subgroup.

For $k\in K$ and $x\in G$,Thus conjugation carries $K$ into itself, and applying the inverse conjugation gives equality: $K$ is a normal subgroup. If $K=\{e,\xi \}$, conjugation cannot send $\xi$ to the identity because it is injective. ConsequentlyThis proves directly the normal subgroup of order two is central principle for kernels, including the kernel facts needed below.

Lagrange's theorem says that, for a finite group $G$ and subgroup $H$,Here $[G:H]$ is the number of left cosets. Every $g\in G$ lies in $gH$. If $gH$ and $g^{\prime }H$ meet, write $gh=g^{\prime }{h}^{\prime }$; then $g^{\prime }=gh(h^{\prime }{)}^{-1}$, so $g^{\prime }H=gH$. Hence the left cosets partition $G$. The bijection $h\mapsto gh$ makes each coset have $|H|$ elements. Adding their cardinalities proves Lagrange's theorem.

False. Use the same subgroups of $S_3$. The double coset $HeH=H$ has two elements, while $H(13)H$ has four. Thus even a fixed pair of subgroups can give different double-coset sizes. Their disjoint union is all six elements of $S_3$.

False. Let $G=S_3$ and $H=K=⟨(12)⟩$. For $g=(13)$, direct multiplication gives the double cosetIt has four elements, and $4\nmid 6=|S_3|$. This is a concrete example of double-coset sizes not dividing group order; Lagrange's theorem applies to subgroups, not to arbitrary double cosets.

Every $g\in G$ lies in $HgK$, since both subgroups contain the identity. If two double cosets intersect, chooseThen $g^{\prime }=h_2^{-1}{h}_{1}g{k}_1{k}_2^{-1}\in HgK$, so every element of $H{g}^{\prime }K$ lies in $HgK$. Reversing the argument gives equality. Thus the double cosets partition the group.

Unlike single cosets, double cosets can have different cardinalities. For finite $G$, decompose $HgK$ into left cosets of $K$, or apply the orbit-stabilizer theorem to the action of $H$ on $gK$. Its stabiliser subgroup is $H\cap gK{g}^{-1}$, givingThis formula explains the counterexamples in the following parts.

The identity belongs to the center of a group. If $z,w$ commute with every $g$, then $zw$ does too; rearranging $gz=zg$ also shows that $z^{-1}$ commutes with $g$. Hence $Z(G)$ is a subgroup. Moreover $gz{g}^{-1}=z$ for $z\in Z(G)$, so it is a normal subgroup.

Suppose $|G|=p^n>1$. The conjugation action partitions $G$ into conjugacy classes. The orbit-stabilizer theorem gives class size $[G:C_G(x)]$, where $C_G(x)$ is the centralizer of $x$. By Lagrange's theorem, this size is a power of $p$; it is one exactly when $x$ is central. Thus the class equation has the formIt follows that $p\mid |Z(G)|$. Since the identity is central, $|Z(G)|\ge 1$, and thereforeThis is the nontrivial-centre property of a finite p-group.

Write the Heisenberg group element as $M(x,y,z)$. Matrix multiplication and inversion giveAll matrices have determinant one. They contain $M(0,0,0)=I$, are closed under multiplication and inverses, and inherit associativity from matrix multiplication; hence they form a subgroup of $G{L}_3(\mathbb{R})$.

Two such matrices commute precisely when $x{z}^{\prime }=x^{\prime }z$. Requiring this for every $x^{\prime },z^{\prime }$ forces $x=z=0$. ConsequentlyThe map $\pi (M(x,y,z))=(x,z)$ is a surjective group homomorphism to the additive group ${\mathbb{R}}^2$, with kernel of a group homomorphism $Z(H)$. Equivalently, it gives the explicit quotient group isomorphismIt is well-defined because multiplying by a central element changes only $y$. It is injective because equal $(x,z)$ imply that the quotient of the two matrices is central, and surjectivity follows by taking $y=0$. This also proves the needed instance of the first isomorphism theorem directly.

Write $K=\mathbb{Q}({\mu }_m)$, $L=\mathbb{Q}({\mu }_n)$. For coprime $m,n$, Bézout's identity and powers of a primitive $mn$-th root give $KL=\mathbb{Q}({\mu }_{mn})$. The degrees of these cyclotomic fields are $\phi (m)$, $\phi (n)$ and $\phi (mn)=\phi (m)\phi (n)$. The natural restriction mapis injective and, by these orders, bijective. Under this product description the subgroups fixing $K$ and $L$ are the two complementary factors. Together they generate the whole group, so their common fixed field is $\mathbb{Q}$. By the Fundamental theorem of Galois theory, $K\cap L=\mathbb{Q}$. This also includes $m=1$ or $n=1$.

The Fundamental theorem of Galois theory gives an inclusion-reversing bijection between subgroups $H\le \mathrm{Gal}(L/K)$ and intermediate fields $M$, by $H\mapsto L^H$ and $M\mapsto \mathrm{Gal}(L/M)$, for a finite Galois extension. One has $[L:L^H]=|H|$ and $[L^H:K]=[\mathrm{Gal}(L/K):H]$. The field $L^H$ is Galois over $K$ exactly when $H$ is normal; then its Galois group is the quotient by $H$.

Let $\alpha =\sqrt[3]{2}$ and $\zeta =e^{2\pi i/3}$. The splitting field is $L=\mathbb{Q}(\alpha ,\zeta )$. A splitting field in characteristic zero is normal and separable, hence Galois. Eisenstein criterion at $2$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=3$. This field is real, whereas $\zeta$ is not, and $\zeta$ satisfies a quadratic polynomial. Thus $[L:\mathbb{Q}]=6$. The faithful permutation action on the three roots identifies the Galois group with $S_3$. The subgroup fixing $\alpha$ is generated by the transposition interchanging the other two roots. Its conjugates are different transposition subgroups, so it is not normal.