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For a number field $K$ and a modulus of a number field $\mathfrak{m}$, let $I_K(\mathfrak{m})$ be the group of fractional ideals prime to its finite part. Let $P_K(\mathfrak{m})$ consist of principal ideals generated by elements congruent to one modulo its finite part and positive at each real place in its infinite part. The ray class group is ${\mathrm{Cl}}_{\mathfrak{m}}(K)=I_K(\mathfrak{m})/P_K(\mathfrak{m})$.

The classification theorem of class field theory asserts the existence of a ray class field $K_{\mathfrak{m}}$ and an Artin map isomorphismIts intermediate fields correspond contravariantly to subgroups of this ray class group. Equivalently, for each congruence subgroup of fractional ideals $P_K(\mathfrak{m})\subseteq H\subseteq I_K(\mathfrak{m})$ there is a unique finite abelian extension with Artin group kernel $H$ and group $I_K(\mathfrak{m})/H$. Every finite abelian extension occurs for some modulus. For a fixed modulus, these are exactly the extensions whose conductor divides that modulus; the same field may be presented using larger moduli, with the appropriate pulled-back group kernel.

We quote the conductor facts needed to specialize this classification. The finite conductor exponent is zero exactly at the unramified primes. A real place occurs in the conductor exactly when it becomes complex. Thus conductor one means unramified at every finite prime and split at every real place. With $\mathfrak{m}=1$ the ray class group is the ordinary ideal class group, so classification gives an extension $H_K/K$ withIt is unramified at all finite primes, and its real places split. Conversely, every finite abelian extension with those properties has conductor one and lies in $K_1=H_K$. This proves the existence and maximality of the ordinary Hilbert class field. Its degree is finite by finiteness of the ideal class group. The real-place convention matters: allowing real places to become complex gives the narrow class field instead; for the imaginary quadratic field in part (ii) this distinction disappears.

A smooth fiber bundle $\pi :E\to M$ with typical fiber $F$ is locally a product: every base point has a neighborhood $U$ and a local trivialization ${\pi }^{-1}(U)\cong U\times F$ commuting with projection to $U$. Its overlap maps act smoothly on $F$. A principal bundle with structure Lie group $G$ has a free right $G$-action, each fiber is one orbit, and the local trivializations are equivariant for right multiplication on $G$. A smooth section of a fiber bundle is a map $s:M\to E$ with $\pi \circ s={\mathrm{id}}_M$.

For a principal bundle, a global section of a fiber bundle gives the mapThe free transitive action on each fiber makes $\Phi$ bijective. In a local trivialization, write $s(x)=(x,a(x))$; then $\Phi (x,g)=(x,a(x)g)$ and its inverse is $(x,k)\mapsto (x,a(x{)}^{-1}k)$. Both are smooth and $\Phi$ is equivariant. Conversely, a product principal bundle has the section $x\mapsto (x,e)$. This proves a principal bundle is trivial exactly when it has a global section. The argument depends on the free transitive action: an arbitrary bundle section need not trivialize its bundle.

For an $n$-dimensional smooth manifold, the frame bundle has fiber over $x$ consisting of ordered bases of $T_{x}M$, equivalently linear isomorphisms $u:{\mathbb{R}}^n\to T_{x}M$. Right composition by $GL(n,\mathbb{R})$ makes it a principal bundle. If a metric has signature $(p,q)$, its pseudo-orthonormal frame bundle consists of the isometries $u:{\mathbb{R}}^{p,q}\to T_{x}M$, with structure group $O(p,q)$. In the positive-definite case this is the usual orthonormal frame bundle. Choosing space and time orientations reduces the Lorentzian structure group to its corresponding connected subgroup.

For Minkowski spacetime, use a fixed origin and a standard Lorentzian frame. Every pair consisting of a point and a pseudo-orthonormal frame is the image of this reference pair under a unique affine isometry. Thus, for the full groups,Here the first expression denotes the total space of the orthonormal frame bundle, not merely the Lorentz group. Projection sends an affine isometry to the image of the reference point, and the right action of the stabilizer changes its frame.

Realize de Sitter spacetime of radius $\ell$ as ${\eta }_{AB}{X}^A{X}^B={\ell }^2$ in ${\mathbb{R}}^{1,4}$. The ambient orthogonal group is transitive on this hyperboloid. Its point stabilizer acts on the Lorentzian tangent space as $O(1,3)$; choosing a tangent orthonormal frame removes that stabilizer completely. ConsequentlyFor oriented, time-oriented frames, replace these groups by $S{O}_0(1,4)$ and $S{O}_0(1,3)$, and similarly use the proper orthochronous Poincaré group for Minkowski spacetime. The isometries are uniquely determined by their action on a point and its frame, which explains these bundle identifications.

Finally, choose a basis $v_1,\dots ,v_n$ of the Lie algebra $T_{e}G$. The left-invariant vector fields $X_i(g)=(d{L}_g{)}_e{v}_i$ form a basis at every $g$, because $L_g$ is a diffeomorphism. They give a global section of a fiber bundle of the frame bundle, so that bundle is trivial. More explicitly,Thus every Lie group is a parallelizable manifold; connectedness is not needed for this construction.

Use relative homology. Every simplex of $X=B\cup C$ lies in $B$ or $C$, and those lying in both are exactly the simplices of $A$. Consequently inclusion induces a chain-complex isomorphismIt is bijective on the respective bases of simplices and commutes with boundary because it comes from inclusion. Thus

For any pair $(Y,Z)$, the long exact sequence in homology shows that inclusion induces isomorphisms in every degree exactly when $H_j(Y,Z)=0$ for every $j$. To verify both directions, if all relative groups vanish, exactness makes each $H_j(Z)\to H_j(Y)$ injective and surjective. Conversely, if the inclusion maps are isomorphisms, the map from $H_j(Y)$ into the relative group is zero, while the relative connecting map is injective and has zero image because the next inclusion is injective; hence the relative group itself is zero. This includes degree zero, with the negative-degree ordinary group zero.

Apply the criterion first to $(B,A)$ and then to $(X,C)$ using the chain isomorphism. It proves

For a linear map $\theta :U\to V$, its rank of a linear map is $r(\theta )=\dim \mathrm{im}\theta$, and its nullity is $n(\theta )=\dim \ker \theta$. Choose a basis $e_1,\dots ,e_s$ of its kernel and extend it to a basis $e_1,\dots ,e_d$ of $U$. Then $\theta e_{s+1},\dots ,\theta e_d$ span its image. They are linearly independent: a linear combination mapped to zero would belong to the kernel, contradicting independence of the extended basis unless all its coefficients vanish. Hence $r(\theta )=d-s$, provingFor endomorphisms of $U$, define $(\theta +\varphi )(u)=\theta (u)+\varphi (u)$ and $(\theta \varphi )(u)=\theta (\varphi (u))$. The image of the sum lies in the sum of the two images. ThusTo prove the product inequality, restrict $\varphi$ to $K=\ker (\theta \varphi )$. This map has kernel $\ker \varphi$ and image $\mathrm{im}\varphi \cap \ker \theta$: every vector in the intersection has a preimage in $K$. Applying the rank-nullity theorem to the restricted map givesNow put $d=\dim U$ and $s=r(\theta )+r(\varphi )$. If both equalities hold, $r(\theta +\varphi )=s\le d$, while $n(\theta \varphi )=2d-s\le d$ forces $s\ge d$. Therefore $s=d$, the sum has full rank, and the product has nullity $d$. The sum is an isomorphism and the product is zero.

Conversely, if $\theta \varphi =0$, then $\mathrm{im}\varphi \subseteq \ker \theta$, giving $s\le d$. If the sum is also an isomorphism, its rank is $d\le s$. Thus $s=d$, and the two inequalities become equalities. This proves the equality in the rank-sum and nullity-product inequalities in both directions.

There are two necessary qualifications to the literal formulation. The whole group $G$ is itself a closed normal subgroup, so the claim about containment in the centre must concern proper closed normal subgroups. Also, the homomorphism conclusion needs a nontrivial target $H$: for example, $SU(2)\to \{1\}$ has surjective differential onto the zero Lie algebra, but its group kernel is all of $SU(2)$ and no quotient by a finite subgroup is a one-point group. We prove the precise assertions, including these exceptional cases.

Write $\mathfrak{g}=\mathrm{Lie}(G)$. The centre $\mathfrak{z}(\mathfrak{g})$ is invariant under the Adjoint representation, so irreducibility makes it either zero or all of $\mathfrak{g}$. In the latter case $\mathfrak{g}$ is Abelian. For connected $G$, the adjoint action is then trivial: the group is generated by exponentials, and ${\mathrm{Ad}}_{\exp X}=e^{\mathrm{ad}X}=I$. A nonzero trivial representation is irreducible only in dimension one. A connected compact one-dimensional Lie group is a circle, so $G\cong U(1)$.

For every other $G$, the Lie algebra of its centre is zero. The centre is a closed compact Lie subgroup; a zero-dimensional compact Lie group is finite. Let $N$ be a closed normal subgroup. The closed-subgroup theorem identifies $\mathfrak{n}=\mathrm{Lie}(N)$ as an adjoint-invariant subspace of $\mathfrak{g}$, hence $\mathfrak{n}=0$ or $\mathfrak{n}=\mathfrak{g}$. If the latter holds, $N$ contains an identity neighbourhood and is open, so connectedness forces $N=G$. If $N$ is proper, it therefore has zero Lie algebra and is finite. For any $a\in N$, the continuous map $g\mapsto ga{g}^{-1}$ takes connected $G$ into the finite set $N$, so it is constant and equal to $a$. ThusProper closed subgroups of the circle are finite as well, and of course central.

Now let $d\Phi$ be surjective. The submersion theorem makes $\Phi (G)$ contain an identity neighbourhood in $H$, so it is an open subgroup. Since $H$ is connected, $\Phi (G)=H$. Its group kernel $F$ is closed and normal. If $H$ is nontrivial, $F$ is proper and the preceding argument gives a finite central group kernel; this also holds for $G\cong U(1)$, whose proper closed subgroups are finite. The Lie-group first isomorphism theorem givesIf $H$ is trivial, the correct statement is instead $H=G/G$.

For the Spin groups, use the real Clifford algebra with $v^2=-\Vert v{\Vert }^2$. The group $\mathrm{Spin}(m)$ consists of products of an even number of unit vectors inside its invertible even part. Conjugation on the embedded ${\mathbb{R}}^m$ defines the standard double covering $\mathrm{Spin}(m)\to SO(m)$ with group kernel $\{\pm 1\}$. The Clifford relations show that a unit-vector reflection, with the usual twisted conjugation, is a hyperplane reflection; the even products give orientation-preserving transformations. The Cartan–Dieudonné theorem states that every orthogonal transformation is a product of reflections, proving surjectivity. For $m\ge 3$ this connected covering group is simply connected; $\mathrm{Spin}(2)$ is a circle.

Here is an explicit exterior-square construction of the requested six-dimensional map. Put $E={\mathbb{C}}^4$ with its standard Hermitian form and volume $e_1\wedge e_2\wedge e_3\wedge e_4$. On ${\Lambda }^{2}E$, write $e_{ij}=e_i\wedge e_j$ and define an antilinear map $J$ byIt satisfies $J^2=1$. Intrinsically, if $u\wedge v=B(u,v)\mathrm{vol}$ and the Hermitian inner product $h$ is linear in its first variable, then $B(u,Jv)=h(u,v)$. Because $SU(4)$ preserves both $B$ and $h$, it commutes with $J$. This is the real structure of the SU4 exterior square.

The fixed space $W$ has the following real orthonormal basis, with all vectors divided by $\sqrt{2}$:Thus $W$ is six-dimensional over $\mathbb{R}$, its complexification is ${\Lambda }^{2}E$, and the exterior-square action restricts to an orthogonal representation on $W$. Its determinant is one because $SU(4)$ is connected. We obtain $R:SU(4)\to SO(W)\cong SO(6)$.

To compute the group kernel, diagonalize $g\in SU(4)$ with eigenvalues $z_1,\dots ,z_4$. If $R(g)=I$, its complexified exterior-square action is the identity, hence $z_i{z}_j=1$ for every $i<j$. Comparing pairs shows that all $z_i$ are equal, and their common value has square one. A unitary matrix with that single eigenvalue is scalar, so $\ker R=\{\pm I_4\}$. Conversely, those two scalar matrices plainly act trivially on the exterior square.

The differential of $R$ is injective because its group kernel is finite. Both Lie algebras have dimension $15$, so the differential is an isomorphism. Its image is consequently open in connected $SO(6)$, hence all of $SO(6)$. ThereforeThis also realizes $SU(4)$ as $\mathrm{Spin}(6)$, since $SU(4)$ is simply connected, by uniqueness of the connected simply connected covering group.

Put $V={\mathbb{C}}^n$. We use complete reducibility of compact-group representations, and the highest-weight classification: the multiplicity of $V_{\lambda }$ in a completely reducible representation is the dimension of its weight-$\lambda$ subspace killed by all positive root vectors. Each irreducible summand contributes just its one-dimensional highest-weight line to that subspace.

Symmetrizing and antisymmetrizing the three tensor factors give invariant subspaces ${\mathrm{Sym}}^{3}V$ and ${\Lambda }^{3}V$ of $V^{\otimes 3}$. The vectors $e_1^{\otimes 3}$ and $e_1\wedge e_2\wedge e_3$ are highest-weight vectors of weights $(3,0,\dots ,0)$ and $(1,1,1,0,\dots ,0)$ respectively. The unitary version of the Weyl dimension formula isIt givesThese are exactly the dimensions of the third symmetric power and exterior power, so each is irreducible and occurs once in those subspaces.

The weight $(2,1,0,\dots ,0)$ in $V^{\otimes 3}$ has basisThe only positive root vector that acts nontrivially on this weight space is $E_{12}$. Acting on the tensor product as the sum of its actions on the three factors, it sends each of $a,b,c$ to $e_1^{\otimes 3}$. Consequently the highest-weight vectors in this space are preciselyThis is a two-dimensional space, so $V_{(2,1)}$ occurs exactly twice. Its dimension isThe dimensions of the summands already found add toThus no other irreducible summand can remain. We have proved the full decompositionThere are four summands counted with multiplicity and three distinct isomorphism types.

For $n=2$, the exterior cube vanishes. The third symmetric power still has highest weight $(3,0)$ and dimension $4$, while the same two-dimensional highest-vector calculation gives two copies of the representation of weight $(2,1)$ and dimension $2$. It is $\det \otimes V$, since a determinant twist shifts the standard highest weight $(1,0)$ to $(2,1)$. Hence

Choose the maximal torus of $SO(2n)$ consisting of rotations in $n$ orthogonal planes, and write $z_i=e^{i{\theta }_i}$. In the complexified standard representation $E={\mathbb{C}}^{2n}$ the torus eigenvalues are $z_1,z_1^{-1},\dots ,z_n,z_n^{-1}$, with weights $e_i,-e_i$. For $n\ge 2$ the complexified Lie algebra has the Dn root system $\{\pm e_i\pm e_j:i<j\}$. The root-space decomposition theorem splits this Lie algebra into its Cartan subalgebra and one-dimensional root spaces. The Cartan subalgebra contributes the zero weight with multiplicity $n$. Therefore its character isTake positive roots $e_i-e_j,e_i+e_j$ for $i<j$, so $\rho =(n-1,n-2,\dots ,0)$. For $n>2$ the root $\theta =e_1+e_2$ is a highest weight: its root vector is killed by every positive root vector, since $\theta +\alpha$ is never a root for positive $\alpha$. To verify irreducibility without relying on an unstated simplicity theorem, apply complete reducibility and compare dimensions. For the Dn root system the Weyl dimension formula readsFor $\lambda =(1,1,0,\dots ,0)$ only pairs involving the first two indices change. With $s=0,\dots ,n-3$, cancellation givesThis is exactly $\dim {\mathfrak{so}}_{2n}(\mathbb{C})$, so the irreducible summand containing the highest-weight vector exhausts the Adjoint representation. Thus for $n>2$ the complexified Adjoint representation is irreducible, with highest weight $e_1+e_2$.

For $n=2$, the roots split into the two independent systems $\{\pm (e_1+e_2)\}$ and $\{\pm (e_1-e_2)\}$. The Adjoint representation splits into two irreducibles of dimension three, with respective highest weights $(1,1)$ and $(1,-1)$. Equivalently these are the self-dual and anti-self-dual exterior two-forms in four dimensions. Their characters areThey add to the displayed adjoint character. This agrees with the Chiral decomposition of the complexified so4 Lie algebra.

For a direct character comparison, put $s(z)={\sum }_i(z_i+z_i^{-1})$. The eigenvalue formula for an exterior square givesThis is precisely ${\chi }_{\mathrm{ad}}$. More intrinsically, let $B$ be the nondegenerate symmetric bilinear form preserved by $SO(2n)$ and defineThis endomorphism is skew with respect to $B$, so $\Psi$ maps ${\Lambda }^{2}E$ to $\mathfrak{so}(E,B)$. In an orthonormal basis the bivectors $e_i\wedge e_j$ map to a basis of skew skew-symmetric matrices, proving that it is an isomorphism. Since $B$ is invariant,Thus the exterior square and the complexified Adjoint representation are naturally isomorphic, explaining the character equality without a weight calculation.

The symmetric square is not irreducible. The inverse metric tensor $\Omega$ is a nonzero invariant vector, and contraction with $B$ splits off its trivial line:The contraction sends $\Omega$ to $2n$, so this is indeed a direct sum. A highest-weight vector $f_1\otimes f_1$, where $f_1$ has torus weight $e_1$, lies in the traceless part because $B(f_1,f_1)=0$, and has highest weight $2e_1$. The same dimension formula gives, for $n\ge 2$,This equals $\dim {\mathrm{Sym}}^{2}E-1=n(2n+1)-1$. Complete reducibility therefore proves that the traceless part is irreducible, including the dimension-nine case for $SO(4)$. In particular,If the circle case $n=1$ is included, the adjoint character is $1$ and the symmetric square instead has three one-dimensional weights $2,0,-2$.

We use the general adjoint functor theorem in its solution-set form. Let $G:\mathcal{D}\to \mathcal{C}$ be a functor between locally small categories, with $\mathcal{D}$ a complete category. Then $G$ has a left adjoint exactly when it preserves all small categorical limits and, for each $C\in \mathcal{C}$, there is a set-indexed familysuch that every $u:C\to GD$ factors as $Gf{u}_i$ for some $i$ and some $f:D_i\to D$. This is the solution-set condition. The indexing family is a set, whereas the entire comma category $(C\downarrow G)$ can be large.

We first prove the initial-object lemma for complete categories with a weakly initial set. Let $\mathcal{E}$ be locally small and complete, with a weakly initial set $(E_i{)}_{i\in I}$. Its product in a category $W={\prod }_i{E}_i$ is weakly initial: follow a projection by a map from the corresponding $E_i$ to any desired object. The hom-set $\mathcal{E}(W,W)$ is a set, so we can form the simultaneous equalizerThis is a small limit; equivalently, equalize the family of endomorphisms with the family of identity maps into a product of copies of $W$. Every object receives a map from $E$, since it receives one from $W$.

To prove uniqueness, let $a,b:E\rightrightarrows B$, and take their equalizer $j:V\to E$. Weak initiality of $W$ gives $t:W\to V$. Since $ejt$ is an endomorphism of $W$, the defining property of $e$ givesThe monomorphism $e$ cancels, yielding $jte=1_E$. Thus $j$ is both a monomorphism and a split epimorphism, hence an isomorphism. Its equalizer property now implies $a=b$. Therefore $E$ is initial.

Fix $C$. Small categorical limits in $(C\downarrow G)$ are obtained from the limits in $\mathcal{D}$: preservation by $G$ supplies the unique map from $C$ to the image of the limiting object. The comma category is locally small, and the solution family is a weakly initial set in it. The lemma supplies an initial object $(LC,{\eta }_C:C\to GLC)$.

For $k:C\to C^{\prime }$, initiality gives a unique arrow $Lk:LC\to L{C}^{\prime }$ satisfying $G(Lk){\eta }_C={\eta }_{C^{\prime }}k$. Uniqueness proves preservation of identities and composition, so $L$ is a functor. Initiality also gives natural bijectionswhich prove $L⊣G$.

Conversely, if $L⊣G$, the singleton ${\eta }_C:C\to GLC$ is a solution set by the adjunction bijection. To see that $G$ preserves a small limit $D={\lim }_j{D}_j$, use the adjunction and the limit property to obtain, naturally in $C$,This is precisely the universal property of the image limit cone. It proves the necessity of both hypotheses.

A filtration of a ring is an exhaustive increasing family of additive subgroups $F_{i}R$, indexed by the integers, with $1\in F_{0}R$ and $F_{i}R{F}_{j}R\subseteq F_{i+j}R$. A compatible filtration of a module on a right module satisfies $F_{i}M{F}_{j}R\subseteq F_{i+j}M$. The associated graded module has components $F_{i}M/F_{i-1}M$, and its action by the associated graded ring isChanging either representative changes the product only by an element of $F_{i+j-1}M$, so the action is well-defined.

The subspace filtration on $N$ is $F_{i}N=N\cap F_{i}M$. The quotient filtration is $F_i(M/N)=(F_{i}M+N)/N$. Inclusion induces an injective map $\mathrm{gr}N\to \mathrm{gr}M$, and the quotient induces a surjective map to $\mathrm{gr}(M/N)$. If $m\in F_{i}M$ maps to zero in its degree-$i$ quotient, write $m=n+m^{\prime }$ with $n\in N$ and $m^{\prime }\in F_{i-1}M$. Then $n\in N\cap F_{i}M$, so the kernel is exactly the image of $\mathrm{gr}N$. We have the short exact sequencewhich is exactness of associated graded modules for induced filtrations.

For the negative case, $F_{i}R=R$ for $i\ge 0$. Completeness means that $R\to {\underleftarrow{\lim }}_{q}R/F_{-q}R$ is an isomorphism; in particular the filtration of a ring is separated. If $x\in F_{-1}R$ and $r\in R$, then $(xr{)}^q\in F_{-q}R$. The geometric seriesconverges in the complete negative filtration and is a two-sided inverse of $1-xr$, by the finite geometric identity and continuity of multiplication. The unit criterion for the Jacobson radical now gives

Let $I$ be any right ideal. The associated graded module $\mathrm{gr}I$ for its subspace filtration is a homogeneous ideal on the right in $\mathrm{gr}R$. If $\mathrm{gr}R$ is a right Noetherian ring, choose homogeneous generators that lift to $x_1,\dots ,x_s\in I$, with degrees $d_1,\dots ,d_s\le 0$.

For $x\in I\cap F_{d}R$, cancel its degree-$d$ symbol by a sum ${\sum }_i{x}_i{a}_{i,0}$ with $a_{i,0}\in F_{d-d_i}R$. The remainder belongs to $I\cap F_{d-1}R$. Repeat the cancellation in each succeeding degree. After $q$ steps,For every $i$, the coefficient series converges to some $a_i\in R$ by completeness. Passing to the limit gives $x={\sum }_i{x}_i{a}_i$. Thus these finitely many $x_i$ generate the actual right ideal $I$; no assumption that $I$ is closed is needed. Since every right ideal is finitely generated, $R$ is right Noetherian, by complete negative filtered-graded transfer of Noetherianity.

A group $G$ is a finitely presented group if there is an epimorphism $F(X)\to G$ from a finite-rank free group whose kernel is the normal closure of finitely many words $R$. Equivalently, $G=⟨X\mid R⟩$ with both $X$ and $R$ finite. Normal closure means that every kernel element is a finite product of conjugates of the chosen relators or their inverses.

For independence of finite presentation from finite generators, suppose $G=⟨y_1,\dots ,y_m\mid r_1,\dots ,r_s⟩$, and let $x_1,\dots ,x_k$ be any other finite generating set. Choose words $u_j(X)$ representing $y_j$ and words $v_i(Y)$ representing $x_i$. Consider the finite presentationSending each $x_i$ to its actual element gives a surjection $P\to G$. Conversely, send $y_j$ to $u_j(X)$; the substituted relators make this a homomorphism $G\to P$. Their composition on $G$ fixes every $y_j$, and their composition on $P$ fixes every $x_i$ by the extra relations. They are inverse isomorphisms. Hence finite presentability is independent of the chosen finite generating set.

For subgroups $A,B\le G$ and an isomorphism $\varphi :A\to B$, adopt the conventiona quotient of the free product $G\ast ⟨t⟩$. The generator $t$ is the stable letter of this HNN extension. Fix right coset transversals $R_{+}$ for $B\backslash G$ and $R_{-}$ for $A\backslash G$, each with $1$ representing the subgroup itself. Thus each $g$ has a unique decomposition $g=br$ with $b\in B,r\in R_{+}$, and a unique decomposition $g=ar$ with $a\in A,r\in R_{-}$.

A normal form is a stringwhere $r_i\ne 1$ whenever ${ϵ}_{i+1}=-{ϵ}_i$. The case $n=0$ is just $g_0$. The prohibition removes adjacent inverse stable letters that would cancel. The following permutation proof of HNN normal form uniqueness proves both existence and uniqueness.

Let $\mathcal{N}$ be the set of these formal strings. A base element $g$ acts by left multiplication on the initial coefficient $g_0$; call this permutation $L_g$. To define $L_t$, decompose the initial coefficient as $g_0=br$, with $b\in B,r\in R_{+}$. Unless $r=1$ and the old first stable letter is $t^{-1}$, replace the initial segment byIn the exceptional case, if the old string begins $b{t}^{-1}{r}_1\cdots$, cancel the new $t$ with that $t^{-1}$ and obtain the string beginning ${\varphi }^{-1}(b)r_1\cdots$. These operations use the defining identity $tb={\varphi }^{-1}(b)t$.

Define $L_{t^{-1}}$ similarly: write $g_0=ar$ with $a\in A,r\in R_{-}$, prepend $\varphi (a)t^{-1}r$, and cancel with an old first $t$ exactly when $r=1$. The resulting strings satisfy the normal-form conditions. These two maps are inverses. If the first operation inserts a letter, the reverse operation decomposes its new initial coefficient in the associated subgroup and cancels that letter. If the first operation cancels a letter, the reverse coset decomposition recovers the removed representative and reinserts it; an additional unwanted cancellation is excluded by the original normal-form condition.

The permutations satisfy $L_g{L}_h=L_{gh}$ andIndeed, in the decomposition $g_0=br$, multiplying $g_0$ on the left by $\varphi (a)$ leaves its representative $r$ unchanged and changes $b$ to $\varphi (a)b$. Its image under ${\varphi }^{-1}$ is $a{\varphi }^{-1}(b)$, exactly the coefficient produced by multiplying the normalized result on the left by $a$. The cancellation case has the same representative and the same conclusion. ThereforeAll defining relations of the HNN extension hold in this action, so it induces a homomorphism from $G{\ast }_{\varphi }$ into the permutations of $\mathcal{N}$.

Starting with the empty string, successive left multiplications normalize any word, and every normalization step uses a defining relation. This proves existence. If two normal words represented the same group element, their induced permutations would agree. Applied to the empty string, each normal word gives exactly its own formal string, since its representatives are already chosen and all forbidden cancellations have been excluded. Thus the strings are equal. This proves the normal form theorem for an HNN extension, including the injectivity of the base group $G\to G{\ast }_{\varphi }$.

A reduced sequence in an HNN extension is a wordwith no pinch $t^{-1}at$ for $a\in A$ and no pinch $tb{t}^{-1}$ for $b\in B$. Equivalently, if adjacent exponents are $-1,+1$, their intervening coefficient is outside $A$; if they are $+1,-1$, it is outside $B$.

Normalize such a word from right to left using the procedure above. Inductively its stable-letter length does not change. To see why, suppose the next letter has exponent $\delta$. Moving an associated-subgroup factor through it contributes to its left an element of ${\varphi }^{-1}(B)=A$ when $\delta =1$, or of $\varphi (A)=B$ when $\delta =-1$. If the preceding exponent is $-\delta$, this factor belongs to that preceding letter's pinch subgroup. Multiplying an intervening coefficient outside that subgroup by an element inside it cannot put the product inside the subgroup. Hence no cancellation occurs; when the two exponents agree, cancellation is impossible anyway. Thus a reduced word with $n\ge 1$ has a normal form still containing $n$ stable letters, which cannot be the identity normal form. This is Britton's lemma:

For the final example, write the centralizing extension asIt is generated by $a,b,t$. Suppose it were finitely presented. By the finite-generator independence already proved, the kernel of $F(a,b,t)\to E$ would be normally generated by finitely many words. Each of these kernel group generators is a finite product of conjugates of the displayed defining relators, so only finitely many indices occur in all their expressions. Choose $N$ at least as large as all of them. The full kernel would then equal the normal closure of the first $N$ relators: one inclusion follows from those expressions, and the reverse inclusion follows because all these relators already belong to the kernel.

But the group with only those first $N$ relators is the identity HNN extension over $H_N=⟨w_1,\dots ,w_N⟩$. Since the $w_j$ are a free basis of $H$, $w_{N+1}\notin H_N$. The word $t^{-1}{w}_{N+1}t{w}_{N+1}^{-1}$ is reduced in this partial extension, so Britton's lemma makes it nonidentity there. It is identity in $E$, contradicting equality of the kernels. ThereforeThis is the free-basis instance of the general fact that finite presentation of a centralizing HNN extension detects finite generation of its associated subgroup.

Let $K=\ker \theta$. The subgroup correspondence for a surjective group homomorphism sends a subgroup $B\le H$ to ${\theta }^{-1}(B)$ and a subgroup $A\le G$ containing $K$ to $\theta (A)$. Inverse images are subgroups containing $K$, and images are subgroups. Surjectivity gives $\theta ({\theta }^{-1}(B))=B$. Conversely, if $K\le A$ and $\theta (g)\in \theta (A)$, choose $a\in A$ with $\theta (g)=\theta (a)$. Then $g{a}^{-1}\in K\le A$, so $g\in A$. Thus ${\theta }^{-1}(\theta (A))=A$. These two assignments are inverse bijections.

For $C={\theta }^{-1}(B)$, the map of left coset setsis well defined and onto. If $\theta (g_1)B=\theta (g_2)B$, then $\theta (g_2^{-1}{g}_1)\in B$, so $g_2^{-1}{g}_1\in C$ and $g_{1}C=g_{2}C$. Hence it is injective, provingNo normality of $B$ is required; these are coset sets, not necessarily quotient groups.

The abelianization is $\overline{G}=G/[G,G]$, where the commutator subgroup is generated by the commutators $[x,y]=x^{-1}{y}^{-1}xy$. Conjugating a commutator gives another commutator, so this subgroup is normal. In the quotient all commutators are trivial, and therefore the quotient is abelian. If $N⊲G$ and $G/N$ is abelian, every commutator lies in $N$, so $[G,G]\le N$. Henceis well defined and satisfies $\pi =\tilde{\pi }\circ q$, where $q:G\to \overline{G}$ is the natural map. It is uniquely determined because $q$ is onto. This is the universal factorization property of abelianization.

Now write $S_i$ for the unique index-$i$ subgroup of $S$. Since conjugation preserves index, as shown in part (ii), $g{S}_i{g}^{-1}=S_i$ for every $g\in S$. Thus every $S_i$ is normal. For each prime $p$ the quotient $S/S_p$ has order $p$ and is cyclic, so there is a surjection $S\to C_p$. Being an abelian quotient, it factors through $\overline{S}$.

The abelianization is finitely generated because $S$ is. By the Fundamental theorem of finitely generated abelian groups,where $T$ is finite. If $r=0$, choose a prime $p>|T|$. The finite group $T$ could not surject onto the order-$p$ group $C_p$, contradicting the preceding surjections. Hence $r\ge 1$, and projection from a free coordinate supplies

Every subgroup of finite index in $\overline{S}$ has an inverse image of the same index in $S$, so there cannot be two different subgroups of any given finite index in $\overline{S}$. If $r\ge 2$, reduction modulo a prime $p$ in the first and second free coordinates gives two distinct surjective maps to $C_p$ with different kernels: the first basis vector lies in one kernel but not the other. Both kernels have index $p$, a contradiction. Therefore $r=1$.

If $T\ne 1$, choose a prime $p$ dividing its order. The decomposition of a finite abelian group into cyclic factors gives a surjection $\chi :T\to C_p$. In $\mathbb{Z}\oplus T$, the mapshave distinct index-$p$ kernels, again distinguished by $(1,0)$. This is another contradiction, so $T=1$. We have proved uniqueness of finite-index subgroups forces infinite cyclic abelianization:

A residually finite group is one in which every nonidentity element has nonidentity image under some homomorphism to a finite group. Equivalently, the intersection of all finite-index normal subgroups is trivial: a finite separating map has such a kernel, and conversely quotienting by such a normal subgroup gives a finite separating map.

Choose the surjection $\sigma :S\to \overline{S}\cong \mathbb{Z}$. For each positive integer $i$, the subgroup ${\sigma }^{-1}(i\mathbb{Z})$ has index $i$ and hence equals $S_i$. Its kernel $[S,S]$ is therefore contained in every finite-index subgroup. Indeed,If $S$ is residually finite, this intersection is trivial. Thus $[S,S]=1$ and $\sigma$ is an isomorphism. The residually finite group with one subgroup of each finite index is infinite cyclic conclusion isResidual finiteness is used exactly to eliminate the commutator subgroup that all finite quotients would otherwise fail to detect.

The Fundamental theorem of finitely generated abelian groups expresses every finite abelian group of this order as a product of cyclic groups of prime-power order. Their exponents correspond to the partitions of an integer $4$:These are pairwise nonisomorphic: the largest order of a group element distinguishes all except the two groups with largest element order $p^2$, and those have respectively $p^2$ and $p^3$ elements annihilated by multiplication by $p$. The classification exhausts the possibilities, so there are five isomorphism classes.

For an open subset $U\subseteq B$, let $E_U={\pi }^{-1}(U)$ and define the degree-preserving module mapUsing the prescribed fiber basis identifies its domain with the tensor product in the question. We prove that ${\Phi }_B$ is an isomorphism, giving the Leray-Hirsch theorem in this setting.