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By the preceding argument take a nonzero integer $m\in \mathfrak{a}$. The ideal property gives $m{\mathcal{O}}_L\subseteq \mathfrak{a}$. If $d=[L:\mathbb{Q}]$, an integral basis identifies the additive group of ${\mathcal{O}}_L$ with ${\mathbb{Z}}^d$, soThe natural projection onto ${\mathcal{O}}_L/\mathfrak{a}$ is surjective. Thus the latter is an abelian group of order at most $|m{|}^d$, provingIn particular the ideal is a full-rank integer lattice; this permits the integral-basis argument in the next part.

A cyclotomic field over $\mathbb{Q}$ is a field $\mathbb{Q}({\zeta }_m)$ generated by an $m$th root of unity of exact order $m$. It contains all roots of $X^m-1$, and is its splitting field. In characteristic zero this polynomial is separable, so the extension is Galois.

Every automorphism sends ${\zeta }_m$ to ${\zeta }_m^a$ with $a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }$, and is determined by that image. Composition multiplies the exponents modulo $m$, yielding an injective homomorphism into an abelian group. Thus the Galois group is abelian. Using irreducibility of the cyclotomic polynomial, every unit exponent occurs andIf “cyclotomic” is used more broadly for a subfield of such a field, it is still Galois and abelian, because every subgroup of an abelian Galois group is normal.

The sheaf of nonzero rational functions on an irreducible variety ${\mathcal{K}}_X^{\ast }$ is a flasque sheaf. Indeed, its sections on any nonempty open subset are $K^{\ast }$: irreducibility makes nonempty open subsets connected and makes any two of them meet, so locally constant rational-function values must agree. Restrictions between such opens are identities; restriction to the empty set is also surjective. The preliminary argument therefore givesApply the long exact sequence in sheaf cohomology to the short exact sequence of sheaves of multiplicative abelian groupsIts relevant part isPart (b) identifies the first map with the principal divisor map $K^{\ast }\to \mathrm{Div}(X)$. Exactness therefore identifies its cokernel with $H^1(X,{\mathcal{O}}_X^{\ast })$. Combining this with part (a) givesThe connecting map sends a divisor to the class of the unit ratios of its local rational representatives; these are precisely the transition data of the associated line bundle, up to the usual choice of inverse transition convention.

Let $r:B\to A$ split the reflexive pair $f,g$. Fix $C$, and take objects $u:C\to B$ and arrows $a:C\to A$ with source $fa$ and target $ga$. Since $\mathcal{C}$ is an additive category, its hom-sets are abelian groups, and composition is additive.

The identity at $u$ is $ru$. For $a:u\to v$ and $b:v\to w$, defineIts source is $fa-v+fb=u-v+v=u$ and its target is $ga-v+gb=v-v+w=w$. The two identity laws follow from $a-rv+rv=a$ and $ru-ru+a=a$. For $c:w\to x$, both parenthesizations of the triple composite are $a-rv+b-rw+c$, so composition is associative.

Define the inverse byIts source is $u+v-u=v$ and its target is $u+v-v=u$. Moreover $a^{-1}\star a=ru$ and $a\star a^{-1}=rv$. Thus these data form a groupoid, proving the represented form of reflexive pair in an additive category is an internal groupoid:

Use countably many copies of $A$, and let $c:S={\coprod }_{i\ge 1}A\to P={\prod }_{i\ge 1}A$ be the assumed isomorphism. Write $p_i={\pi }_{i}c:S\to A$, let $\delta :A\to P$ be the diagonal with every component $1_A$, and put $q=c^{-1}\delta$. Thus $p_{i}q=1_A$ for every $i$. Let $s:S\to S$ shift the injections, $s{\nu }_i={\nu }_{i+1}$, and let $\sigma :S\to A$ be the fold map, $\sigma {\nu }_i=1_A$.

The coordinate relations give $p_{1}s=0$ and $p_{i+1}s=p_i$, while $p_1{\nu }_1=1_A$ and $p_{i+1}{\nu }_1=0$. Therefore $q$ and ${\nu }_1+sq$ have identical $p_i$-coordinates. Since $c$ is invertible, these coordinates jointly distinguish arrows into $S$, soThe coproduct property gives $\sigma s=\sigma$, because both have value $1_A$ on each injection. Consequently $z=\sigma q$ satisfiesThis proves the absorbing-endomorphism assertion without assuming an infinite addition operation on hom-sets.

If the category is an additive category, each hom-set is an abelian group, so cancellation gives $1_A=0$. Any arrow $f:A\to B$ is then $f{1}_A=f0=0$. Every hom-set has exactly one arrow, and every object is isomorphic to the zero object. This is countable biproducts in an additive category force triviality:

For the First Chern class identity, use finite-dimensional classifying maps. A complex line bundle on a compact Hausdorff space embeds in a finite-dimensional trivial vector bundle, so it is a pullback vector bundle of the tautological bundle on some Complex projective space. For the two complex line bundles, the universal tensor product of vector bundles is classified by the mapIts restriction to either factor pulls the tautological bundle back to the tautological bundle on that factor. The degree-two cohomology of the product is the direct sum of the degree-two cohomology of its factors, so these two restrictions determine its First Chern class. Pulling back to $X$ provesThis proves the integral identity, including possible torsion elements in the cohomology of $X$; an argument using only curvature forms would not detect those torsion elements.

The splitting principle for complex vector bundles says that for a vector bundle $E\to X$ there is an iterated projective bundle $p:F(E)\to X$ for which $p^{\ast }$ is injective on integral cohomology, and $p^{\ast }E$ is a direct sum of complex line bundles. The same assertion holds with rational coefficients, and a common such space can split any finite collection of vector bundles. The injectivity follows at each stage from the projective bundle formula for complex vector bundles; a Hermitian metric splits the resulting filtration into a direct sum.

Write $x_i=c_1(L_i)$ for the Chern roots of the pulled-back vector bundle. Define the degree-$2j$ component of its Chern character byThe Fundamental theorem of symmetric polynomials expresses each power-sum symmetric polynomial ${\sum }_i{x}_i^j$ uniquely as an integral polynomial in the elementary symmetric polynomials, which are the Chern classes of $E$. Thus the definition makes sense on $X$ itself, without choosing actual Chern roots there. On a compact Hausdorff space, each vector bundle is classified by a finite-dimensional Grassmannian; its high-degree Chern character components therefore vanish. Hence this defines a class in the direct sum $H^{2\ast }(X;\mathbb{Q})$ even if $X$ has no finite dimension. On more general spaces an infinite Chern character is instead interpreted in the product of the even-degree cohomology groups.

On a common splitting space, the Chern roots of a direct sum are the combined root lists, whereas the First Chern class identity makes the Chern roots of a tensor product of vector bundles the pairwise sums $x_i+y_j$. ConsequentlyThe injectivity in the splitting principle for complex vector bundles descends both identities to $X$. Additivity extends the Chern character to the Grothendieck group by $\mathrm{ch}([E]-[F])=\mathrm{ch}(E)-\mathrm{ch}(F)$. Multiplicativity and $\mathrm{ch}(1)=1$ then show that the Chern character is a unital ring homomorphism.

For Complex K-theory of complex projective space, let $\gamma$ be the tautological bundle, put $t=1-[\gamma ]$, and set $h=-c_1(\gamma )$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^{n+1})$. The CW filtration gives a cofibrationThe K-theory six-term exact sequence and Complex K-theory of a sphere give, inductively,Here $b_n$ is the image of a Bott element from $S^{2n}$, normalized so that $\mathrm{ch}(b_n)=h^n$. This normalization follows from the multiplicative form of Bott periodicity: the Chern character of the degree-two Bott element is an integral generator, and multiplication yields the top-degree generator on $S^{2n}$. In particular, the abelian group $K^0({\mathbb{CP}}^n)$ is a free abelian group of rank $n+1$.

Assume inductively that $1,t,\dots ,t^{n-1}$ form an integral basis on ${\mathbb{CP}}^{n-1}$. Their lifts, together with $b_n$, form an integral basis on ${\mathbb{CP}}^n$. Sincethe Chern characters of this basis are linearly independent over $\mathbb{Q}$: their first nonzero degrees are $0,2,\dots ,2n$. Thus the Chern character is injective on $K^0({\mathbb{CP}}^n)$. By induction $t^n$ restricts to zero on ${\mathbb{CP}}^{n-1}$, so $t^n=a{b}_n$ for an integer $a$. Comparing Chern characters gives $a=1$. Also $\mathrm{ch}(t^{n+1})=0$, so $t^{n+1}=0$. This proves the integral answer, rather than merely its rationalization:The induction begins with ${\mathbb{CP}}^0$ a point and $t=0$.

All Topological K-theory groups below are complex and are indexed modulo two using Bott periodicity. For $n\ge 1$, Complex K-theory of a sphere givesFor an even-dimensional sphere, choose a generator $u$ of Reduced topological K-theory. Its square vanishes: the Chern character sends $u$ to a top-degree cohomology class, whose square is zero, and the Chern character is injective on this torsion-free abelian group. Thus the ring answers areThe exceptional zero-dimensional sphere consists of two points: $K^0(S^0)=\mathbb{Z}\times \mathbb{Z}$ with coordinatewise multiplication and $K^{-1}(S^0)=0$. In particular its reduced generator is idempotent rather than square-zero.

For the Euler characteristic identity, induct over the cells of a finite CW complex. The starting CW complex consisting of $v$ vertices has $K^0\cong {\mathbb{Z}}^v$, $K^{-1}=0$, and $\chi =v$. If $X$ is obtained from $A$ by attaching one $d$-dimensional cell, the quotient is $S^d$, and the K-theory six-term exact sequence isThese abelian groups are finitely generated by induction. Tensoring the exact sequence with $\mathbb{Q}$ preserves exactness. The alternating sum of dimensions in a cyclic six-term exact sequence is zero, soTherefore

There is a genuine convention issue in the nilpotence assertion. With the usual definition of Reduced topological K-theory as the kernel of restriction to one basepoint, the assertion needs $X$ to be connected. For $X=S^0$ based at its first point, $(0,1)$ lies in ${\tilde{K}}^0(X)$ and satisfies $(0,1{)}^m=(0,1)$ for every $m\ge 1$. Thus it is not a nilpotent element. For an arbitrary finite CW complex, the correct assertion uses the rank map in topological K-theory on every connected component:For a connected space, $I(X)={\tilde{K}}^0(X)$.

Here is an induction proving the corrected assertion. On the vertices, $I(X)=0$. For a positive-dimensional cell attachment $A\subset X$, the ideal $J=\ker (K^0(X)\to K^0(A))$ is square-zero. Indeed, lift two elements of $J$ to relative topological K-theory $K^0(X,A)$. Their product is induced by the reduced diagonalThis map is null-homotopic, since $d<2d$ and $S^{2d}$ is $(2d-1)$-connected. The relative product in topological K-theory is therefore zero, giving $J^2=0$. If $a\in I(X)$, its restriction belongs to $I(A)$, so induction gives $(a{|}_A{)}^m=0$ for some $m$. Then $a^m\in J$ and $a^{2m}=0$. We have proved every Topological K-theory class of rank zero on every connected component is a nilpotent element, and hence the requested result for connected $X$.

Use the standard complex K-theory Thom class convention. For a rank-$r$ complex vector bundle, its K-theory Euler class is the pullback of the K-theory Thom class along the zero section, namelyHere ${\Lambda }^j$ denotes an exterior power, and a Hermitian metric identifies the dual bundle with the conjugate vector bundle. Choosing ${\lambda }_{-1}(E)$ instead changes the K-theory Euler class by the unit $(-1{)}^r[\det E]$, so its kernel and cokernel computations are unchanged.

For the disk bundle $D(E)$ and sphere bundle $S(E)$, the K-theory Thom class identifies $K^i(D(E),S(E))$ with $K^i(X)$. The exact sequence of this pair becomes the K-theory Gysin sequence of a sphere bundle:The groups are two-periodic by Bott periodicity.

View Real projective space as the circle bundle of ${\gamma }^{\otimes 2}$ over Complex projective space, where $\gamma \to {\mathbb{CP}}^n$ is the tautological bundle. Indeed, $S^{2n+1}/\{\pm 1\}={\mathbb{RP}}^{2n+1}$ maps to $S({\gamma }^{\otimes 2})$ by $[v]\mapsto v\otimes v$; in each fiber squaring identifies exactly $v$ and $-v$.

Use Complex K-theory of complex projective space in the formThe K-theory Euler class of ${\gamma }^{\otimes 2}$ is $1-[\gamma {]}^{-2}$. Multiplying it by the unit $-[\gamma {]}^2$ gives $1-[\gamma {]}^2=2t-t^2$. The K-theory Gysin sequence of a sphere bundle therefore givesUnder the projection $q:{\mathbb{RP}}^{2n+1}\to {\mathbb{CP}}^n$, the pullback vector bundle $q^{\ast }\gamma$ is the complexification of a real vector bundle $L$ of the tautological bundle of Real projective space: the fiber map sends $[v,z]$ to $zv$. Thus $t$ pulls back to $1-[L]$. Put $\mu =[L]-1$. The relation $L^{\otimes 2}\cong 1$ gives ${\mu }^2=-2\mu$, and $t^{n+1}=0$ gives $2^n\mu =0$. HenceFor completeness, these relations give precisely the abelian group $\mathbb{Z}\oplus \mathbb{Z}/2^n$. In the basis $t,\dots ,t^n$, the image of multiplication by $2t-t^2$ is generated by $2t-t^2,2t^2-t^3,\dots ,2t^n$; its relation matrix is triangular with determinant $2^n$, and the relations express every $t^j$ as $2^{j-1}t$. Thus there is exactly one cyclic torsion subgroup of order $2^n$.

If $n\ge 1$ and $a=a_0+a_{1}t+\cdots +a_n{t}^n$ is in the kernel, successive coefficients of $(2t-t^2)a$ give $2a_0=0$, $2a_1-a_0=0$, and so on, up to $2a_{n-1}-a_{n-2}=0$. Hence $a_0=\cdots =a_{n-1}=0$, with $a_n$ arbitrary. For $n=0$ the multiplier is zero on $R=\mathbb{Z}$. In all cases,

Finally let $E=L^{\oplus k}$ with $k\ge 1$, over $B={\mathbb{RP}}^{2n+1}$. Since $L^{\ast }\cong L$, multiplicativity of the K-theory Euler class givesOn $K^0(B)=\mathbb{Z}\oplus (\mathbb{Z}/2^n)\mu$, the image of multiplication by this K-theory Euler class is the subgroup generated by $2^{k-1}\mu$: multiplication by an integer produces every such multiple, while multiplication by $\mu$ only adds multiples of $2^k\mu$. ThereforeMultiplication by $\mu$ is zero on $K^{-1}(B)\cong \mathbb{Z}$, since $2^n\mu =0$ and this abelian group has no nonzero torsion elements. Thus the K-theory Gysin sequence of a sphere bundle gives a short exact sequenceIt splits as an exact sequence of abelian groups because its quotient is a free abelian group. ConsequentlyThe convention $\mathbb{Z}/1=0$ covers $n=0$ and $k=1$. For example, when $k=1$ the sphere bundle is $(S^{2n+1}\times S^1)/\{(v,z)\sim (-v,-z)\}$, a mapping torus of the antipodal map; its Topological K-theory has no nonzero torsion elements, agreeing with the formula. If $k=0$ is allowed, the rank-zero vector bundle has an empty unit sphere bundle, so $K^0(S(E))=0$ instead.

Now let $n=m=p$ be prime and put $\alpha =\sqrt[p]{p}$ and $A=\mathbb{Z}[\alpha ]$. The Eisenstein criterion applied at $p$ gives $[\mathbb{Q}(\alpha ):\mathbb{Q}]=p$, while the discriminant of elements of a number field of its power basis iswhere $f(X)=X^p-p$. By the discriminant-index formula for an integral lattice, every prime dividing $[{\mathcal{O}}_L:A]$ must therefore be $p$.

Suppose $A\ne {\mathcal{O}}_L$. The finite abelian group ${\mathcal{O}}_L/A$ is then a nontrivial finite p-group, so it contains an element of order $p$. Consequently there is an $x\in {\mathcal{O}}_L\setminus A$ such thatwith some $c_k$ not divisible by $p$.

The polynomial $X^p-p$ is Eisenstein, so total ramification from an Eisenstein polynomial gives a unique prime ideal $P$ above $p$ with normalized discrete valuationLet $k_0$ be the least index for which $p\nmid c_{k_0}$. The term $c_{k_0}{\alpha }^{k_0}$ has valuation $k_0<p$; every earlier term has valuation at least $p$, and every later term with coefficient prime to $p$ has a distinct, larger valuation. The non-Archimedean valuation therefore givesThis contradicts $x\in {\mathcal{O}}_L$, since an algebraic integer has nonnegative valuation at every prime ideal. Thus $A={\mathcal{O}}_L$, and

Cauchy's theorem for finite groups states that if a prime number $p$ divides the order of a finite group $G$, then $G$ contains an element of order $p$.

ConsiderThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$ is divisible by $p$. The cyclic group $C_p$ acts on $X$ by cyclically shifting the coordinates; the shifted tuple remains in $X$ because $x_2\cdots x_p{x}_1=x_1^{-1}(x_1\cdots x_p)x_1=e$. Every orbit has size one or $p$. The fixed points are exactly $(x,\dots ,x)$ with $x^p=e$. Their number is therefore divisible by $p$. Since the identity is one fixed point, another exists; its order is $p$.

Now let the abelian group $G$ have order $pq$ for distinct primes $p,q$. Cauchy's theorem supplies $x,y$ of orders $p,q$. Their cyclic subgroups intersect trivially, and commutativity shows that $xy$ has order $pq$. Thus $G$ is cyclic andThe corresponding assertion for order $p^2$ is false: $C_{p^2}$ is not isomorphic to $C_p\times C_p$, since the former has an element of order $p^2$ while every nonidentity element of the latter has order $p$.

The group $G$ is a nontrivial finite p-group, so the nontrivial center of a finite p-group gives $Z(G)\ne \{e\}$. If $Z(G)\ne G$, then $Z(G)$ is already a nontrivial proper normal subgroup, so $G$ is not a simple group. If $Z(G)=G$, then $G$ is abelian. By Cauchy's theorem for finite groups, it has a subgroup of order $p$; every subgroup of an abelian group is normal, and this subgroup is nontrivial and proper because $|G|=p^4$. Thus in every case

Every singular simplex has compact image, and a singular chain is a finite sum of simplices. The image of any chain is therefore compact and lies in some $X_a$. Directedness puts any finite collection of chains into one common $X_b$, soBecause filtered colimits of abelian groups are exact, kernels and images commute with this colimit. Taking homology gives the homology of a directed union:

For an open $U\subseteq {\mathbb{R}}^N$, use the directed family of finite unions of closed rational cubes contained in $U$. Every compact subset of $U$ lies in one such finite polyhedron, and each polyhedron has finitely generated cellular homology. There are only countably many of them, so their direct limit is countable. Thus every $H_i(U;\mathbb{Z})$ is countable.

Cohomology behaves differently because $\mathrm{Hom}$ turns a direct sum into a direct product. The connected open sethas one independent loop around each puncture, so $H_1(U;\mathbb{Z})\cong {⨁}_{n\ge 1}\mathbb{Z}$. Since $H_0(U;\mathbb{Z})=\mathbb{Z}$, the universal coefficient theorem for cohomology giveswhich is uncountable. This is the first cohomology of the countably punctured plane.

For a finite generating set $S$ of $G$, the Cayley graph ${\mathrm{Cay}}_S(G)$ has vertex set $G$ and an edge from $g$ to $gs$ for every $g\in G$ and $s\in S\cup S^{-1}$; one may retain orientations and labels, or forget them.

As unlabelled undirected graphs, the Cayley graph of the infinite cyclic group $\mathbb{Z}$ with generator $1$ and the Cayley graph of the infinite dihedral groupwith generators $a,b$ are both the two-way infinite line. The first group is an abelian group and the second is not, so they are not isomorphic. Thus the requested example is the infinite line as a Cayley graph of $\mathbb{Z}$ and $D_{\infty }$.

We induct on the nilpotency class $s$. For $s=1$, the group is Abelian, so take $C_1=A$ and no $X_i$. For $s>1$, part (i) writeswith $X$ small and every $⟨B_i⟩$ of class at most $s-1$. Apply the induction hypothesis to each $B_i$. Since $B_i\subseteq A^{O(1)}$ and its approximation parameter is $K^{O(1)}$, all resulting small sets lie in $A^{O_s(1)}$, all their sizes are at mostand all resulting approximate groups lie in $A^{O_s(1)}$ and have approximation parameter $K^{O_s(1)}$. Their generated subgroups are abelian groups.

There are $O({\log }^{O(1)}(2K))$ factors at each of at most $s$ induction levels. Absorbing the resulting products of the bounds into the $O_s$ notation givesKeeping the factors in the order supplied by the induction yields the required product of the $X_i$ and $C_j$ containing $A$.

A group is solvable when its derived serieseventually reaches the trivial group; equivalently, it has a subnormal series with abelian quotients.

The symmetric group $S_4$ is solvable becausehas abelian quotients $V_4$, $C_3$, and $C_2$, where $V_4$ is the Klein four-group. Thus

For subgroups $H,P\le G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the double cosets $HxP$ partition $G$.

Let $H$ act by left multiplication on the set of left cosets $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer isThe orbit-stabilizer theorem therefore gives

Suppose $P$ is a Sylow subgroup of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xP{x}^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xP{x}^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the Sylow subgroup of a subgroup from double cosets argument.

To count the general linear group over a finite field $G{L}_n({\mathbb{F}}_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. ConsequentlyNone of the factors $p^j-1$ is divisible by $p$, so the upper unitriangular group is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The permutation matrices form a subgroup isomorphic to the symmetric group $S_n$.

By Cayley theorem, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $G{L}_{|G|}({\mathbb{F}}_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.

The counting part of the Sylow theorems says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies

Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts giveso $n_p=1$ and the Sylow $p$-subgroup is a normal subgroup. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their direct product and is abelian. In the nonabelian case one must therefore have $n_q=p$, whenceor equivalently $q\mid p-1$, as recorded by nonabelian group of order pq.

Suppose $\hat{\Gamma }\cong {\mathbb{Z}}_p$. This completion is an abelian group, so every finite quotient of $\Gamma$ is abelian. Consequently the quotient map to the abelianization $A={\Gamma }^{\mathrm{ab}}$ inducesThe group $A$ is a finitely generated abelian group. If its free rank is zero, $\hat{A}=A$ is finite. If its free rank is positive, $A$ and hence $\hat{A}$ have a nontrivial quotient $\mathbb{Z}/q\mathbb{Z}$ for every sufficiently chosen prime $q$. But ${\mathbb{Z}}_p$ has no nontrivial finite quotient of order coprime to $p$. Both cases are impossible, so $\hat{\Gamma }\ncong {\mathbb{Z}}_p$.