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    Solutions to Exercises from

    Cohomology of Groups

    Christopher A. Gerig, Cornell University (College of Engineering)

    August 2008 - May 2009

    I claim there are no errors, but maybe youll come up with a counterexample. I wouldappreciate it if you could email me any errors (and corrections): [email protected].

    Any errors would be due to solely myself, or at least the undergraduate-version ofmyself when I last looked over this. Remark made on 10/22/13.

    Professor Kenneth S. Brown, Cornell University Department of Mathematics

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    PREFACE

    This is ultimately a solutions manual to Ken Browns graduate textbook, Cohomology of Groups.Only chapters one through six are fully covered, with a few solutions not provided (the problems labelledomitted were not solved because they required certain theories that I have not touched yet); someexercises from the remaining chapters have also been solved. The ending section contains additional

    exercises (with solutions), some resulting from questions Prof. Brown and I posed throughout the study,and others taken from various texts.

    I currently work with Prof. Brown in an independent study (of which this paper is a byproduct),learning group cohomology in order to dive into other interesting topics and to begin research (possiblyconcerningEssential Cohomology).

    Errata to Cohomology of Groups :

    pg62, line 11 missing a paranthesis ) at the end.pg67, line 15 from bottom missing word, should say as an abelian group.pg71, last line of Exercise 4 hint should be on a new line (for whole exercise).pg85, line 9 from bottom incorrect function, should be

    gC/Hg

    1 gm.pg114, first line of Exercise 4 misspelled endomorphismwith an extra r .pg115, line 5 from bottom missing word, should say 4.4 is achain map.pg141, line 4 from bottom missing a hat

    on the last H.

    pg149, line 13 the ideal Ishould be italicized.pg158, line 20 there should be a space between S L2(Fp) and (p odd).pg160, line 5 from bottom missing the coefficient in cohomology,H(H, M).pg165, line 26 the comma in the homology should be lowered, Hq(C,p).

    major: Engineering Physics (B.S.)status: Sophomore Undergraduate

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    Contents

    1 Chapter I: Some Homological Algebra 4

    2 Chapter II: The Homology of a Group 11

    3 Chapter III: Homology and Cohomologywith Coefficients 20

    4 Chapter IV: Low-Dimensional Cohomologyand Group Extensions 30

    5 Chapter V: Products 40

    6 Chapter VI: Cohomology Theoryof Finite Groups 46

    7 Chapter VII: Equivariant Homology andSpectral Sequences 54

    8 Chapter VIII: Finiteness Conditions 57

    9 Chapter IX: Euler Characteristics 58

    10 Additional Exercises 60

    11 References 75

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    1 Chapter I: Some Homological Algebra

    2.1(a): The setA = {g 1 | g G, g= 1} is linearly independent because

    g(g 1) = 0

    gg=g =

    g1 +

    0g, and since ZG is a free Z-module,

    g has a unique expression, yielding

    g = 0 g A. To show that I = span(A), let

    gg I and hence

    g = 0. Thus we can write

    gg =g g 0 =gg g =g(g 1). Since A is a linearly independent set whichgeneratesI, it is a basis for I.2.1(b): Consider the left ideal A = ({s 1 | s S}) overZG.A I since(A) =

    s

    (

    grgg)(s 1)

    =

    s (

    grgg)(s 1) =y 0 = 0.Ifx I then x =

    igi

    j gj such that

    i =

    j.

    x= (

    igi

    i) (

    j gj

    j ) + (

    i

    j )Thusx =

    i(gi 1)

    j (gj 1) and it suffices to show g 1 A for any g G so that x A and

    IA. Since G= S, we have a representation x = s11 s1n . By using ab 1 =a(b 1) + (a 1)

    and c1 1 = c1(c 1), the result follows immediately.

    2.1(c): Suppose S G | I = ({s 1| s S}). Then every element of I is a sum of elements of

    the form g g | g = gs1. For g G, g 1 I, so we have a finite sum g 1 =

    n1i=1(gi g

    i).

    Since this is a sum of elements in G where G is written multiplicatively, i0 | gi0 = g, say i0 = 1.Thus g 1 =g g1+n1

    i=2(gi gi) g

    1 1 =

    n1i=2(gi g

    i). Another iteration yields g2 =g

    1 and

    g= g1= g1s11 =g2s

    11 =g

    2s12 s

    11 . At the last iteration, g

    n2 1 =gn1 g

    n1 g

    n1= 1,g

    n2=

    gn1 = gn1s

    1n1 = 1s

    1n1. Through this method we obtain a sequence g1, g2,...,gn1, gn |gi = gi+1s

    1i

    and g1 = g, gn = 1. g has a representation in terms of elements ofSand so G = S since g wasarbitrary.

    2.1(d): If G is finitely generated, then by part(b) above, I is finitely generated. For the converse,suppose I= (a1,...,an) is a left ideal overZG. Noting from part(a) that I=g 1gG as aZ-module,eachai can be represented as a finite sumai =

    jzj (gj 1). Since eachai is generated by finitely many

    elements, and there are finitely many ai, I is finitely generated as a left ideal by elements s 1 wheres G. Therefore, we apply part(c) to have G= s1, . . . , sk and thus G is a finitely generated group.

    2.2: Let G = t with |G| = n and let t be the image of T in R = Z[T]/(Tn 1) = ZG. The el-ement T1 is prime in Z[T] because Z[T]/(T1) = Z which is an integral domain (An ideal P ofa commutative ring R is prime iff the quotient ring R/Pis an integral domain; Proposition 7.4.13[2]).By Proposition 8.3.10[2] (In an integral domain a prime element is always irreducible), T1 is irre-ducible in Z[T]. We also could have obtained this result by applying Eisensteins Criterion with thesubstitutionT =x + 3 and using the prime 2. Since Z[T] is a Unique Factorization Domain, the specificfactorizationTn 1 = (T 1)(Tn1 +Tn2 + +T+ 1) with the irreducible T 1 factor is unique,

    considering the latter factorn1

    i=0 Ti as the expansion of its irreducible factors [note: ifn is prime thenn1

    i=0 Ti = n(T), a cyclotomic polynomial which is irreducible in Z[T] by Theorem 13.6.41[2]]. Thus

    everyfR is annihilated by t 1 iff it is divisible by N=n1

    i=0 ti (and vice versa), and so the desired

    free resolution ofM= Z = ZG/(t 1) is:

    R

    t1

    R

    N

    R

    t1

    R M0 .

    3.1: The right cosetsH gi are H-orbits ofG with theH-action as group multiplication. SinceG =

    Hgiwheregi ranges over E,ZG=

    Z[Hgi]=

    Z[H/Hgi ]. G is a free H-set because hg = g hgg

    1 =gg1 h = 1, i.e. the isotropy groups Hg are trivial. Therefore,ZGis a freeZH-module with basisE.

    3.2: Let S = H G and consider Z[G/H]. Now x Z[G/H] has the expression x =

    zi(giH),and there exists an element fixed by H, namely, x0 = g0H = H where g0 H. H is annihilated byI= Ker= ({s 1}) since (s 1)H=sH H=H H= 0 s S, and soI x0= 0. We have g 1 Isince(g 1) =(g) (1) = 1 1 = 0. Hence (g 1)x0= 0 gx0= x0g G Gx0= x0. Finally,GH=HG H. G= H=S.

    4.1: Orienting each n-cell en gives a basis for Cn(X). If X is an arbitrary G-complex, then with

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    ie

    ni Cn(X), g G can reverse the orientation ofe

    n by inversion (fixing the cell). Thus G need notpermute the basis, and hence Cn(X) is not necessarily a permutation module.

    4.2: Since X is a free G-complex, it is necessarily a Hausdorff space with no fixed points under theG-action. First, assume G is finite and take the set of elements in Gx0, which are distinct pointsgix0 with 1x0 = x0 for an arbitrary point x0 X. Applying the Hausdorff condition, we have open

    sets Ugi containing gix0 where each such set is disjoint from U1 containing x0. Form the intersectionW = (

    i g

    1i Ugi) U1 which contains x0. SincegkW Ugk, we have gkW W = for all nonidentity

    gk G, and so W is the desired open neighborhood ofx0 [This result does not follow for arbitrary Gsince an infinite intersection of open sets need not be open].Assume the result has been proved for G infinite.Let : X X/G be the quotient map, which sends the disjoint collection ofgiWs to (W). Since1(W) =

    i giW, giW (W) is a bijective map (restriction of) and thus it is a homeomorphism

    (continuous and open). This covering space is regular becauseG acts transitively on 1(Gx) by def-inition. Elements ofG are obviously deck transformations since Gg x= Gx, hence G Aut(X). GivenAut(X) with (a) = b, those two points are mapped to the same orbit in X/G (since = ),and so g G sendinga to b. By the Lifting Lemma (uniqueness) we have =g, hence Aut(X) G G is the group of covering transformations.

    IfX is contractible then G=1(X/G)/1(X)

    =1(X/G)/0 =1(X/G) and is the universal cover ofX/G, so X/G is a K(G, 1) with universal cover X.

    It suffices to show that the G-action is a properly discontinuous action on Xwhen G is infinite:Every CW-complex with given characteristic maps fj,n : (B

    nj, S

    n1j ) (

    nj,

    nj ) admits a canonical

    open cover {U} indexed by the cells, where Ua and Ub are disjoint open sets for distinct cells of equaldimension (for instance, if X is a simplicial complex we can take U = St() which is the open starof the barycenter of in the barycentric subdivision of X). More precisely, for one cell nk X

    n ineach G-orbit of cells define its barycenter as nk fk,n(bk), where bk = 0 B

    n is the origin of then-disk; for the rest of the cells {ni =gi

    nk } in each G-orbit define their barycenters as

    ni fi,n(bi),

    where bi Bni is chosen so that

    ni = gi

    nk . Considering the 0-skeleton X

    0, its cells 0j are by defini-

    tion open and so the canonical open cover ofX0 is the collection {U0j = 0j }. Proceeding inductively

    (with U open in Xn1), consider the n-skeleton Xn = Xn1

    jnj and note that the preimage under

    fj,n of the open cover of Xn1

    is an open cover of the unit circle Sn1j . Take an open set f

    1j,n (U)

    in Sn1j and form the open sector Wj, Bnj which is the union of all line segments emanating

    from bj and ending in f1j,n (U), minus bj and f

    1j,n (U); each U determines such a Wj,. As fj,n is a

    homeomorphism of Int(Bnj) with nj , we have such open sectors fj,n(Wj,) in the n-cell. Noting the

    weak topology on X, the set U = U

    jfj,n(Wj,) is open in Xn iff its complement in Xn is closed

    iff (Xn U) ik

    ik (U

    ik) is closed in

    ik for all cells in X

    n. For i < n, U i = U

    i

    because fj,n(Wj,) nj which is disjoint from the closure of all other cells, and the complement of this

    intersection in i Xn1 is closed because U is open by inductive hypothesis. Therefore, it suffices toshow that nk (U

    nk ) is closed in

    nk k, which is equivalent under topology of cells for U

    nk to be

    open in nk . For arbitraryk we haveU

    nk = (U

    nk )

    j(fj,n(Wj,) nk ) = (U

    nk )

    fk,n(Wk, )

    and taking the preimage we have Y = f1k,n(U

    nk ) = f

    1k,n(U)Wk, . The n-disk is compact, the

    CW-complex Xn is Hausdorff, a closed subset of a compact space is compact (Theorem 26.2[6]), the

    image of a compact set under a continuous map is compact (Theorem 26.5[6]), and every compact subsetof a Hausdorff space is closed (Theorem 26.3[6]); thusfk,n is a closed map and hence a quotient map fornk (by Theorem 22.1[6]). It suffices to check thatY B

    nk is open, for then fk,n(Y) =U

    nk is open

    in nk by definition of a quotient map. By construction, Y = {x Bnk bk | r(x) f

    1k,n(U)} where

    r :B nk bk Bnk S

    n1k is the radial projection r(x) =

    xbk||xbk||

    . As r is continuous and f1k,n(U) is

    open in B nk , we have Y = r1(f1k,n(U)) open in B

    nk bk and hence in B

    nk (by Lemma 16.2[6]). Our

    new collection for Xn is the open sets U (where Xn1) plus the open n-cells Unj =

    nj; this is the

    canonical open cover ofXn and hence completes the induction.Any point x X will lie in an i-cell which lies in the open set U, and we take this as our desiredneighborhood ofx: since any open set of our constructed cover is bounded by barycenters, and g Gmaps barycenters to barycenters by construction, we have gU = U

    g which is disjoint from U

    by

    construction for all g = 1.

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    4.3: Given G = Z Z we have the torus T with 1T = G and its universal cover : R2 T.After drawing the lattice Z2 R2, pick a square and label that surface L, with the bottom left cor-ner as the basepoint x0 and the bottom side as the edge es and the left side as the edge et (so thecorners are x0, sx0,stx0, tx0 going counterclockwise around L, and the top and right sides ofL are re-spectively tes and set). Following Browns notation [in this section], x0 generates C0(R2) and es, etgenerate C1(R2) with 1(es) = (s 1)x0 and 1(et) = (t 1)x0. Lastly, L generates C2(R2) with2(L) =es+ set tes et= (1 t)es (1 s)et. Thus the desired free resolution ofZ overZG is:

    0 ZG 2 ZG ZG

    1 ZG Z 0 .

    4.4: Omitted.

    5.1: The homotopy operator h in terms of the Z-basis g[g1| |gn] for Fnis h(g[g1| |gn]) = [g|g1| |gn].

    5.2: Using G= Z2 ={1, s}, the elements of the normalized bar resolution F =F/D are [s|s| |s],and each element forms a basis for the corresponding dimension, giving the identification Fn = ZG.Denotingsi = s i,

    di[s1|s2| |sn] =s[s| |s] , i= 0[s1| |si1|sisi+1| |sn] = 0 , 0< i < n

    [s| |s] , i= n

    The middle equation resulted from sisi+1= s2 = 1 (so the element lies in D). The boundary operator

    n then becomess 1 fornodd ands + 1 for n even. the normalized bar resolution is:

    ZG s1

    ZG s+1 ZG s1

    ZG Z 0 .

    5.3(a): [[Geometric Realization of a Semi-Simplicial Complex]]For each (n+1)-tuple = (g0, . . . , gn), let be a copy of the standardn-simplex with verticesv0, . . . , vn.Let di = (g0, . . . , gi, . . . , gn) and let i : di be the linear embedding which sends v0, . . . , vn1to v

    0, . . . , v

    i, . . . , v

    n. Consider the disjoint union X

    0= (topologize it as a topological sum) and

    define the quotient space X def= X0/ using the equivalence relation generated by (, ix) (di, x),

    where we rewrite as for clarity of the relation properties.We assert that the geometric realization Xis a CW-complex with n-skeleton Xn = (

    dimn

    )/.

    X0 is the collection of vertices and hence a 0-skeleton, and so we proceed by induction on n [sketch]:

    In Xn the equivalence relation identifies a point on a boundary (n) with a point in Xn1, and

    it doesnt touch the interior points of (n) . This means that the n-cells are{

    (n) } with the attaching

    maps induced bydi i. Refer to Theorem 38.2[4] for the analogous construction with adjunction spaces,providing Hausdorffness of Xn and weak topology w.r.t. {Xi}i

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    G-complex X as above, and since X X/G is a regular covering map by the result of exercise I.4.2above, the orbit spaceX/Gis aK(G, 1), called the classifying space.

    5.3(b): For the normalized standard resolution F, we simply follow part(a) while making these furtheridentifications in X0to collapse degenerate simplices. For each = (g0, . . . , gn) let si= (g0, . . . , gi, gi, . . . , gn),and when forming the quotient X0 X we also collapse si to via the linear map Li which

    sends v0, . . . , vn+1 to v0, . . . , vi, vi, . . . , vn (so the only simplices of X are those whose associated tu-ples have pairwise distinct coordinates). Thus the equivalence relation in part(a) is also generated by(, Lix) (si, x). During the inductive process for X

    n, if is a degenerate simplex then Xn1 will

    already contain it and so those simplices need not be considered as n-cells. No problems arise when usingthe homotopy because for tuples of the form = (1, g0, . . . , gn)| g0 = 1, the cone v0 = h = has the identity map 0 still being nulhomotopic [note: we actually have a deformation retraction since(1) remains fixed instead of looping around (1,1) as in part(a)].

    6.1: Given a finite CW-complex X with a map f : X X such that every open cell satisfiesf()

    = where dim dim, we have the condition f() = and there are no fixed

    points. Viewing the open n-cell on the chain level in Hn(X(n), X(n1)) = Cn(X), f() does not

    consist of and so the respective matrix has the value 0 at the row-column intersection for. Therefore,

    tr(f, Cn(X)) = 0 n since the diagonal of the matrix for the basis elements is zero. By the Hopf TraceTheorem,

    (1)itr(fi, Hi(X)/torsion) =

    (1)itr(f, Ci(X)) and so the Lefschetz number (f) = 0.

    If X is a homology (2n 1)-sphere, then Hi(X) is nontrivial only in dimensions 0 and 2n 1 (inwhich case it is isomorphic to Z). Thus, by the Lefschetz Fixed Point Theorem, (1)0tr(f0, Z) +(1)2n1tr(f2n1, Z) = 1 d= 0 d= 1 and f : H2n1(X) H2n1(X) is the identity.

    6.2: The group action G Homeo(S2n) yields a degree map

    : G Aut(H2n(S2n))=Aut(Z) ={1}= Z/2Z

    which sends g G to the degree d = deg(g) of its associated homeomorphism g : S2n S2n [note:deg(g) deg(g1) = deg(g g1) = deg(id) = 1 |d|= 1].Consider nontrivial G = Z/2Z and assert that this is not injective:Ker = 0 3 |G| = |Ker| |Im| = 1 |Im|. Since Im Z/2Z, |Im|= 1 or 2 . In either casewe arrive at a contradiction (since 1, 2 < 3). g Ker | g = id degg = 1. Now assume thisaction is free, and use the notation fi : Hi(S2

    n) Hi(S2n). By the Lefschetz Fixed Point Theorem,

    (1)0tr(f0) + (1)2ntr(f2n) = 1 +d = 0 degf = d = 1 f = id. Our contradiction has now been

    reached (taking f=g from above).

    7.1: Given the finite cyclic groupG = t, the free resolution F ofZ over ZG with period two (chaincomplex with rotations of S1), and the bar resolution F, we obtain a commutative diagram wheref :FF is the desired augmentation-preserving chain map:

    ZG

    f3

    t1 ZG

    f2

    N ZG

    f1

    t1 ZG

    f0

    Z

    idZ

    0

    F

    3

    F

    2

    F

    1

    F

    0

    Z

    0We definefinductively asfn+1= knfn, wherek is a contracting homotopy for the augmented complexassociated toF, and each map is determined by where it sends the basis element:f0(1) =k1idZ(1) =k1idZ(1) =k1(1) = (1) [ ]f1(1) =k0f0(1) =k0f0(t 1) =k0{tf0(1) f0(1)}= k0{t[ ] [ ]}= [t] [1]

    f2(1) =k1f1(1) =k1f1(N) =k1{n1

    i tif1(1)}=

    n1i ([t

    i, t] [ti, 1])

    f3(1) =k2f2(1) =k2f2(t 1) =n1

    i ([t, ti, t] + [1, ti, 1] [t, ti, 1] [1, ti, t])

    7.2: Here is the axiomatized version, Lemma 7.4:Under the additive category A, let (C, ) and (C, ) be chain complexes, let r be an integer, and let(fi :Ci C

    i)ir be a class of morphisms such that

    ifi =fi1i for i r. IfCi is projective relative

    to the classEof exact sequences for i > r, andCi+1 Ci C

    i1 is in E fori r, then (fi)ir extends

    to a chain map f :C C and fis unique up to homotopy.

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    (Theorem 7.5 follows immediately)

    7.3(a): Given an arbitrary category C, let A Ob(C) be an object and let hA = HomC(A, ) : C (Sets) be the covariant functor represented by A, with uA hA(A) as the identity map A A. LetT : C (Sets) be an arbitrary covariant functor. Any natural transformation : hA T yields thecommutative diagram (with f :A B in hA(B) arbitrary):

    hA(A) hA(f) hA(B)

    T(A) T(f) T(B)

    For any v T(A) suppose we have the natural transformation with (uA) = v. By commutativity,T(f)(v) = ( hA(f))(uA) =(f uA) =(f), and hence the transformation is unique [determined bywhere it sends the identity]. For existence of the natural transformation(f) =T(f)(v), we assert thatit satisfies the commutative diagram, using arbitrary g : B C (andf as above):

    T(g)[(f)] =T(g)[T(f)(v)] =T(g f)(v) =(g f)[hA(g)(f)] =(g f) =T(g)[(f)]

    Thus, HomF(hA, T)=T(A) whereFis the category of functors C (Sets), and we have finished provingYonedas Lemma.

    7.3(b): An M-free functor F : C Ab is isomorphic to

    ZhA , where A M [M is a sub-class of Ob(C)] and ZhA: C Ab is the composite ofhA and the functor (Sets) Ab which associatesto a set the free abelian group it generates. Given the additive categoryAwhose objects are covariantfunctors C Ab and whose maps are natural transformations of functors, let Ebe the class ofM-exactsequences inA. Consider the mapping problem (for all rows in E):

    F

    0

    Ti

    Tj

    T

    By Yonedas Lemma (part(a) above), each component ZhA ofFwith any natural transformation in theabove diagram is completely determined by the identity uA , and so these identities form a basis forF. In particular, for the identity uA we obtain the exact sequence T

    (A)T(A)T(A) of abeliangroups from the above mapping problem since the associated row lies in E with A M. Thus, for eachidentity we have (uA) Kerj = Imi, which implies x T

    (A) | i(x) = (uA), and so weform by(uA) =x. This means that F (an M-free functor) is projective relative to the class E ofM-exact sequences.

    7.3(c): There is a natural chain map in A from M-free complexes to M-acyclic complexes, and itis unique up to homotopy [using the categorial definitions from parts (a) and (b)]. This statement is aresult of the combination of part(b) and Exercise 7.2 above, and is precisely the Acyclic Model Theoremin a rephrased form (M is the set of models).

    7.4: Under the category of R-modules, let (C, ) and (C, ) be cochain complexes, let r be an inte-ger, and let (fi :

    Ci Ci)ir be a class of morphisms such that fi1i1 =i1f

    i for i r . IfC

    i isinjective relative to the class E of exact sequences for i > r [yielding a cochain complex of injectives],and Ci1 Ci Ci+1 is inE fori r [an acyclic cochain complex], then (fi)ir extends to a cochainmap f : CC andf is unique up to homotopy.(The analogous Theorem 7.5 follows immediately)

    8.1: Obviously the trivial group is one, since Z[{1}] = Z and Z is a projective Z-module; so assumeG isnontrivial. GiveZ the trivial module structure (so that for r ZG, r a= (r)a a Z). Consideringthe short exact sequence of modules 0 I ZG Z 0, we must find a splitting : Z ZG forZto possibly beZG-projective. Any such map is determined by where 1 Z is sent; say (1) =x. Then,for nontrivial = rigi,( 1) = (1) = x= x. But( 1) =( ri) =(1) ri = ( ri)x,

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    so x= (

    ri)x. Thus (

    ri)x= 0

    rigi =

    ri

    ri(gi 1) = 0 [this sum can be viewedas having nogi = 1, and hence it lies in I]. Restricting our choice of nontrivial to one which is not aninteger, there is some nontrivialri0 associated togi0 = 1 and hence we must have gi = 1 i (by freenessofI). But then G is the trivial group, and we are done.

    8.2: Assume P is a projective ZG-module and consider the subgroup H G. Then F = P K

    whereF is a free ZG-module, by Proposition I.8.2[1]. Byrestriction of scalars fromZG to its subringZH [r n = f(r)n with homomorphism f : ZG ZH preserving identities], F has an inherent ZH-module structure. Since Pis a direct summand of such a free module, it is a projective ZH-module, byProposition I.8.2[1]. Alternatively, we can note from Exercise 3.1 above that ZG=

    ZHand soF is a

    direct sum ofH-modules, henceZH-free.

    8.3(a): GivenFas a non-negative acyclic chain complex of projective modules Pn (over an arbitrary

    ringR), it will be contractible if each short exact sequence 0 Zn PnnZn1 0 splits (where is

    induced by the boundary ), by Proposition I.0.3[1]. The casen= 0 is trivial using the zero map, andso arguing inductively we assume that n1 splits. Since Kern1 = Zn1 is a direct summand of theprojective modulePn1 and a projective module is a direct summand of a free module by PropositionI.8.2[1],Zn1is necessarily a direct summand of a free module, and hence is projective by Lemma I.7.2[1].

    Therefore, n must have a splitting in the aforementioned short exact sequence by Proposition I.8.2[1].

    8.3(b): If R is a principal ideal domain, then submodules of a free R-module are free by TheoremII.7.1[5]. Since a projective module is a direct summand of a free module by Proposition I.8.2[1], it isin particular a submodule and hence is free (over R as a PID). Therefore, submodules of a projectivemodule over a PID are free and hence projective (free modules are projective by Lemma I.7.2[1]). Thenon-negativity hypothesis of Corollary I.7.7 can be dropped if we then restrict ourselves to PIDs, becausewe can follow part(a) above but notuse induction since Zn1 is already projective, being a submoduleof the projective chain module Pn1 (i.e. we dont need any starting point in the resolution to obtainthe desired splitting).

    8.4: Every permutation module admits the decomposition QX =

    Q[G/Gx] and a direct sum of

    projective modules is projective iff each summand is projective (by Lemma XVI.3.6[5]). Thus it sufficesto show thatQ[G/Gx] is a projectiveQG-module, whereG is an arbitrary group and Gx is finite. Notethat HomQG(Q[G/Gx], ) is a left-exact functor (Corollary 10.5.32[2]); it is given by MMGx becauseany homomorphism is determined by (Gx), and (Gx) = (g Gx) = g (Gx) for any g Gx.Thus, we must show that HomQG(Q[G/Gx], ) takes surjective homomorphisms M M to surjectivehomomorphisms MGx MGx [because then the functor is exact and then Q[G/Gx] is projective bydefinition]. If m MGx , lift mto m M. Then 1|Gx|

    gGx

    gm is also a lifting of mwhich lies in MGx

    (because 1|Gx|

    gm 1|Gx|

    gm= 1|Gx| |Gx|m= m), and so MGx MGx is surjective. Thus,QX is

    a projectiveQG-module, where X is a G-set and Gx is finite for all x X.

    8.5: If G is finite and k is a field of characteristic zero, consider any short exact sequence of kG-

    modules of the form 0 M M P 0. Sincek-vector spaces are free modules over k, and free

    modules are projective (by Lemma I.7.2[1]), P is k-projective and hence the sequence [as k-modules]splits by Proposition I.8.2[1]. Choosing a splitting f : P M for the underlying sequence ofk-vectorspaces, we form the homomorphism : PM byx 1|G|

    gG gf(g

    1x).

    Sincef is a k-module homomorphism, it suffices to show that is equivariant (compatible with G-action)for it to thus be a kG-module homomorphism:

    (g0x) = 1|G|

    gf(g1g0x) =

    1|G|

    g0hf(h

    1x)

    [where h= g10 g]g0(x) =g0(

    1|G|

    gf(g1g0x)) =g0(

    1|G|

    hf(h1x)) =(g0x)

    [because

    gG h=

    gG g, as g10 permutes the elements of G]

    By Proposition 10.5.25[2] it suffices to show that = idP for to thus be a splitting (noting thatf=idP):

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    ((x)) =( 1|G|

    gG gf(g1x)) = 1|G|

    gG (gf(g

    1x)) = 1|G|

    gG g(f(g1x)) =

    1|G|

    gG gg

    1x= 1|G|

    gG x= 1|G| |G|x= x

    Since is a kG-splitting, P is kG-projective by Proposition I.8.2[1].

    8.6: Suppose P is an R-module such that : P R P HomR(P, P) is surjective, where this

    map is given by (u m)(x) = u(x) m. Then in particular we have idP = ( fi ei), so thatx = idP(x) =

    (fi ei)(x) =

    fi(x)ei. Thus, P is projective by Proposition I.8.2[1], and it is

    finitely generated by theei elements (noting that the summation is finite for tensor products).

    8.7: Assume P is a finitely generated projective R-module and M is any (left) R-module, and takethe canonical isomorphism : P RM HomR(P, M) from Proposition I.8.3[1] which is given by(u m)(x) = u(x) m. For any z P RP we define the map z : HomR(P, M) P

    RM asz(f) = (P

    f)(z), which is a homomorphism since tensors are distributive over sums.Method 1 : View 1 as a natural transformation HomR(P, ) P

    R , where HomR(P, )and P R are exact covariant functors from the category R-mod to the category Ab by Corollary10.5.41[2] and Corollary 10.5.32[2], noting that P is projective by Proposition I.8.3[1] and hence is aflat module by Corollary 10.5.42[2]. By Yonedas Lemma, 1 is uniquely determined by 1(idP) =z,

    and the proof of the lemma (Exercise I.7.3(a) above) states that 1

    (f) = (P

    f)(z) =z(f). Thus,the inverse homomorphism 1 is a map of the form z (independent ofM).Method 2: By Proposition I.8.2[1] we can choose elements ei P and fi P

    such that for everyx P, x =

    fi(x)ei and fi(x) = 0 for cofinitely many x. Set z =

    fi ei. We have z = id

    because z((u m) ) = (P u m)(z) =

    fi u(ei) m =

    fi u(ei) m = u m [note:

    u(x) = u(

    fi(x)ei) =

    fi(x) u(ei) as u is an R-module homomorphism]; we also have z = idbecause (z(f)) =[(P

    f)(z)] =(

    fi f(ei)) =

    fi f(ei) =f [note: f(x) =f(

    fi(x)ei) =fi(x) f(ei) asfis also an R-module homomorphism]. Thus, the inverse

    1 is a map of the formz(independent ofM).

    8.8: SincePis finitely presented, we can form the obvious exact sequence F1 F0 P0 withF0andF1 free of finite rank (the generators and relators, respectively). By Theorem 10.5.33[2], HomR(, D)is a left exact contravariant functor, and so we obtain an exact sequence 0 P F0 F

    1 of right

    R-modules [notation: M = HomR(M, R)]. Since Pis a flat module, we can tensor this exact sequencewith P to obtain the exact sequence 0 P R P F

    0 R P F

    1 R P. Since Fi [i = 0, 1] is

    free, it is necessarily projective (by Lemma I.7.2[1]) and hence Fi RP =HomR(Fi, P) by Proposition

    I.8.3[1]. SinceF1 is contained in the quotient ofF0 which gives P, HomR(F1, P) = 0 and thus we havethe isomorphism P RP =HomR(F0, P).Now HomR(, P) as a functor on the original presentation sequence gives rise to the exact sequence 0 HomR(P, P) HomR(F0, P) HomR(F1, P) = 0, and thus we have the isomorphism HomR(P, P) =HomR(F0, P).

    Since the discovered isomorphism P RP =HomR(P, P) is surjective, P is a projective R-module by

    Exercise I.8.6 above.

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    2 Chapter II: The Homology of a Group

    2.1: For San arbitrary G-set,ZS=

    Z[G/Gs]. By the Orbit-Stabilizer Theorem we have a bijectionbetween Gs and G/Gs. When passing to the quotient for the group of co-invariants, the subset Gs Sis sent to the element Gs S/G sinces is identified with gsin S/G (giving trivialG-action). Therefore(ZS)G=(Z[G/Gs])G

    = (Z[G/Gs])G= Z[Gs]

    = Z[S/G].

    2.2: A weaker hypothesis LetXbe an arbitraryG-complex without inversions suffices. ThenC(X)=Z[Xi] is a direct sum of permutation modules where Xi is the basis set of i-cells, so by the previous

    exercise,C(X)G =

    (Z[Xi])G =

    Z[Xi/G] = C(X/G) which has a Z-basis with one basis elementfor each G-orbit of cells ofX.

    2.3(a): With theG-module Mand normal subgroup H G,MH=M/IMwhereI is the augmentationideal ofZH. The inducedG/H-action is given by gH(m + IM) =gm + IM. The properties of an actionare obviously satisfied, and it is well defined because ifg2 is another coset representative ofg1H, theng2= g1handg2m+IM =g1hm+IM =g1hmg1m+g1m+IM= (h1)g1m+g1m+IM=g1m+IM.

    2.3(b): We can form the group homomorphism : MG (MH)G/Husing part(a) by m m + IM,

    which is well-defined because m = gm gm + IM = gH(m + IM) = m + IM; it is a homomor-phism since(m1 m2) =(m1m2) =m1m2+ IM= (m1+ IM)(m2+ IM) = (m1+ IM)(m2+ IM) =(m1)(m2). For the inverse we use m + IM m which is well-defined since given the equivalentelementsm + IM andgm + (h 1)m + IM in (MH)G/H,

    (gm + (h 1)m + IM) =gm+hm1m =m+mm =m = (m + IM); it is a homomorphism be-cause (m1+ IM m2+ IM) = (m1m2+ IM) =m1m2 =m1 m2 = (m1+ IM)(m2+ IM). Thus,we have the isomorphism MG=(MH)G/H.

    2.3(c): LetZ[G/H] ZGM be aG/H-module, whereZ[G/H] is the obvious (G/H,G)-bimodule whichforms the tensor product and gives it the desired module structure. The map Z[G/H] MMH givenby (a, m) am is clearly G-balanced, and so by the universal property of tensor products (Theorem10.4.10[2]) there exists the group homomorpism : Z[G/H] ZGMMH given by a mam, andit is clearly a G/H-module homomorphism. There is a well-defined map : MH Z[G/H]ZG Mdefined as m1 mbecause of the identity 1H hm= 1Hh m= 1H m, and it is a G/H-modulehomomorphism because(gH m) =(gm) = 1H gm = H g m= gH m= gH1H m= gH (m)[noting that gH =H g since H is normal, and G/Hacts on MH by part(a) above]. Since and areinverses of each other, they are isomorphisms and we obtain MH= Z[G/H] ZGM.

    3.1: Letg1, . . . gn G be pairwise-commutative elements and considerz =

    (1)sgn[g(1)| |g(n)]Cn(G), where ranges over all permutations of{1, . . . , n}. The sign of a permutation is defined here tobe the number of swaps between adjacent integers to bring the permuted set back to the identity. Lookingat the boundary z where=

    (1)idi, a particulardj withj = 0, nwill provide elements in Cn1(G)

    of the form [ |gkgk | ]. Each of these appears twice becausegkgk =gkgk, but the paired elementswill have opposite signs and hence will cancel each other. For j = 0, n we have elements of the form[gi| ] with one gk missing from each, and each of these elements also appears twice because d0 will

    take off gk from the beginning of some element while dn will take off gk from the end of some otherelement. The paired elements differ by the sign (1)n due to the boundary map, and they also differby the sign (1)n1 due to the permutation which takes the first slot and sends it to the last slot; since(1)n(1)n1 = (1)2n1 =1, these paired elements will also cancel each other. Thus, z = 0 and zis a cycle in Cn(G).

    3.2: Suppose Z admits a projective resolution of finite length over ZGwhereG = Zn. Then i0| HiG=0 i > i0, and we make note that HiG is independent of the choice of resolution (see Section II.1[1]).Yet by an earlier calculation II.3.1[1] (using an infinite resolution), HiG= Zn for all positive odd integersi. Thus we have arrived at a contradiction.

    3.3: If G has torsion, say Zn G, and Z admits a projective resolution of finite length over ZG,

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    and hence Rab = Ker1 is always in the chain complex for Y. IfY is a K(G, 1) then Y is acyclic and(*) is exact by Proposition I.4.2[1], so C2(Y) =Rab and hence Rab is a free ZG-module.Now suppose Y is not acyclic, so that Y is not a K(G, 1). Then H2Y is nontrivial (since HiY = 0 fori >2 by (*) andH1Y= 0 by the simply-connected property ofY) which implies that the boundary map2 is not injective, and by exactness we can refer to this non-injective map as C2(Y) Ker1 = Rab.Therefore, there exists a nontrivial ZG-relation amongst the words ri in Ker2, so {[ri]} is not ZG-

    independent in Rab and hence does not generate Rab freely.

    5.2(c): Let G = S; r be an arbitrary one-relator group and write r = un F = F(S), wheren 1 is maximal. By a result of Lyndon-Schupp, the image t ofu in Ghas order exactly n, and we letC =t = Zn. Ifn > 1 then the relation module Rab is not freely generated by r mod[R, R] since thisgenerator is fixed by C, but a result of Lyndon shows that no other relations hold (i.e. the projectionZ[G/C] Rab is an isomorphism).Let Y be a bouquet of circles indexed by S, and let Y be the connected regular covering space of Ycorresponding to the normal subgroup R of F = 1Y. Choosing a basepoint v Y lying over thevertex of Y, we identify G with the group of covering transformations of Y; as explained in [1] onpg15, Y is a (1-dimensional) free G-complex. Since u ends at tv, the lifting r is the composite path

    tv

    tu

    vu

    t2v tn1v

    tn1u

    Thus the map S1 Ycorresponding to r is compatible with the action ofC, whereCacts on S1 as agroup of rotations (i.e. tk is multiplication bye2ik/n). Consider the 2-complexXobtained by attaching2-cells to Y along the loops gr, where g ranges over a set of representatives for the cosets G/C. Forx X (2-cell), lift it to its preimagex0 under the characteristic map f. Projectx0 radially ontoS1 = B2 from the origin 0 B 2 associated to. Denote this radial projection by the functionp, andletL = x0p(x0) be the relative distance ofx0 away from 0. Consideringf(p(x0))

    Y and applying

    the G-action from Y to obtain g f(p(x0)) , define gx to be the point whose preimage under f

    lies on the radial line with endpoints {f1 (g f(p(x0))), 0} and satisfies the same relative distance [i.e.gx = f(L(g p(x0))) X];

    is the closed cell which contains g p(x) and g p(y), where y is a point

    in which does not lie on the radial line that contains x (this ensures continuity of the action). ThisG-action makesX aG-complex since the permutations of 2-cells are determined by the permutations oftheir boundary loops. Thus, if is the 2-cell attached along rthen onlyCG will fix , sinceti simplyrotates the loop (i.e. G =C). Let = (

    s S

    1) re2 be the standard 2-complex associated to the

    presentation ofG; its universal cover is = Y

    gG g whereg is attached along g r, and C2() = ZG.

    ThusXis the quotient of by identifyingg withgti for alli (for eachg), andC2(X) = Z[G/C]=Rab.Ifn= 1 then C ={1} and X =. Lyndons theorem about one-relator groups says thatRab is freelygenerated by the image of r, provided r is not a power, which is equivalent to being a K(G, 1) bypart(b) above, and hence equivalent to X being contractible (Xis the Cayley complex associated to thepresentation ofG).

    5.3(a): With G = F/R and following Kenneth Browns proof of Theorem II.5.3, let F = F(S), let

    Ybe a bouquet of circles indexed by S, and let Y be the connected regular covering space of Y cor-responding to the normal subgroup R ofF = 1Y. Choosing a basepoint v Y lying over the vertexof Y, we identify G with the group of covering transformations ofY. For any f F we regard f asa combinatorial path in the CW-complex Y and we denote by f the lifting of f to Y starting at v.This path f ends at the vertex fv, where f is the image off in G. Define the functiond : F C1Yby letting df be the sum of the oriented 1-cells which occur in f. Since the lifting off1f2 is the path

    vf1

    f1vf1f2 f1f2v, we have d(f1f2) =df1+ f1df2 for all f1, f2 F. Thus, if we regard the G-module

    C1Y as an F-module via the canonical homomorphism q : F G, then d is a derivation [since theF-action is given by restriction of scalars: f1 df2= f1df2= q(f1)df2].

    5.3(b): For any free group F we can apply part(a) above with R = {1} to get the desired deriva-tion d : F whereG= F /1 =Fand =C1Y = ZF(S) which is the free module with basis (ds)sS

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    [note: ds1 =s1ds].The above note is a result ofd(1) =d(1 1) =d(1) + 1d(1) = 2d(1) d(1) = 0.We write the total free derivative df offas the sum df =

    sS(f/s)ds, where f/s ZF is the

    partial derivative offwith respect to s [the coefficient ofds when df is expressed in terms of the basis(ds)].It is immediate that /s : F ZF is a derivation because d is a derivation: f f d(f f) =

    df+ f df = [(f/s) + f(f/s)]ds(f/s) + f(f/s). Fort Swe have dt= (t/s)ds=s=t0ds + 1dt, and so t/s= s,t.

    Example: S= {s, t} (ts1ts2)/s = ts1/s + ts1(ts2/s) =[t/s + t(s1/s)] + ts1[t/s + t(ss/s)] =

    0 + t[s1(s/s)] + ts10 + ts1t[s/s + s(s/s)] =ts11 + ts1t[1 + s1] = ts1ts + ts1t ts1

    5.3(c): Consider any free group F =F(S) and derivation d : FM whereMis an F-module. By therepresentationf=s11 s

    1n and the definition of a derivationd(gh) =dg + gdh, we have the equation

    df=

    sSwsdsthrough trivial induction onn, wherews= f/sby definition of the partial derivative.

    5.3(d): Consider in the exact sequence 0 Rab

    ZG(S) ZG Z 0 of G-modules, whereZG(S) is free with basis (es)sS andes = s 1 [bar denotes image in G], and consider : R ZG(S)

    given byr sS(r/s)es where (r/s) is the image ofr/sunder the canonical mapZF ZG.In order to show that is induced by we must verify exactness of the above sequence, and so we startby calculating the partial derivatives of the representation r = sb11 s

    bnn Rab (wheresi =sj=i). Since

    sb/s =b1

    j=0 sj forb >0 and sb/s =

    bj=1 s

    j forb 0

    sb11 sbi1i1

    bij=1 s

    ji bi < 0

    Injectivity of |Rab follows immediately from the freeness ofZG(S) and the fact that any nontrivial r

    has some nontrivial bi (hence r/si = 0). It suffices to show that(r) =

    (r/s)(s 1) = 0 for

    r Rab. For a particulari, (r/si)(si 1) =sb11 s

    bi1i1 ( s

    1ji s

    ji ) =s

    b11 s

    bi1i1 (s

    bii 1),

    and (r/si)(si 1) + (r/si+1)(si+1 1) =sb11 sbi

    1i1 + sb11 sbii sbi+1i+1 [suppressing the ]. Thus,(r/s)(s 1) =1 + 0 + + 0 + sb11 s

    bnn =1 + r= 1 + 1 = 0, and exactness is satisfied.

    IfR is the normal closure of a subset T F, then the projection ZG(T) Rab given by g et g [t]and the above exact sequence provides us with a partial free resolution:

    ZG(T) 2

    ZG(S) 1 ZG Z 0

    Rab

    where the matrix of2 is the Jacobian matrix (t/s)tT,sS.

    5.4: [[Sketch of explicit formulas for Hopfs isomorphism Kenneth Brown]]

    LetG = F /Rand use the same notation as presented in Exercise 5.3(a) above. Consider the chain mapin dimensions2 from the bar resolution B to the partial resolution given in the diagram

    B2

    2

    2 B1

    1

    1 B0

    0

    Z 0

    ZG(R) C1(Y) C0(Y) Z 0

    We have the identification C1(Y) = ZG(S) = IR where I is the augmentation ideal ofZF (so IR is afree G-module on the images s 1), and ZG(R) is the free G-module with basis (er)rR which mapsonto Rab = H1Y C1(Y). The specific chain map is given by 2 : [g1|g2] g1g2 r(g1, g2) and1 : [g] f(g) 1 and 0 : [ ] 1, where r(g1, g2) R and f(g) F such that f(g) = g G andf(g)f(h) = f(gh)r(g, h). Applying the coinvariants functor, the group homomorphism C2(G) Rabgiven by [g|h]r(g, h)mod[R, R] induces the isomorphism : H2G R [F, F]/[F, R] by passage to

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    subquotients.A specific chain map in the other direction is given by 2 : r h1(r 1) and 1 : s 1 [s] and0 : 1 [ ], where h: B1 B2 is the contracting homotopy h(g [h]) = [g|h]. RegardingC2(G) as an

    F-module via f [g|h] = [f g|h], the map D : F ZF I IR1B1

    hB2 (B2)G C2(G) is a

    derivation such that Ds = [1|s]. Moreover, Df =

    sS[f/s|s], where the symbol [|] is Z-bilinear.

    Then D|R : R C2(G) is a homomorphism (since R acts trivially on C2(G)) which induces 1 by

    passage to subquotients.Let a1, . . . , an, b1, . . . , bn F such that r =

    ni=1[ai, bi] R. The formula

    1(r mod[F, R]) =ni=1{[Ii1|ai] + [Ii1ai|bi] [Ii1aibia

    1i |ai] [Ii1|bi]}, where Ii = [a1, b1] [ai, bi], is proven in the

    universal example where F is the free group on a1, . . . , an, b1, . . . , bn and R is the normal closure ofr.Using the constructed formula for 1 and the product rule for derivations, the desired formula arisesfromDr=

    i Ii1 D[ai, bi] and D[a, b] = [1|a] + [a|b] [aba

    1|a] [aba1b1|b].

    5.5(a): With the presentationG = s1, , sn | r1, , rmwe associate the 2-complex Y = (

    s S1)r1

    e2 rme2 so that1Y =G. By computing the Euler characteristic (Y) two different ways (by The-

    orem 22.2[4]) we obtain the equation

    (1)irkZ(HiY) =

    (1)ici, whererkZ is the rank and ci is thenumber ofi-cells. Then 1 rkZ(Gab) + rkZ(H2Y) = 1 n+m, and so rkZ(H2Y) = m n+r wherer= rkZ(Gab) = dimQ(Q Gab). NowH2Y= Ker2 is a free abelian group (subgroup of cellular 2-chain

    group), and by applying Theorem II.5.2[1] we get a surjection H2Y H2G(from the exact sequence inthe theorem). Thus H2G canbe generated by m n + r elements.

    5.5(b): Since Gab = 0 and the number of generators equals the number of relations (in the finitepresentation), m n + r = m m + 0 = 0. Thus by part(a), H2G can be generated by at most 0elements, and so H2G= 0.

    5.5(c): Given Gab = 0, H2G = Z2 Z2, and n as the number of generators, let m be the numberof relations in the presentation ofG. Then H2G is generated by the two elements (0, 1) and (1, 0), andr= 0, so by part(a), 2 m n + 0 m n + 2. Therefore, any n-generator presentation must involveat least n + 2 relations.

    5.6(a): From the group extension 1 N G

    Q 1 we have G/N =Q, and from Hopfs formulawe have H2G = R [F, F]/[F, R] and H2Q = S[F, F]/[F, S], where G = F/R and Q = F/S with

    R S F (soN=S/R). (H1N)Q=(Nab)Q = Nab/{(q1) n[N, N]}= Nab/{gng1n1[N, N]}=

    (N/[N, N])/([G, N]/[N, N]) = N/[G, N], where the Q-action on Nab is induced by the conjugationaction of G on N, and the latter isomorphism follows from the Third Isomorphism Theorem. Now[G/N, G/N] = {g1N g2N g

    11 Ng

    12 N} = {g1g2g

    11 g

    12 N} = [G, G]N/N, so we have H1Q

    = Qab =(G/N)/[G/N, G/N]=G/(N[G, G]), where the latter isomorphism follows from the Third IsomorphismTheorem.Thus, the desired 5-term exact sequence is obtained by showing the exactness of the sequence

    R [F, F]/[F, R] S [F, F]/[F, S]

    N/[G, N]

    G/[G, G]

    G/(N[G, G]) 0

    where and are induced by the injection and surjection of the group extension. From : g [G, G]

    g[G, G]N = g N[G, G] we have Ker= N/[G, G] and Im= G/(N[G, G]). From : n[G, N] n[G, G]we have Ker = N [G, G]/[G, N] and Im = N/[G, G] = Ker. As deduced above, N/[G, N] =(S/R)/[F/R, S/R] = S/(R[F, S]) and so from : s[F, S] s[F, S]R = sR[F, S] we have Ker =R [F, F]/[F, S] and Im = S[F, F]/(R[F, S]) = N [G, G]/[G, N] = Ker. Finally, from :r[F, R]r[F, S] we have Im= R [F, F]/[F, S] = Ker.

    5.6(b): Applying part(a) to the group extension 1 R F G 1 we obtain the exact se-quence

    H2F

    =

    H2G

    (H1R)G

    =

    H1F

    =

    H1G

    =

    0

    0 R/[F, R] F/[F, F] G/[G, G]

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    where the first vertical isomorphism follows from Example II.4.1[1] and the second vertical isomorphismarises in the solution to part(a). By the First Isomorphism Theorem and exactness of the sequence,H2G=H2G/Ker=Im=Ker= R [F, F]/[F, R].

    5.7(a): S3 is a closed orientable 3-manifold, and it has a group structure under quaternion multpli-cation (S3 < H as the elements of norm 1). The finite subgroupG ofS3 provides a multiplication action,

    and it is free because the only solution in H to the equation gx = x for nontrivial x is g = 1. Froma result in the solution to Exercise I.4.2, the G-action is a properly discontinuous action and so thequotient map : S3 S3/G is a regular covering space.Following a proof by William Thurston,S3/G is Hausdorff: considering two points , ofS3 in distinctorbits, form respective neighborhoods U and U such that they are disjoint and neither neighborhoodcontains any translates of or (this can be done by the Hausdorff property ofS3). Taking the unionK of these two neighborhoods, I nt(K

    g=1 gK) yields a neighborhood of and a neighborhood of

    which project to disjoint neighborhoods ofG and G inS3/G[note: we needed to refine Kbecause ifU intersects g U , then after projecting to the orbit space, (gU) =(U ) intersects (U)].S3/Gis a closed connected 3-manifold since it has S3 as a covering space (Theorem 26.5[6] provides thecompactness, Theorem 23.5[6] provides the connectedness, and the property of evenly-covered neighbor-hoods provides the nonboundary and manifold structure).

    Since the actions g : S3

    S3

    are fixed-point free, we have deg(g) = (1)3+1

    = 1 by Theorem 21.4[4]and hence Gacts by orientation-preserving homeomorphisms [note: orientation is in terms of local ori-entationsx H3(S

    3, S3 x)=H3(S3)= Z, as defined in [3] on pg234]. Local orientations Gx = (x)

    ofS3/G= (S3) are given by the images x(x) = gx(gx) under the isomorphisms of local homologygroups x :H3(S

    3, S3 x) H3((S3), (S3) (x)) which arise from the Excision Theorem and the

    local-homeomorphism property of covering spaces; the local consistency condition follows in the samerespect (where a ballB containingGx and Gy has as preimage under a union of balls, each of which ishomeomorphic toB, and such a homeomorphic ball containing g1xandg2yprovides the local consistencycondition for S3). Therefore, S3/Gis orientable.Since S3 is simply-connected, 1S

    3 = 0 and so G=1(S3/G)/1(S

    3)=1(S3/G). Applying Poincare

    Duality and the Universal Coefficient Theorem we obtain H2(S3/G)=H1(S3/G)=Hom(H1(S3/G), Z)=Hom(Gab, Z) = 0 [noting that G is finite]. A theorem of Hopf (Theorem II.5.2[1]) gives us an exact se-quence which includes the surjection H2(S3/G) H2G 0, hence H2G= 0.

    5.7(b): The binary icosahedral group G (of order 120) is the preimage in S3 which maps onto thealternating group A5 under S

    3 SO(3) [up to isomorphism with the group of icosahedral-rotationalsymmetries] as explained in [3] on pg75. By part(a), H2G= 0. Consider the group extension 1 KG A5 1 where K is the central kernel of order 2 (corresponding to G mapping onto A5). Theassociated 5-term exact sequence becomes 0 H2(A5) (H1K)A5 0 because H1G = Gab = 0,and thus we obtain the isomorphism H2(A5) = (H1K)A5 . The A5-action on H1K = Kab = K = Z2is induced by the conjugation G-action on K =Z2 which is the trivial action (since K Z(G)), andthereforeH2(A5)=H1K= Z2.

    5.7(c): Consider the abstract group G = x,y,z ; x2 = y3 = z5 = xyz which is a finite presenta-tion with the same number of generators as relations. We show that G is perfect (G= [G, G]) so that

    H1G = Gab = 0 and H2G = 0 by Exercise 5.5(b) above. Now G/[G, G] is an abelian group with therelations 2x = 3y = 5z = x+ y + z. From this we see that x = y + z, so we need not look at x.Subsequently, 2y + 2z = 3y y= 2z and so we need not look at y. Finally, 5z = 3(2z) = 6z z = 0and so all generators vanish, i.e. G/[G, G] = 0.Alternatively, the commutator subgroup is [G, G] G, and to prove the opposite inclusion it suffices toshow thatx,y, z all lie in [G, G] G.

    x2 =xyz

    x= yz x2 =yzyz= y3 z1yzy1 =z2y

    z2[z1, y] =yThe two overbraced equations allow us to finish by showing that z G.

    yzyz= y3 y2 =zyz [y, z] =z1yz1 y = z[y, z]zWe then have z 5 =xyz z 3 =xyz1 =yz z[y, z]z z1 =z[y, z]z z2[y, z] z 2 = [y, z]z3[y, z] =

    z3z3 [y, z]z3[y, z] z1 =g[y, z] z = [y, z]1g1 G withg = z3[y, z]z3 G by normality of16

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    the commutator subgroup.

    With the cyclic subgroup C=xyz we have G/C=A5 and hence the group extension 1 CG A5 1 [A5 = x,y,z ; x2 = y3 = z5 = xy z = 1 of order 60]. The associated 5-term exact sequencenow yields the isomorphism H2(A5)=H1C= C, noting the trivialA5-action (sinceCis generated by acentral element) and noting the cyclicity Cab=C. Thus, by part(b) we deduce that |C|= 2 and hence

    |G|= 2 60 = 120.In fact, G= SL2(F5) is the binary icosahedral group!

    6.1: Given N G, let Fbe a projective resolution ofZ overZG and consider the complex FN. Since aprojective ZG-module is also projective as a ZH-module for any subgroupHG (by Exercise I.8.2), Fis a projective resolution ofZ overZN and so H(FN) =HN. By Exercise II.2.3(a),FN is a complexofG/N-modules and so H(FN) inherits aG/N-action, with FNFNgiven by x(gN)x= gx. ForCorollary II.6.3 we have the conjugation action : NNgiven byn gng1 (forg G), and we havethe augmentation-preserving N-chain map :FFgiven by x gx[it commutes with the boundaryoperatorofF sinceis equivariant, and it satisfies the condition(nx) =gnx= gng1gx = (n)(x)].By Proposition II.6.2[1], if is conjugation by g N then H() is the identity (hence trivial action),and so :FNFNis given by

    (x) = (gN)x= gx which agrees with the above map.

    6.2: For any finite set A let (A) be the group of permutations ofA. For|A| |B|, choose an injectioni: A B and consider the injection (A)(B) obtained by extending a permutation on A to be theidentity on B iA. In order to show that the induced mapH(A)H(B) is independent of thechoice ofi, it suffices to show that any two injectionsi1and i2give conjugate maps i1, i2: (A)(B),because the conjugation map (B) (B) induces the identity map on homology H(B) H(B)by Proposition II.6.2[1]. Let be the permutation which takes i1(a) to i2(a) for all a A and is anarbitrary permutation (B i1A) (B i2A). Then for a permutation i1() = (B), the per-mutation 1 is equal to i2(). Thus i1and i2are conjugates of each other by , and the result follows.

    6.3(a): Given the homomorphism : G G, the n-tuples in Cn(G) are sent to the n-tuples in Cn(G)

    coordinate-wise via , where [gh] [(gh)] = [(g)(h)]. Thus H1() maps g to (g), where g de-notes the homology class of the cycle [g]. We also have the explicit isomorphism H1G Gabgiven by g

    g mod[G, G]. It is immediate that we have the commutative diagram H1G

    H1()

    = Gab

    H1G = Gab

    with :g mod[G, G](g)mod[G, G], which is precisely the map obtained from by passage to thequotient. Thus, the isomorphism H1( )=( )ab is natural.

    6.3(b): Suppose G = F/R and G = F/R with F = F(S) and F = F(S) free, and suppose : G G lifts to : F F. Let Y and Y be associated to the presentation of G as in ExerciseII.5.3(a), and similarly forY and Y withG. Now yields a mapY Y that sends the combinatorialpath s = S1s to the combinatorial path (s). ***Incomplete***

    7.1: Consider G = G1 A G2 with k : A Gk not necessarily injective, and let G1 = 1(G1),G2= 2(G2), A= 11(A) =22(A) be the images ofG1, G2, A in G[wherek :Gk G arises fromthe amalgamation diagram for G]. Form the amalgam H= G1A G2 and the commutative diagram:

    A

    i1

    i2 G2

    j2

    2

    G1

    1

    j1H

    Gwherek is the natural inclusion and ik is the obvious injection [subsequently, we have 1i1= 2i2 and

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    0

    =

    Z2

    =

    Z4 Z2

    =

    H2n1(Z4Z2Z2= Z4)= Z4

    0

    =

    0

    =

    0 Z2 Z4 Z2 H2n1(SL2(Z))(2) 0 0

    Then by the Five-Lemma, is an isomorphism and soH2n1(SL2(Z))(2)= Z4.

    Thus, H2n1(SL2(Z))= Z4 Z3= Z12.

    Hi(SL2(Z))=

    Z i= 0

    Z12 iodd

    0 ieven

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    3 Chapter III: Homology and Cohomology

    with Coefficients

    0.1: LetFbe a flat ZG-module andMaG-module which is Z-torsion-free (ie. Z-flat), and consider thetensor productF Mwith diagonal G-action. Since (F M) G = (F M )G = FG(M ),it suffices to show that X=F

    G(M ) is an exact functor so that F M isZG-flat (by Corollary

    10.5.41[2]). But by the same corollary M is Z-exact (and a G-module) andFG is ZG-exact, soXis exact and the result follows.

    0.2: Let F be a projective ZG-module and M a Z-free G-module, and consider the tensor productF Mwith diagonalG-action. Since HomG(F M, )=Hom(F M, )G=Hom(F, Hom(M, ))G=HomG(F, Hom(M, )) where the second isomorphism is adjoint associativity (Theorem 10.5.43[2]), itsuffices to show that X = HomG(F, Hom(M, )) is an exact functor so that FM is ZG-projective(by Corollary 10.5.32[2]). But by the same corollary Hom(M, ) is Z-exact (and a G-module) andHomG(F, ) isZG-exact, so Xis exact and the result follows.

    1.1(a): For a finite group G and a G-module Mwe have the norm map N :MG MG induced from

    the map MMwhich is multiplication by the norm element N = gG g. Noting that N m= |G|mfor both m MG and m MG, we see that |G| KerN = 0 (as N m = N m/ = 0 by definition ofkernel) and |G| CokerN= 0 (as CokerN=MG/NM andNm modN M= 0).

    1.1(b): Suppose M is an induced module (M = ZG A) where A is an abelian group and G actsbyg (r a) = gr a. ThenMG = (ZG)G A =Z A and MG = (ZG)G A =Z N A, whereN is the norm element. The norm map N : MG MG is now given by z a zN a (for z Z)which is clearly a bijection. It is an isomorphism because N[z1 a1+ z2 a2] =N z1 a1+ Nz2 a2=z1N a1+ z2N a2= N[z1 a1] + N[z2 a2].

    1.1(c): Let M be a projective ZG-module; it is a direct summand of a free module F=

    i ZG. Byapplication of part(b) above withA =

    i Z we see thatNis an isomorphism for FbecauseF=

    i ZG=

    i(ZG Z) = ZG (

    iZ). Now (M N)G= Z ZG (M N)=(Z ZG M) (Z ZG N)=MG NG,

    and (M N)G

    =MG

    NG

    under the coordinate-wiseG-action (ofF) since g (m, n) = (g m, g n) =(m, n) g m= m , g n= n. ThusMG NG =M

    G NG, and since the norm map is bilinear wehave MG =M

    G.

    1.2: Using the standard cochain complex, an element of C1(G, M) is a function f : G M, andunder the coboundary map it is sent to (f)(g, h) = g f(h) f(gh) +f(g). The kernel of this mapconsists of functions which satisfy f(gh) = f(g) +g f(h), and these are derivations, so Z1(G, M) =Der(G, M). Since an element ofC0(G, M) is simply m M, and under the coboundary map it is sentto (m)(g) = g m m, which is a principal derivation, we have B1(G, M) = PDer(G, M). Thus,H1(G, M)=Der(G, M)/PDer(G, M).IfG acts trivially on Mthen PDer(G, M) = 0 and Der(G, M) = Hom(G, M), so H1(G, M)=Hom(G, M) =Hom(Gab, M) = Hom(H1G, M), where the second-to-last equality comes from the fact that any grouphomomorphism from G to an abelian group factors through the commutator subgroup [ G, G] by Propo-sition 5.4.7[2]. In particular, H1(G) = 0 for any finite group G.

    1.3: Let A be an abelian group with trivial G-action, and let F Z be a projective resolutionof Z over ZG. Then F G A = (F A)G = Z G (F A), and since the diagonal G-action onF A is simply the left G-action on F (since G acts trivially on A), we can apply tensor associa-tivity (Theorem 10.4.14[2]) to obtain ZG (F A) = (Z G F) A = FG A. Thus there is auniversal coefficient sequence 0 Hn(G) A Hn(G, A) Tor

    Z1 (Hn1(G), A) 0 by Proposition

    I.0.8[1]. Also,HomG(F, A) = Hom(F, A)G = Hom(FG, A), where the last isomorphism arises because

    (gu)(m) = g u(g1m) = u(g1m) and so we must have g1m = m F for gu = u. Thus there isalso a universal coefficient sequence 0 Ext1Z(Hn1(G), A) H

    n(G, A) Hom(Hn(G), A) 0 byProposition I.0.8[1].

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    1.4(a): Let f : C C be a weak equivalence between arbitrary complexes, and let Q be a non-negative cochain complex of injectives. Then the map HomR(f, Q) :HomR(C, Q) HomR(C

    , Q) is aweak equivalence.To prove this, note that the mapping cone C =C C offis acyclic by Proposition I.0.6[1], where(C)p = C

    p1is the 1-fold suspension ofC. The mapping cone ofHomR(f, Q) is HomR(C

    , Q) because

    HomR(C

    , Q)n HomR(C, Q)n = qHomR(Cq , Qq+n) qHomR(Cq, Qq+n1) =qHomR(C

    q Cq+1, Qq+n) =HomR(C

    1C, Q)n = HomR(C, Q)n

    noting that C 1C= (Cq Cq+1) = (Cq1 Cq) = C

    C =C. Thus it suffices to show thatHomR(C

    , Q) is acyclic (by Proposition I.0.6[1]), i.e., that Hn(HomR(C, Q)) [C, Q]n = 0 n Z.

    By the uniqueness part of the result of Exercise I.7.4, [C, Q]n [nC, Q] is indeed 0, since any map

    on Q is zero in negative dimensions (so all extensions off of that zero map are homotopy equivalent).

    1.4(b): Let : F Z be a projective resolution and let : M Q be an injective resolution.By part(a) above and noting that is a weak equivalence (regarded as a chain map with Mconcentratedin dimension 0), we have a weak equivalence HomR(F, Q) HomR(Z, Q). Similarly, by Theorem I.8.5[1]we have a weak equivalence HomR(F, M) HomR(F, Q).In particular,H(G, M) = H(QG) because H(G, M) = H(HomG(F, M)) and H

    (HomG(F, M)) =H(HomG(Z, Q)) =H(Hom(Z, Q)G), noting that HomZ(Z, Q)=Q.

    2.1: Given projective resolutions F M and P N of arbitrary G-modules M and N, there is

    an isomorphism of graded modules : FGP =PGF given by f p(1)deg

    fdegpp f, wherewe consider diagonal G-action onFi Pj =Pj Fi [generally, degx= n for x Cn]. If we show that (a degree 0 map) is a chain map, then it is a homotopy equivalence (hence a weak equivalence) and soTorG(M, N) =H(FGP)=H(PGF) = Tor

    G(N, M). Denote by d and d

    the boundary operatorsof F and P, respectively, and denote by D and D the boundary operators of FG P and P G F,respectively. Then

    D(f p) =D[(1)degfdegpp f] = (1)degfdegp(dp f+ (1)degpp df)= (1)degfdegpdp f+ (1)degp(degf+1)p df

    andD(f p) = [dfp + (1)degff dp]= (1)degp(degf1)p df+ (1)degf+degf(degp1)dp f

    = (1)degp(degf+1)p df+ (1)degfdegpdp f

    Thus D = D and so is a chain map.(This simultaneously provides a solution to Exercise I.0.5)

    3.1: Let P be a projective R-module, and let C be a short exact sequence of S-modules which canbe regarded as R-modules via restriction of scalars. SinceP is projective, HomR(P, C) is a short exact

    sequence, and so it suffices to show that the isomorphism of functors HomS(SRP, ) =HomR(P, )

    is natural [because then HomS(SR P,C) is a short exact sequence which implies that SRP is aprojectiveS-module]. Given a module homomorphism : MN, we must check commutativity of the

    diagramHomS(SRP, M)

    1

    HomS(SRP, N)

    2

    HomR(P, M)

    HomR(P, N)

    where and are given by f f, and i (i = 1, 2) is given by f fi under the universalmapping property with i: P SRP , i(p) = 1 p. Now 2[(F)] =2[ F] = ( F) i and[1(F)] =[F i] = (F i) = ( F) i. Therefore, 2= 1 and the result follows:Extension of scalars takes projectiveR-modules to projectiveS-modules.

    3.2: LetQ be an injectiveR-module, and let C be a short exact sequence ofS-modules which can be re-garded asR-modules via restriction of scalars. Since Qis injective, HomR(C, Q) is a short exact sequence,

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    and so it suffices to show that the isomorphism of functors HomS(, HomR(S, Q)) = HomR(, Q) is

    natural [because then HomS(C, HomR(S, Q)) is a short exact sequence which implies that HomR(S, Q)is an injective S-module]. Given a module homomorphism : M N, we must check commutativityof the diagram

    HomS(N, HomR(S, Q))

    1

    HomS(M, HomR(S, Q))

    2

    HomR(N, Q)

    HomR(M, Q)

    where and are given by f f, and i (i = 1, 2) is given by f funder the universalmapping property with : HomR(S, Q) Q , (f) =f(1). Now 2[(F)] =2[F ] = (F )and [1(F)] =[ F] = ( F) = (F ). Therefore, 2= 1 and the result follows:Co-extension of scalars takes injectiveR-modules to injectiveS-modules.

    3.3: GivenSwhich is flat as a right R-module, let Q be an injective S-module, let C be a short exactsequence of S-modules, and consider HomR(C, Q) where C and Q are regarded as R-modules via re-striction of scalars. Since S is R-flat, SR C is a short exact sequence, and since Q is S-injective,HomS(SR C, Q) is a short exact sequence. It suffices to show that the isomorphism of functors

    HomS(SR , Q)

    =

    HomR(, Q) is natural [because then HomR(C, Q) is a short exact sequence whichimplies thatQ is an injective R-module]. Given a module homomorphism : M N, we must checkcommutativity of the diagram

    HomS(SRN, Q)

    1

    HomS(SRM, Q)

    2

    HomR(N, Q)

    HomR(N, Q)

    where is given byf f (SR ), is given byf f ,1 is given by f f iNfor the naturalmap iN :NSRN, and 2 is given similarly by f f iM. Now 2[(F)] =2[F (SR)] =(F (SR )) iM =F ((SR ) iM) and [1(F)] =[F iN] = (F iN) = F (iN ). Also,iN[(m)] = 1 (m) = (SR)(1 m) = (SR)[iM(m)]. Therefore, 2 = 1 and the resultfollows:

    Restriction of scalars takes injectiveS-modules to injectiveR-modules ifSis a flat rightR-module.

    3.4: GivenS which is projective as a left R-module, let P be a projective S-module, let C be a shortexact sequence ofS-modules, and consider HomR(P, C) where C and Pare regarded as R-modules viarestriction of scalars. Since S is R-projective, HomR(S,C) is a short exact sequence, and since P isS-projective, HomS(P, HomR(S,C)) is a short exact sequence. It suffices to show that the isomorphism

    of functors HomS(P, HomR(S, )) =HomR(P, ) is natural [because then HomR(P, C) is a short exact

    sequence which implies that P is a projective R-module]. Given a module homomorphism : MN,we must check commutativity of the diagram

    HomS(P, HomR(S, M))

    1

    HomS(P, HomR(S, N))

    2

    HomR(P, M) HomR(P, N)

    where is given by f f for (g) = g (g : S M), is given by f f, 1 is givenby f M funder the universal mapping property with : HomR(P, M) M , M(f) = f(1),and 2 is given similarly by f N f. Now 2[(F)] = 2[ F] = N ( F) = (N ) Fand [1(F)] =[M F] = (M F) = ( M) F. Also, N[(f)] =N[ f] = ( f)(1) =[f(1)] =[M(f)]. Therefore, 2= 1 and the result follows:Restriction of scalars takes projectiveS-modules to projectiveR-modules ifSis a projective leftR-module.

    4.1: LetR = Z/nZ, and note that the ideals ofR are the ideals Ix = (x)mod(n) forx|n Z by the 4th

    Isomorphism Theorem. It suffices to show that every map : IR extends to a mapR R so thatRis self-injective by Baers Criterion (Proposition III.4.1[1]). GivenIx and (xmod(n)) =r mod(n), wecan write n= yx so that (yx mod(n)) = 0. But (yx mod(n)) =y(xmod(n)) mod(n) =yr mod(n)

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    and thus yr = mn = myx r = xm. Then, since (x mod(n)) = [x mod(n)][m mod(n)] we canextend to the domain R by setting (1 mod(n)) =m mod(n).

    (a): Let A be an abelian group such that nA = 0, and let C A be a cyclic subgroup of order|C| = n. We can regard A as an R-module because xn = x0 R and R acts on A by xi a = ia.As C is a [self-injective] subgroup of A, we have an inclusion C A of an injective R-module intoan R-module. By definition of injective module (pg.782-783 of [5], statement XX.4.I1), every exact

    sequence of modules 0 Q MM 0 splits for injective Q, hence Q is a direct summand ofM.Therefore,Cis a direct summand ofA.

    (b): Given A as above (i.e. an arbitrary abelian group of finite exponent), we regard A as an R-module. Since n is minimal (to annihilate A) there exists an element of order n and hence a cyclicsubgroup C = R of order n in A. If we can show thatA = A A for A of smaller exponent thenby induction on n we have that A is a direct sum of cyclic groups; thus it remains to show that A

    is a direct sum of modules (each isomorphic to R) and is a direct summand of A. An application ofZorns Lemma on the set of direct sums with summands in A isomorphic to R (noting that the set isnonempty because it containsC) provides the maximal elementA; we can use this lemma because anupper bound for any chain would be the direct sum of those elements (the direct sums) in that chain. Aring is Noetherian iff every ideal is finitely generated (by Theorem 15.1.2[2]); thus Z is Noetherian (beinga Principal Ideal Domain). Now R = Zn is also Noetherian, because a quotient of a Noetherian ring by

    an ideal is Noetherian (by Proposition 15.1.1[2]). It is a fact that a ring is Noetherian iff an arbitrarydirect sum of injective modules (over that ring) is injective. Thus A = i R is R-injective, so A is adirect summand ofA, and A is a direct sum of cyclic groups.This result is known as Prufers Theorem for abelian groups.

    5.1: For any H-module M consider the G-module IndGHM =

    gG/HgM where this equality fol-

    lows from Proposition III.5.1[1]. The summandgM is agHg1-module and hencegHg1 is the isotropygroup of this summand in IndGHM. By Proposition III.5.3[1], Ind

    GHM=Ind

    GgHg1gM.

    In particular, by Proposition III.5.6[1] we have the K-isomorphismgEInd

    KKgHg1Res

    gHg1

    KgHg1gM=

    ggEIndKKggH(gg)1Res

    ggH(gg)1

    KggH(gg)1ggM

    Thus theK-module IndK

    KgHg1Res

    gHg1

    KgHg1gMdepends up to isomorphism only on the class ofg E

    in K\G/H.

    5.2(a): For any H-module M and G-module N consider the tensor product NIndGHM which hasthe diagonal G-action. By Proposition III.5.1[1] and the fact that tensor products commute with directsums, we have N IndGHM =N (

    gG/HgM)

    =

    gG/H(N gM) which has N M as a direct

    summand in the underlying abelian group. Treating this as an H-module ResGHNM with a diago-nal action so that H is its isotropy group, we have N IndGHM =Ind

    GH(Res

    GHN M) by Proposition

    III.5.3[1].In particular, forM= Z we have N Z[G/H]=IndGHRes

    GHN.

    5.2(b): For any H-module M and G-module N consider U = Hom(IndGHM, N) which has the di-agonal G-action given by (gu)(m) = g u(g1m). By Proposition III.5.1[1] and the fact that theHom-functor commutes with direct sums/products, we have U=Hom( gM,N)= Hom(gM,N),where the indices on the sum/product symbols are implicitly the coset representatives in G/H. ThusUadmits a direct product decomposition (g : UHom(gM,N)). Using the denotation g (u) = ug,we have [(g g0)(u)](m) = [g(g0u)](m) = g0 ug(g

    10 m) = g0 ug10 g

    (m) = g0 [g10 g(u)](m), and so

    gg0 g10 g, with m IndGHM. This decomposition has U Hom(M, N) as one of the surjections

    in the underlying abelian group. Treating this surjection as an H-module 1 : U Hom(M, ResGHN)

    so that H is its isotropy group, we have Hom(IndGHM, N) =CoindGHHom(M, Res

    GHN) by Proposition

    III.5.8[1].An analogous proof will provide Hom(N, CoindGHM)=Coind

    GHHom(Res

    GHN, M).

    5.3: LetFbe a projective G-module andM a Z-freeG-module, and consider the tensor productF Mwith diagonalG-action. By Corollary III.5.7[1], ZGMis a freeG-module sinceMis free as a Z-module.

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    Since F is projective, it is a direct summand of a free G-module F=

    i ZG, so that F= F K. Asthe tensor product commutes with arbitrary direct sums (Corollary XVI.2.2[5]), F M=

    i(ZG M)

    and hence is a free G-module. Finally, F M= (F K) M = (F M) (K M) and so F M isa direct summand ofF M, hence G-projective.Note that this gives a new proof of the result of Exercise III.0.2above.

    5.4(a): If |G : H| = then there are infinitely many distinct coset representatives in G/H. NowIndGHM =

    gG/HgM by Proposition III.5.1[1], and it has a transitive G-action which permutes the

    summands. Consider an arbitrary [nontrivial] element x =N

    i=1 gimi with all mi = 0 (this also refersto the sum over all coset representatives with cofinitely many mi = 0). We may take the summandentry g1m1 ofx and a summand g

    Mwhich doesnt appear in the representation ofx (i.e. g =gj for1 j N), and there then exists g G such that g g1m1 = g

    m because the G-action is transitiveon the summands. Thus x is not fixed by G (since g x=x), and so (IndGHM)

    G = 0.Note that if the index is finite, then this result does not hold. For instance, take 2Z Z =xwhich

    has index |Z : 2Z| = |Z2| = 2. Then IndZ2ZM = M xM where x is the coset representative ofx(2Z)

    which generates Z2. Sincex2i (m1, xm2) = (x2im1, x(x2im2)), we must have (IndZ2ZM)

    Z M2ZxM2Z.Since x2j+1 (m1, xm2) =x (m1, xm2) = (m2, xm1), we must have m1 = m2 and hence (Ind

    Z2ZM)

    Z ={(m, xm)| m M2Z}. For M = Z4 with 2Z-action defined as x2i m= (1)im, the largest submodule

    on which 2Z acts trivially [so that m= m] isZ2. Thus (IndZ2ZZ4)Z = 0 is our desired example.

    5.4(b): Omitted.

    5.4(c): Assume statement (i), so that there is a finitely generated subgroupG G such that|G :G gH g1|= for allg G(withHG). Using the analogue of Proposition III.5.6[1] for coinduction and

    passing to G-coinvariants, we obtain (ResGGCoindGHM)G =(

    gECoind

    G

    GgHg1ResgHg1

    GgHg1gM)G=

    gE(CoindG

    GgHg1M)G where this latter isomorphism follows from commutativity of the tensor

    product with direct sums (and E is the set of representatives for the double cosets KgH). It is afact that if G is finitely generated and |G : H| = then (CoindGHM)G = 0 for any H-module M[part(b) above], so by statement (i) we have (ResGGCoind

    GHM)G =

    0 = 0 for any H-module M.

    Noting that restriction of scalars gives the action g x = (g)x where is a map G G, we have

    (CoindGHM)G (ResGGCoindGHM)G . Thus (CoindGHM)G = 0 and (i) implies (ii).Assume statement (ii), so that (CoindGHM)G = 0 for all H-modules M. Then in particular (forM= Z)there is only one element of (CoindGHZ)G, and that element must be zero, so (ii) implies (iii).Assume statement (iii), so that the element of (CoindGHZ)G represented by the augmentation map CoindGHZ is zero. We first note that [in general] if n0 = 0 NG for some n0 N then it isalso zero in NG for some finitely generated subgroup G

    G; this is because NG =N/gn n and son0 can be written as a finite Z-linear combination of elements of the form gn n, which implies thatwe can take G to be the subgroup generated by those specific gs. In particular, = 0 (CoindGHZ)Gfor some finitely generated subgroup G ofG. Using the double coset formula (analogue of PropositionIII.5.6[1]) and treating Z and other modules appropriately over specific groups (to ignore restriction),we must have g = 0 (Coind

    G

    GgHg1Z)G for all g G, where g denotes the component of theaugmentation map in the specific summand of coinduction. Now if |G : G gH g1| < then

    CoindG

    GgHg1Z = IndG

    GgHg1Z = Z[G/G gHg1], giving g =gKg g(g) = ( g) 1whereKis the set of coset representatives for the quotient G /G gH g1. Then g g =g g

    G,

    so G = 0 (CoindG

    GgHg1Z)G . Thus we must have |G :G gH g1|= , so (iii) implies (i).

    5.5: Let G be a finite group and let k be a field, and consider the free module kG. We have kG =kG k k= Ind

    G{1}k

    =CoindG{1}k= Homk(kG,k), where the second-to-last equation follows from the ana-logue overk of Proposition III.5.9[1] since|G|< . Then HomkG(, kG)=HomkG(, Homk(kG,k))=Homk(, k) where the last isomorphism follows from the universal property of co-induction. It is a factthat everyk-vector space is an injective k-module [if the vector spaceWwith basis B is a subspace of avector spaceV, then we can extend B to a basis ofVand then V =W UwhereUis the vector spacespanned by the additional basis vectors extended from B]; thus k is injective Homk(, k) is exact HomkG(, kG) is exactkGis self-injective as a kG-module.

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    Hn(G, M) Hn(G1, ResGG1M) H

    n(G2, ResGG2M)

    Hn(A, ResGAM)

    7.1(a): Consider the exact sequence C11 C0

    M 0, where each Ci is H-acyclic. We

    can apply the dimension-shifting technique using the short exact sequences 0 Ker C0 M 0and 0 Keri Ci Keri1 0 to obtain the isomorphism Hn(G, M) = H1(G, Kern2) =Ker{H0(G, Kern1) H0(G, Cn1)}= Ker{(Kern1)G (Cn1)G}. Now consider the diagram be-

    low concerning CG

    (Cn+1)Gn+1

    1

    (Cn)Gn

    (Cn1)G

    (Zn)G

    (Zn1)G

    2

    with Zi = Keri, noting that the composition is exact [right-exactness of ()G on 0 Kern Cn Kern1 0] and Ker2=Hn(G, M). Thus Imn+1= Im= Ker(since1is surjective) and soHn(CG) = Kern/Ker. Noting that (Zn1)G = (Cn)G/Kerby the First Isomorphism Theorem, takethe kernel of2 [denoted K/Ker] and take its preimage under to obtain K in (Cn)G. Since Kmapsto zero under 2= n we have K Kern. We also have Kern K[hence they are equal] becauseby commutativity of the maps Kern maps toK/Kerand hence lies in K. Thus Kern/Ker= Ker2

    and soH(G, M)=H(CG).

    7.1(b): Consider the exact sequence 0 M C0 0 C1

    1 , where each Ci is H-acyclic. Wecan apply the dimension-shifting technique using the short exact sequences 0 MC0 Ker0 0and 0 Keri1 Ci Keri 0 to obtain the isomorphism H

    n(G, M) = H1(G, Kern2) =Coker{H0(G, Cn1) H

    0(G, Kern1)} = Coker{(Cn1)G (Kern1)

    G}. Using a similar approachas in part(a) above, we see that H(G, M)=H(CG).

    7.2: This will reprove Proposition III.2.2[1] on isomorphic functors.Method 1 : Let M be Z-torsion-free, so that M is an exact functor (M is Z-flat). Then

    H(G, M ) is a homological functor because given a short exact sequence of modules C, M Cis a short exact sequence and FG(M C) is a short exact sequence of chain complexes (F is projec-tive, hence flat), so the corresponding long exact homology sequence gives us the desired property (byLemma 24.1[4] and Theorem 24.2[4]). Similarly, TorG(M, ) = H(F G) is a homological functor,whereF Mis a projective resolution. Both functors are effaceable [erasible] in positive dimensions,since the chain complexes F G (MP) and F

    G P are exact for P projective. In dimension 0,H0(G, MN) = Z G (MN) = (MN)G = MG N = Tor

    G0(M, N). Therefore, by Theorem

    III.7.3[1] we have an isomorphism of-functorsH(G, M )=TorG(M, ). [The case for cohomology

    is similar].Method 2: The chain complex associated to the group TorG(M, N) is given by F

    0GN

    MGN0, whereF Mis a projective resolution. This can be rewritten as (F0N)G (M

    N)G 0 which yields TorG(M, N) =H(CG), whereCis the chain complex (F

    i N). This is indeed an

    exact sequence because the universal coefficient theorem yields H(FN) =H(F)N=N(nontrivialonly in dimension 0). NowFi is projective and hence a summand of a free moduleF= F

    i K=

    j ZG.

    ThenH(G,FN)=H(G, Fi N)H(G, KN) andH(G,FN)

    = jH(G, ZGN) = 0, notingthat induced modules areH-acyclic by Corollary III.6.6[1]. ThusH(G, Fi N) = 0 and eachFi N isH-acyclic. We can now apply Exercise III.7.1(a) which implies Tor

    G(M, N) =H(CG)=H(G, MN).

    [The case for cohomology is similar].

    7.3: For dimension-shifting in homology, we can choose the induced module M = ZGZ M whichmaps ontoM by(r m) =rm; it is an H-acyclic module by Corollary III.6.6[1]. This map isZ-splitbecause it composes with the natural map i : MMto give the identity, m 1 m1m= m.

    For dimension-shifting in cohomology, we can choose the coinduced module M= HomZ(ZG, M) whichprovides the embedding M M given by m (r rm); it is an H-acyclic module by Corollary

    III.6.6[1]. This map is Z-split because it composes with the natural map : MMto give the identity,m(rrm)[rrm](1) = 1m= m.

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    gE(g

    , gg) =

    gE(gg

    1, (gg)) =

    gE(1, (g

    g)) =

    gE[(gg)], where E is a set of rep-

    resentatives for the right cosets Hg and we note then that g = g. Note that for homology classes,[g1] + [g2] = [g1g2] because of the boundary map 2[g1|g2] = [g2] [g1g2] + [g1]; thus (as a homology class)

    gE[(gg)] = [

    gE(g

    g)]. Using the isomorphism H1(G) Gab given by [g]g mod [G, G], the

    transfer map Gab Hab is computed as gmod [G, G]

    gEg

    g(gg)1 mod[H, H].

    Example: The transfer map Z nZ is multiplication byn, sincex[1x(1x)1][xx(xx)1] [xn2x(xn2x)1][xn1x(xn1x)1] = 1 1 1 xn(1)1 =xn, wherex is

    the generator ofZ.

    10.1: The symmetric group G = S3 on three letters is the group of order 3! = 6 whose elements are thepermutations of the set {1, 2, 3}. The Sylow 3-subgroup is generated by the cycle (1 2 3), and a Sylow2-subgroup is generated by the cycle (1 2). Noting the semi-direct product representationS3= Z3 Z2whereZ2 acts onZ3 by conjugation, we haveH(S3) =H(S3)(2) H(S3)(3)=H(S3)(2) H(Z3)Z2

    by Theorem III.10.3[1]. NowS3 is the unique nonabelian group of order 6, so D6 =S3 and we can useExercise AE.9 which implies that theZ2-action on H2i1(Z3)=H2i(Z3) is multiplication by (1)i (wecan pass this action to cohomology by naturality of the UCT). Thus Hn(Z3)Z2 is isomorphic to Z3 forn= 2iwherei is even, and is trivial forn odd and n = 2iwherei is odd. Taking any Sylow 2-subgroupH =Z

    2, Theorem III.10.3[1] states that H(S

    3)

    (2) is isomorphic to the set ofS

    3-invariant elements of

    H(H). In particular we have the monomorphism H2i1(S3)(2)H2i1(H) = 0, so H2i1(S3)(2)= 0.

    An S3-invariant element z H2i(H) = Z2 must satisfy the equation resHKz = res

    gHg1

    K gz, where Kdenotes H gH g1. Ifg H then gH g1 = Hand the above condition is trivially satisfied for all z(hz = z by Proposition III.8.1[1]). If g / H then K = {1} because H is not normal in S3 and onlycontains two elements, so the intersection must only contain the trivial element. But then the image ofboth restriction maps is zero, so the condition is satisfied for all z ; thus H2i(S3)(2)= Z2. Alternatively,a theorem of Richard Swan states that ifG is a finite group such that Sylp(G) is abelian and M is a

    trivialG-module, then Im(resGSylp(G)) =H(Sylp(G), M)

    NG(Sylp(G)). It is a fact that NS3(Z2) = Z2(refer

    to pg51[2]), so taking G= S3 and H= Syl2(S3)= Z2 and M = Z we have Im(res

    S3H) = (Z2)

    Z2 = Z2 inthe even-dimensional case. Since any invariant is in the image of the above restriction map (by TheoremIII.10.3[1]), the resultH2i(S3)(2)= Z2 follows.

    Hn(S3)=

    Z n= 0Z6 n 0 mod 4 , n= 0

    Z2 n 2 mod 4

    0 otherwise

    10.2(a): LetHbe a subgroup ofGof finite index, letCbe the double cosetHgH, and letT(C) be the en-domorphismH(G, f(C)) ofH(H, M) wheref(C) is theG-endomorphism of IndGHMgiven by 1 m

    cC/Hc1 cm. To show that T(C)z = corHHgHg1res

    gHg1

    HgHg1gz, it suffices to check this equation

    in dimension 0 (by Theorem III.7.5[1]). The right side maps m MH to

    hH/HgHg1h(gm) =

    hgHgH/H(hg)m =

    gC/Hg

    m MH, where g = hg as a coset representative. Now T(C) indimension 0 is given by the composite map

    H0(H, M) H0(G, CoindGHM)

    H0(G, IndGHM)

    f

    H0(G, IndGHM)1

    H0(G, CoindGHM)1

    H0(H, M)

    whereis the Shapiro isomorphismm(sm), andis the canonical isomorphismF

    xG/Hx

    F(x1), and f is induced by f(C), and 1 is the inverse F F(1), and 1 is the inverse x m1[x m](s x) which is sxm ifsx Hand is 0 ifsx /H. This composite is given by

    T(C) :m (sm)

    xG/Hx m

    x

    cC/Hx(c

    1 cm) =

    x

    c xc

    1 cmx

    c

    1[xc1 cm](s xc1)

    x

    c

    1[xc1 cm](xc1) MH

    To simplify this last term, note that the image of1[xc1 cm] is nontrivial iff xc1 H x C, and for each x there is at most one c such that xc1 H. Thus the double sum reduces to

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    xG/H , xCxc

    1 cm=

    xC/Hxm, and this is precisely the image of the right-side map.

    Note: was determined by noting that any element fof (CoindGHM)G must satisfy f(xg) = f(x) and

    hencefis determined byf(1) =m. Sofis given byg m, but it must also commute with theH-actionwhich means thathghm and hence hm= m, i.e. m MH. Thus (m) = (sm), s G.

    10.2(b): If z H(H, M) is G-invariant where H G is of finite index as above, then T(C)z =

    corHHgHg1resgHg1

    HgHg1gz = corHHgHg1res

    HHgHg1z = |H :H gH g

    1|= a(C)z, where the second-to-

    last equality follows from Proposition III.9.5[1].

    10.2(c): Let X = {z H(H, M) | T(C)z = a(C)z C} where C is any H-H double cosetand a(C) = |C/H| = |H : Hg Hg1|. Since the image of the restriction map resGH lies in theset of G-invariant elements of H(H, M), and such elements lie in X by part(b) above, we haveIm(resGH) X. In the situation of Theorem III.10.3[1] and Proposition III.10.4[1], consider the ele-ment w = corGHz H

    n(G, M) where z is an arbitrary element ofX. Then either Hn(H, M) is anni-hilated by |H| [H = Sylp(G)] in which case w H

    n(G, M)(p), or |G : H| is invertible in M [hence

    in Hn(G, M)]. Using Proposition III.9.5[1] we obtain resGHw =

    gH\G/HcorHHgHg1res

    gHg1

    HgHg1gz =

    gH\G/HT(C)z =

    gH\G/Ha(C)z = (

    gH\G/H|H : HgHg1|)z = |G : H|z. Since either

    |G : H| is prime to p or is invertible in M, it follows that z = resGHw where w = w/|G : H|. ThusXIm(resGH) X= Im(res

    GH).

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    the elements s 1 generate Ias a left ZG-ideal by Exercise I.2.1(b), where S is a set of generators for

    G= F(S)/R, and so we have the desired exact sequence ZG(T) 2 ZG(S) 1I0.

    Note that this reproves the second part of Exercise II.5.3(d).

    2.4(d): Since the augmentation ideal I ofZFis free (by Exercise IV.2.3(b)), we have a free resolution0 I

    ZF

    Z0 of

    Zover

    ZF. TakingR-coinvariants and noting that (

    ZF)

    R

    = Z

    [F/R] =Z

    GbyExercise II.2.1, we obtain a complex IR ZG Z 0 whose homology is HR because H1(R, Z)=Ker(IR ZG) by the dimension-shifting technique; ZFis anH-acyclic module by Proposition III.6.1[1]since it is free as a ZR-module by Exercise I.8.2, and it maps onto Z with kernel I. Since I is free with

    basis (s 1)sS, I= ZF(S) and we have the exact sequence 0 H1R ZG(S)

    1 ZG Z 0, wherewe note that taking coinvariants is a right-exact functor.Now we can map the standard (bar) resolution ofZ overZR to the aforementioned free resolution:

    F22

    2

    F11

    1

    F00

    0

    Z

    id

    0

    0 I i ZF Z 0

    where n>1 = 0, 1[r] = r 1 = Dr, and 0[ ] = 1. This is a commutative diagram because

    01[r] = 0(r[ ] 1) = r 1 = i(r 1) = i1[r] and 12[r1|r2] = 1(r1[r2] [r1r2] + [r1]) =r1(r2 1) (r1r2 1) + r1 1 =r1r2 r1 r1r2+ 1 + r1 1 = 0. By applying the coinvariants functorand noting that1|R= D|R and H1R= Ker(1)R/Im(2)R=Rab, we have a commutative diagram

    R

    D|RI

    H1R

    ZG(S)

    where the vertical arrows are quotient maps. The compositeF DIZG(S) is a derivation such that

    ses becauses1s2(s1 1) + s1 (s2 1)es1+ s1es2 . Thus the map is given by rmod [R, R]

    sS(r/s)es, and we have the desired exact sequence 0 Rab

    ZG(S) ZG Z 0 where

    es = s 1 and (r mod [R, R]) = sS(r/s)es.Note that this reproves the first part of Exercise II.5.3(d).3.1(a): Let 0 A E

    G 1 be an extension and let : G G be a group homomorphism, and

    consider the pull-back (fiber-product)EGG = {(e, g) E G | (e) = (g)}. The kernel of the

    canonical projection p : EGG G corresponds to g = 0 (g) = 0 = (e) Ker = A, and

    thus we have an extension 0 A EGG pG 1 which by definition fits into the commutative

    diagram

    0 A E G 1

    0 A EGG p

    G

    1

    This extension is classified up to equivalence (by fitting into the above commutative diagram) because

    given another such extension [corresponding to E] ofG byA, commutativity of the right-hand squareimplies there is a unique map : E EGG

    by the universal property of the pull-back, and thisgives commutativity of the right-half of the diagram below:

    0 A i1 E

    G 1

    0 A i2EGG

    p

    G

    1

    0 A i3 E

    G

    1Note that for the E-extension yields the identity map G G. It suffices to show that the left-side

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    of the diagram also commutes, for then we can apply the Five-Lemma which states is an isomorphism(E =EGG

    ). Now i3(a) =i2(b) for some b A because i3(a) maps to 0 G by exactness of the

    bottom row and hence lies in the kernel ofEG G which is contained ini2(A). Theni3(a) =i2(b),

    and i3(a) =i1(a) by commutativity of the outer left-


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