CLASSIFICATION OF GROUP EXTENSIONS AND H 2

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1 CLASSIFICATION OF GROUP EXTENSIONS AND H 2 RAPHAEL HO Abstract. In this paper we will outline the oundations o homoloical alebra, startin with the theory o chain complexes which arose in alebraic topoloy. Buildin upon that we shall then introduce the derived unctors Ext and Tor and their various properties. Finally, we will apply the theory built thus ar to the case o roup cohomoloy to classiy equivalence classes o roup extensions. Throuhout this paper we will assume basic knowlede o module theory and amiliarity with elementary cateorical lanuae. Contents 1. Chain Complexes 1 2. Ext and Tor Projective Resolutions and Tor Injective Resolutions and Ext 8 3. Group Extensions and Group Cohomoloy Group extensions Group Cohomoloy 13 Acknowledments 17 Reerences Chain Complexes Let R be a unital and assoctiative rin. Unless precised otherwise, R-modules shall reer to riht R-modules. Deinition 1.1. A chain complex C o R-modules is a collection {C n } n Z toether with R-module maps d n : C n C n 1, called boundary operators or dierentials, satisyin the relation d n+1 d n =. In particular, this implies that im(d n 1 ) ker(d n ). I ker(d n ) = im(d n 1 ), we then say the resultin sequence is exact. Denotin im(d n+1 ) by B n and ker(d n ) by Z n, we always have B n Z n C n. Elements in B n are called n-boundaries, while elements in Z n are called n-cycles. This eometric terminoloy comes rom alebraic topoloy where chain complexes irst arose. This leads us to consider the quotient Z n /B n, which we call the n-th homoloy module o C and write as H n (C ). We oten drop the subscript on C and simply write H n (C). Example 1.2. Simplicial homoloy. Given a simplicial complex K o dimension n, we may look at the set K k o k-dimensional simplices ( k n). We orm a chain complex by takin the ree Z-module on the set K k. For k > n we set Date: DEADLINE AUGUST 26,

2 2 RAPHAEL HO C k =. The boundary homomorphisms k : C k C k 1 are iven by sendin a basis element σ = v, v 1,..., v k to k (σ) = k i= ( 1)i v,..., v i 1, v i+1,..., v k These maps are obviously homomorphisms, and one can easily compute that the composite k 1 k is zero. The homoloy o this chain complex is called the simplicial homoloy o K with coeicients in Z. Deinition 1.3. Let C be a chain complex o R-modules. We say a chain complex B is a subcomplex o C i or each n, B n C n and the dierential on B is the restriction to B n o the dierential iven by C. Deinition 1.4. A map : B C between chain complexes is a amily o maps n : B n C n such that the ollowin diaram commutes:... d B n+1 B d n B n 1... n+1 n n 1... C n+1 d C n d C n In other words, d n+1 = n d or all n. In particular, sends cycles to cycles and boundaries to boundaries. Thus, induces a well-deined map : H n (B) H n (C) or all n. Considerin chain complexes as objects and maps between complexes as morphisms, we may deine a cateory Ch(Mod-R) o chain complexes. The arument above shows that or each n, H n is a unctor rom Ch(Mod-R) to Mod-R. Deinition 1.5. Usin the notational chane C n = C n, we can deine the notion o a cochain complex in a similar ashion to chain complexes. A cochain complex C is a amily o R-modules {C n } n Z with maps d n : C n C n+1 such that d n+1 d n =. The kernel Z n o d n is the module o n-cocycles, while B n = im(d n 1 ) is called the module o n-coboundaries. Once aain, we may consider the quotient H n (C ) = Z n /B n, which we unsurprisinly dub the n-th cohomoloy module o C. Maps between cochain complexes are deined in a similar manner to maps o chain complexes. Now that we have deined what a map o chain complex is, we can consider short exact sequences o chain (or cochain) complexes i.e maps such that or all n, A B C A n Bn Cn is exact and the maps commute with the boundary operators. Lemma 1.6. (Snake lemma) Consider the ollowin diaram o R-modules: A i B j C A p B q h C.

3 CLASSIFICATION OF GROUP EXTENSIONS AND H 2 3 I the top and bottom row are exact, then there exists an exact sequence ker() ker() ker(h) δ coker() coker() coker(h). I, urthermore, the rows are short exact sequences, the above sequence can be extended to an exact sequence ker() ker() ker(h) δ coker() coker() coker(h). Proo. The maps between kernels are simply the restriction o the maps o the top row. Commutativity o the diaram ensures that they are well deined. The maps between cokernels are deined by takin a coset represented by one element to the coset represented by the imae o that element under the correspondin bottom row map. To deine the map δ : ker(h) coker(), we must do a bit more diaram chasin. Let γ ker(h). We can lit γ to an element b B, and take its imae (b) in B. Now h(γ) = hj(b) = q(b) =, so (b) ker(q) = im(p), thus we can lit (b) to x A. Since p is injective, this element is unique. Our candidate or the map δ is now the composite δ(γ) = p 1 j 1 (γ)+im(). The only ambiuity in this deinition is the choice o lit j 1 (γ). Suppose we chose j(b 1 ) = j(b 2 ) = γ. Then by the diaram chase above, there exists unique x 1, x 2 A such that p(x 1 ) = (b 1 ) and p(x 2 ) = (b 2 ). Now j is a homomorphism, so b 1 b 2 ker(j) = im(i), hence there exists a A such that i(a) = b 1 b 2, hence j((a)) = (i(a)) = j(x 1 x 2 ). Injectivity now ives us that (a) = x 1 x 2, hence x 1 + im() = x 2 + im(), i.e δ is well deined. The proo that the sequence we obtain by pastin these maps toether is exact at each term is a straihtorward (i somewhat lenthy) diaram chase. Furthermore, i i is injective and q is surjective, the correspondin maps ker() ker() and coker() coker(h) are also injective and surjective respectively rom the way they are deined, ivin us the aumented sequence. Theorem 1.7. Given a short exact sequence o chain complexes there exists a lon exact sequence L M N, H n+1 (N) H n (L) H n (M) H n (N) H n 1 (L). The maps : H n+1 (N) H n (L) are called the connectin homomorphisms. Dually, i L M N is a short exact sequence o cochain complexes, there exist connectin homomorphisms : H n 1 (N) H n (L) such that the ollowin sequence is exact: H n 1 (N) H n (L) H n (M) H n (N) H n+1 (L). Proo. We supply the proo or the case o chain complexes. exactly the same way or cochain complexes. Consider the short exact sequence at the (n + 1)-th term. L n+1 M n+1 N n+1 The proo ollows d L L n M n d M d N N n

4 4 RAPHAEL HO Given any R-module homomorphism A B, we may orm the exact sequence ker() A B coker(). Applyin this construction to the columns o the iven diaram then applyin the snake lemma ives us the ollowin commutatin diaram: Z n+1 (L) Z n+1 (M) Z n+1 (N) L n+1 M n+1 N n+1 d L n+1 L n M n d M n+1 N n d N n+1 L n /B n (L) M n /B n (M) N n /B n (N). By deinition o the boundary operators, the maps d X n : X n X n 1 actor throuh X n+1 /B n+1 (X) Z n (X), thus the above diaram ives the ollowin diaram, in which the rows are exact: L n+1 /B n+1 (L) M n+1 /B n+1 (M) N n+1 /B n+1 (N) d L n+1 d M n+1 d N n+1 Z n (L) Z n (M) Z n (N). Note that the kernel o the vertical maps are precisely the (n + 1)-th homoloy module o that complex, while the cokernels are the n-th homoloy. Thus, applyin the snake lemma to the above sequence ives us an exact sequence H n+1 (L) H n+1 (M) H n+1 (N) H n (L) H n (M) H n (N). By pastin these sequences toether, we obtain the desired lon exact sequence. Deinition 1.8. A map : C D o chain complexes is said to be nullhomotopic i there exists maps s n : C n D n+1 such that n = d n+1 s n + s n 1 d n We say is homotopic to i their dierence is null-homotopic. For convenience, we oten drop the subscripts and simply write = ds + sd. Proposition 1.9. Let, : C D be chain maps. I and are chain homotopic, then they induce the same map on the homoloy modules H n (C) H n (D)

5 CLASSIFICATION OF GROUP EXTENSIONS AND H 2 5 Proo. It suices to prove that i is null-homotopic, then the induced map on homoloy is the zero map. Suppose = ds + sd. Let x be an n-cycle representin an element in H n (C). Then (x) = ds(x) + sd(x) = ds(x) since x ker(d). Thus (x) is an n-boundary in D n, hence is trivial in the module H n (D). 2. Ext and Tor 2.1. Projective Resolutions and Tor. Let R be a ixed rin. Deinition 2.1. An R-module (let or riht) P is projective i iven a surjective map : M N o R-modules and a map : P N, there exists a map : P M such that =. Pictorially, we have the ollowin diaram: P M N. Lemma 2.2. Free R-modules are projective. Proo. Since is surjective, or any n im(), there exists an element m M such that (m) = n. Let A be a set o enerators or the ree R-module P, and deine a set map φ : (A) N M that takes (p) N to an element m p in its pre-imae under. By the universal property o ree objects there exists an R-module homomorphism : P M such that (p) = m p. In other words ( )(p) = (p). Proposition 2.3. An R-module P is projective i and only i it is a direct summand o a ree module. Proo. Let Q be an R-module such that P Q = F is ree. I π : F P is the projection map, π ives a map F M. By the previous lemma, F is projective, thus composin F M with the inclusion P F ives us the desired map : P M such that =. Conversely, i P is projective, let F (P ) be the ree R-module on the underlyin set o P and choose (as we always can) a surjection π : F (P ) P. Takin the identity map on P and usin the deinition o a projective module ives us a map i : P F (P ) such that π i = id, i.e P is a direct summand o F (P ). Deinition 2.4. Let L and M be riht R-modules, and let L ψ M be exact. A riht R-module A is lat i the sequence A L 1 ψ A M is exact. We have a symmetrical deinition or lat let R-modules. Corollary 2.5. Projective modules are lat. Proo. This ollows rom the act that ree modules and more enerally direct summand o ree modules are lat (c D&F, chap. 1.5, Cor. 42) [2]. Deinition 2.6. A let resolution o an R-module M is a chain complex P such that P i = or i <, toether with a map ɛ : P M such that the aumented complex d P 2 d P1 d P ɛ M

6 6 RAPHAEL HO is exact. I each P i is a projective module, we say it is a projective resolution. It is oten convenient to view M as a chain complex concentrated in its last term and ɛ consider P M as a map o chain complexes. Lemma 2.7. Every R-module has a projective resolution. Proo. Startin with an R-module M, choose a ree (hence projective) module P such that M is a quotient o P. The natural projection ɛ : P M is a surjection. We may now choose a ree module P 1 such that the module ker(ɛ) is a quotient o P 1 and deine the map d : P 1 P to be the composite o the natural projection P 1 ker(ɛ) with the inclusion ker(ɛ) P. Since any R- module is always the quotient o a ree module, we may proceed inductively and orm surjections P i+1 ker(d i ) with P i+1 ree. Pictorially we have the ollowin diaram: ker(d 1 ) d 2 P 1 d 1 ker(ɛ) P ɛ M. Theorem 2.8. (Comparison Theorem) Let M, N be R-modules, and suppose : M N is a map o R-modules. Given a ɛ map o chain complexes P M such that each Pi is projective and an arbitrary η let resolution Q N, there exists a map o chain complexes : P Q litin in the sense that η = ɛ. This map is unique up to chain homotopy. P 2 P 1 ɛ P M Q 2 Q 1 Q η N Proo. Since P is projective and η is a surjection, we can lit η to a map : P Q such that η = ɛ. Inductively, i we have a map i 1 such that d i 1 i 1 = i 2 d i 1 (thinkin o as 1 and d as η or ɛ), then d i 1 i 1 d i =, hence i 1 d i maps P i to ker(d i 1 ), which is equal to im(d i ) by exactness. Thus by projectivity o P i we et a map i such that d i i = i 1 d i. Now suppose we have another map : P Q. Then η( ) = hence sends P to ker(η) = im(d 1 ). Since P is surjective, we et a map s : P Q 1 so

7 CLASSIFICATION OF GROUP EXTENSIONS AND H 2 7 that ( ) = ds = ds + s 1 d, where s 1 is the zero map. Inductively, i we have s i 1 such that ( i 1 i 1 ) = d i s i 1 +s i 2 d i 1, then d i ( i i s i 1 d i ) =, so aain by projectivity o P i there exists a map s i : P i Q i+1 such that d i+1 s i = i i s i 1 d i. Lemma 2.9. (Horseshoe Lemma) Suppose we have a short exact sequence M M M o R-modules. Let P, P be projective resolutions o M and M respectively. Then there exists a projective resolution P o M such that P i π P P is a (split) short exact sequence o chain complexes. P 2 P 1 P ɛ M M P 2 P 1 P ɛ M. Proo. The proo o this lemma will be omitted due to space constraints. Readers searchin or completion may ind it illed out in detail in Weibel [1]. The idea however, is to deine P n to be the direct sum P n P n and then use projectivity to build the required maps. Since R preserves exactness o the sequence on the riht but not on the let, we say it is a riht exact unctor. We now introduce the torsion product, which measures how by how much R ails to be exact (i.e both let and riht exact). Deinition 2.1. Let M be a riht R-module and N be a let R-module. Choose ɛ a projective resolution P M. Deine Tor R n (M, N) = H n (P R N). This means that or all n, Tor R n (M, N) is a unctor rom Mod-R R-Mod to Ab. The irst thin we need to check is that Tor R n (M, N) does not depend on the choice o resolution o M. η Lemma I Q M is another projective resolution, then we have an isomorphism H n (Q R N) = H n (P R N), hence Tor R n (M, N) does not depend on the choice o resolution or M. Proo. Consider the identity map id M : M M. Usin the comparison theorem, we et a map o chain complexes : P Q commutin with the boundary maps on P and Q. We thus et maps on the homoloy modules H n (P ) H n (Q ). As we ve seen in the proo o the theorem, this map is unique up to homotopy. Now Q is also projective, so similarly we et a map : H n (Q ) H n (P ). The maps

8 8 RAPHAEL HO and id P are both chain maps P P litin id M, thus = () = (id P ) = id Hn(P ). Similarly, and id Q both lit id M, so is the identity on H n (Q ), hence and are isomorphisms. We may now list a ew useul properties about Tor. Proposition Tor R (M, N) is naturally isomorphic to M R N. Furthermore, i either M or N is projective, then Tor R n (M, N) = or n 1 Proo. Since R is riht exact, the sequence P 1 R N P R N M R N is exact, thus Tor R (M, N) = H (P R N) = M R N. I M is projective, the identity map M M can be seen as a projective resolution. Since projective modules are lat, the the tensor unctor is exact, thus the homoloy roup o the chain is trivial or i >. Theorem For every short exact sequence o riht R-modules M M M, i N is a ixed let R-module then there exists a lon exact sequence Tor R n (M, N) Tor R n (M, N) Tor R n (M, N) Tor R n 1(M, N) Proo. Given M M M choose projective resolutions P M and P M. Usin the Horseshoe Lemma, there exists a projective resolution P M such that the each P n P n P n is split exact. Since R is an additive unctor, P n R N = (P n P n ) R N = (P n R N) (P n R N), hence the sequence P n R N P n R N P n R N is also split exact. This means that (P R N) (P n R N) (P n R N) is a short exact sequence o chain complexes. The correspondin lon exact sequence on the homoloy o these chain complexes is sequence promised in the statement o the theorem. Tor is an example o a let derived unctor, namely the let derived unctor o the tensor product. More enerally, one can deine the let derived unctor o a riht exact unctor between two abelian cateories, provided the domain cateory has enouh projectives. What we mean by this is that or each object A in the cateory there exists a surjection P A with P satisyin the same universal property as layed out in the deinition o a projective module. The let part o the terminoloy comes rom the act that ater applyin our riht exact unctor to a short exact sequence, we wish to prolon this to a lon exact sequence on the let. Axiomatizin the properties we ve seen above or Tor, with a little work it can be shown that the let derived unctor is universal with respect to extendin short exact sequences to lon exact sequences on the let. In particular, this means that we could reverse the roles o M and N above and choose a projective resolution o N instead and still obtain the same unctor.

9 CLASSIFICATION OF GROUP EXTENSIONS AND H Injective Resolutions and Ext. There is a dual notion to projectivity in an abelian cateory which is called injectivity. Althouh the Ext unctor which we shall introduce in this section can be deined usin only projective modules, we shall include some basic acts about injective modules or completion. Deinition Let R, S be rins. An (R, S)-bimodule M is simultaneously a let R-module and a riht S-module such that (rm)s = r(ms). Proposition Let L be a riht R-module, M an (R, S)-bimodule and N a let S-module. We can impose a riht R-module structure on Hom S (M, N) by the rule (r)(m) = (rm). The unctors Hom S (M, ) rom Mod-S to Mod-R and R M rom Mod-R to Ab orm an adjunction, meanin that or every riht R-module L and let S-module N there is a natural isomorphism Hom S (L R M, N) = Hom R (L, Hom S (M, N)) Proo. Startin with a map : L R M N, deine τ : Hom S (L R M, N) Hom R (L, Hom S (M, N)) by lettin (τ)(l) be the map m (l R m) or each l L. Conversely, i we have a map : L Hom S (M, N), let τ 1 () be the bilinear orm l R m (l)(m). The map m (l R m) is an S-module map since ms+m s (l (ms+m s )) = (s(l m)+s (l m )) = (s)(l m)+(s )(l m ) We also have that is an R-module map since (τ)(rl+r l ) is the map m ((rl+ r l ) m) = r(l m)+r (l m), which is the same as takin τ(r)(m)+τ(r )(m ). Furthermore, τ 1 ()(r(l m) + r (l m)) = (rl + r l )(m) = ((rl) + (r l ))(m) so τ 1 () is also an R-module map. The maps τ and τ 1 are clearly inverse o each other, thus τ is an isomorphism. The proo that this isomorphism is natural is let to the reader. Takin S = Z and M = R as a (Z, R)-bimodule in the above proposition ives us the ollowin specialized version. Corollary Let L be a let R-module and N be an abelian roup. We then have a natural isomorphism o abelian roups Hom Z (L, N) = Hom R (L, Hom Z (R, N)) Proo. This ollows immediately rom the act that L R R = L and the above proposition. Deinition An R-module I is said to be injective i or any injection e : M N o modules and each map : M I, there exists a map : M I makin the ollowin diaram commute: M e N. Proposition (Baer s Criterion) A riht module I is injective i and only i or every riht ideal J o R, every map J I can be extended to a map R I. I

10 1 RAPHAEL HO Proo. Since riht ideals are R-modules, the only i direction is clear rom the deinition o injectivity. Conversely suppose every homomorphism : J I can be lited to a map G : R I. Let e : M N be injective, so that e(m) is a submodule o N. Given a map : M I, we can look at the set S o all extensions : A I o to an intermediate submodule e(m) A N. We can deine a partial order on this set by lettin (, A ) (, A ) i L L and = on L. By Zorn s Lemma, there exists a maximal extension : A I in S. Suppose there exists some n N not in A. We can then look at the riht ideal J = {r R mr A }. We thus have a map J m A I, hence by assumption it extends to a map G : R I. Deine A M to be the submodule A + mr and let : A I by (a + mr) = (a) + G(r). I mr A mr, (mr) = G(r) so this map is well deined, and extends, contradictin the maximality o. We thus have that A = N, thus completin the proo. Corollary I R is a PID, an R-module I is injective i and only i it is divisible, meanin that or every non-zero r R and every x I, x = yr or some y I. Proo. I R is a PID, then every ideal is o the orm J = (r) or some r R, hence any R-module homomorphisms : J I is uniquely determined by (r) = i I. We can extend this homomorphism to a map G : R I i and only i there exists an element i I with G(1) = i such that i = (r) = G(r) = ri. Hence by Baer s criterion I is injective i and only i ri = I. Lemma 2.2. Every R-module embeds as a submodule o an injective R-module. Proo. We prove this or Z-modules irst, then eneralize to an arbitrary rin. Since abelian roups can be rearded as Z-modules and Z is a PID, by the above corollary an abelian roup A is injective i and only i A is divisible. I M is any Z-module, let G be a set o enerators or M. Let F be the ree module on the set G. We can then identiy M = F/K where K is a Z-module. Now take Q to be the ree mathbbq-module on G. Q is a direct sum o copies o Q, hence divisible since Q is. Now K F Q, thus M = F/K Q/K. Since the quotient o divisible roups is aain divisible, this show M is a submodule o an injective Z-module. Now let R be an arbitrary rin, and let N be a let R-module. Deine a map j : N Hom Z (R, N) by j(n)(r) = rn. Clearly this is a homomorphism between abelian roups, however recall that Hom Z (R, N) has a let R-module structure iven by (r)(n) = (rn). Thus i r, s R and n N, we have that j(sn)(r) = r(sn) = (rs)n = j(n)(rs) = s(j(n))(r) It ollows that j is in act a map o R-modules. Since j(n)(1) = n, j = i and only i n =, hence j is a injection. Now take any injective abelian roup map i : N D where D is divisible and let i : Hom Z (R, N) Hom Z (R, D) be the induced map. Then i is an injection, thus the composite i j : N Hom Z (R, D) is an injection o R-modules. Since D is an injective Z-module, applyin Hom Z (, D) to an exact sequence M N ives rise to an exact sequence Hom Z (N, D) Hom Z (M, D). Now usin Corollary 2.16, this tells us that the sequence Hom R (N, Hom Z (R, D)) Hom R (M, Hom Z (R, D)) is exact, thus provin that Hom Z (R, D) is an injective R-module.

11 CLASSIFICATION OF GROUP EXTENSIONS AND H 2 11 Deinition An injective resolution o an R-module N is a cochain complex I and a map η : I such that is exact and each I i is injective. N η I d I 1 d I 2 d The ollowin three statements have analoues in the previous section. Their proo is identical up to replacin projective with injective and reversin the direction o the arrows, and thus will not be supplied. Lemma Every R-module has an injective resolution. Theorem (Comparison Theorem) Let M, N be R-modules, and suppose : M N is a map o R-modules. Given N ɛ I an injective resolution and M η J an arbitrary riht resolution, there exists a o chain complexes : J I litin in the sense that η = ɛ. This map is unique up to chain homotopy. M η J J 1 J 2 N ɛ I I 1 I 2 Lemma (Horseshoe Lemma) Suppose we have a short exact sequence N N N o R-modules. Let (I ), (I ) be injective resolutions o N and N respectively. Then there exists an injective resolution I o N such that I i I π I is a short exact sequence o chain complexes. As we ve seen above, the torsion product is riht exact. Buildin upon this unctor we then deined a new unctor Tor which allowed us to continue the sequence on the let. We will now introduce a let exact unctor and similarly deine Ext to be the unctor extendin exactness to the riht. I we start with a short exact sequence o R-modules M M M and an R-module N, the unctor Hom R (, N) is let exact, i.e we et an exact sequence Hom R (M, N) Hom R (M, N) Hom R (M, N) Similarly, one can start with a short exact sequence N N N and a module M, then apply let exact unctor Hom R (M, ). Deinition Let M, N be riht R-modules, and let P ɛ M be a projective resolution. Deine Ext R(M, N) = H (Hom R (P, N)) Alternatively, one can instead choose an injective resolution N η I and apply the let exact unctor Hom R (M, ) then take the homoloy o that complex to deine Ext R(M, N). The proo that this in act ives us the same roup is rather technical and will be omitted due to space constraints. Analoously to Tor, Ext n R(M, N) is a unctor rom (Mod R) op Mod R to Ab or each n.

12 12 RAPHAEL HO The properties o Ext are quite similar to those o Tor and proved analoously by takin a projective resolution o the irst variable, or by takin an injective resolution in the second variable. We will state them without proo to avoid unnecessary redundancy. Proposition Ext R(M, N) is naturally isomorphic to Hom R (M, N). I M is projective, then Ext n R(M, N) = or n. Theorem Let M M M be a short exact sequence o R-modules and N be an R-module. Then the ollowin sequence is exact: Ext n R(M, N) Ext n R(M, N) Ext n R(M, N) Ext n+1 R (M, N) I we now have an exact sequence N N N and a module M, we also have an exact sequence Ext n R(M, N ) Ext n R(M, N) Ext n R(M, N ) Ext n+1 R (M, N ) 3.1. Group extensions. 3. Group Extensions and Group Cohomoloy Deinition 3.1. Let A an abelian roup and G be any roup. We say a third roup E is an extension o G by A i there exists a short exact sequence 1 A i E π G 1 Note that A (more precisely, i(a)) is a normal subroup o E and that the quotient E/A is isomorphic to G. Lemma 3.2. An extension E o G by A deines a G-action on A. Proo. For each G, choose an element e π 1 () E. This deines a map o sets σ : G E called a section o π. Since A is abelian and i is injective, the subroup i(a) E is also abelian, and the element e acts on an element i(a) i(a) by conjuation. As we ve said beore, E/i(A) = G, thus any other element in E mappin to under π is o the orm e i(a 1 ) or some a 1 A. Usin that i is a homomorphism and A is abelian, we see that conjuation by e i(a 1 ) is the same as conjuation by e, hence the action is independent o our choice o e. This means that our G action is well deined on i(a), hence on A as well since i is injective. This action o G on A turns A into a G-module, meanin that G acts on A as automorphisms. We can now look at when two extensions are essentially the same. Deinition 3.3. We say two extensions are isomorphic i there is an isomorphism o short exact sequences 1 A α i E π G 1 A i E π G 1. In other words, this diaram commutes and the vertical maps are isomorphisms. I we require the stroner condition that the maps α and γ be the identity, we then say that E and E are equivalent extensions. β γ 1

13 CLASSIFICATION OF GROUP EXTENSIONS AND H 2 13 Proposition 3.4. Equivalent extensions deine the same G-module structure on A. Proo. I E and E are two equivalent extensions as in the deinition, let e be a representative o π 1 in E, and let e = β(e ). This ives two actions o on a, one sendin a to i 1 (e i(a)e 1 ) and the other to i 1 (e i (a) e 1 ). Since i, i, β are injective, these two actions are equal i and only i (β i)(i 1 (e i(a)e 1 )) = e i (a)e 1, which is true by the deinition o e and the commutativity o the diaram Group Cohomoloy. We will now use an alternative deinition o a G- module. Deinition 3.5. Let G be a roup. The interal roup rin Z[G] is the ree abelian roup with basis the set G, on which we put a rin structure with addition deined ormally on inteer linear combinations o elements o G, and multiplication deined by the roup operation in G, extended Z-linearly to Z[G]. A G-module is now the same as a (let) Z[G]-module. A trivial G-module is an abelian roup A on which G acts trivially, meanin that a = a or all G and or all a A. By convention we will always take Z to be a trivial G-module unless speciied. Deinition 3.6. The aumentation map ɛ : Z[G] Z is deined to be the map takin n to n, n Z. The kernel o this map is called the aumentation ideal, which we denote by I. Since Z[G] has basis {1} { 1 1} as a ree Z-module, it ollows that I has basis { 1 1}. We now introduce two canonical resolutions o the trivial G-module Z that are o reat theoretical importance. Deinition 3.7. Let B u n be the ree Z[G]-module on the set o all symbols [ 1 n ] with i G. We let B u be the ree Z[G]-module on a sinle enerator, which we shall denote by [ ]. Deine maps d : B u n B u n 1 by d([ 1 n ]) = 1.[ 2 n ] ( 1) i [ 1 i i+1 i+2 n ] n 1 + i=1 + ( 1) n [ 1 n 1 ] Theorem 3.8. Given {B u n} n N and maps d as deined above, the chain complex Z ɛ B u d 1 B u d 2 1 is a projective resolution o the trivial G-module Z, where ɛ is the map takin [.] to 1. We call this resolution the unnormalized bar resolution. Proo. The irst thin we need to prove is that this is indeed a chain complex. Deine maps d ([ 1 n ]) = 1.[ 2 n ] d i ([ 1 n ]) = [ 1 i i+1 i+2 n ], 1 i n 1 d n ([ 1 n ]) = [ 1 n 1 ].

14 14 RAPHAEL HO Then d : B u n B u n 1 can be written as n i= ( 1)i d i. A direct computation shows that or i j 1, d i d j = d j 1 d i. When computin d d, the terms d i d j and d j 1 d i appear with opposite sins, hence everythin cancels out pairwise, ivin us that d d is indeed. We shall now ive the splittin maps s : B u n B u n+1 and check that the identity map on B u n is null-homotopic, and hence that the chain complex is (split) exact. Let s 1 : Z B u and s n : B u n B u n+1, n be maps deined respectively by s 1 (1) = [.] s n ([ 1 n ]) = [ 1 n ]. Clearly we have that ɛs 1 = 1 and ds + s 1 d is the identity on B u. For n 1, we see that ds n ([ 1 n ]) =d([ 1 n ]) =[ 1 n ] [ 1 n ] ( 1) i 1 [ 1 i i+1 i+2 n ] n 1 + i=1 On the other hand, we also have that + ( 1) n+1 [ 1 n 1 ]. s n 1 d([ 1 n ]) =s n 1 ( 1 [ 2 n ]) ( 1) i [ 1 i i+1 i+2 n ] n 1 + i=1 + ( 1) n [ 1 n 1 ] =[ 1 n ] ( 1) i [ 1 i i+1 i+2 n ] n 1 + i=1 + ( 1) n [ 1 n 1 ]. Thus we see that ds n + s n 1 d is the identity on Bn, u hence that the complex is split exact. Since each Bn u is ree, it ollows immediately that it is a projective resolution. There exists a normalized version o the bar resolution. We deine B n to be the ree Z[G] module on the set o symbols [ 1 n ], 1 G. By convention, we set [ 1 n ] to be equal to i i = 1 or some i. For n =, we let B = Z[G]. The dierential maps are deined in the same way as or the unnormalized version. It is sometimes advantaeous to identiy B n as the quotient module o B u n by S n, which is the submodule enerated by elements o the orm [ 1 n ] where i = 1 or some i. Theorem 3.9. The chain complex Z ɛ d B 1 d 2 B1 is a projective resolution o the trivial Z[G]-module Z. Proo. The proo o this theorem is virtually identical to the previous one, and thus will be omitted.

15 CLASSIFICATION OF GROUP EXTENSIONS AND H 2 15 Application 3.1. (Group Cohomoloy) Let A be a let G-module. The n-th cohomoloy roup H n (G; A) is deined or all n to be the cohomoloy o Hom G (B u n, A), i.e the roup Ext n Z[G](Z, A). Elements in Hom G (B u n, A) are called n-cochains, and are simply set maps G n A. I φ is a n-cochain, the dierential dφ is an (n + 1)-cochain deined by dφ(,, n ) = φ( 1,, n ) ( 1) i φ(, i i+1, ) n 1 + i=1 + φ(,, n 1 ). I dφ =, we say that φ is a n-cocycle. Elements dφ in Hom G (Bn, u A) are called n-coboundaries. The roup o n-cocycles and o n-coboundaries are denoted by Z n (G; A) and B n (G; A) respectively. We thus recover the amiliar ormula H n (G; A) = Z n (G; A)/B n (G; A). One can also take the cohomoloy o Hom G (B, A) instead and consider normalized cochains, meanin that φ( 1, ) = i any o the i = 1. For the rest o this paper we shall in act take this latter approach, since it will make computations slihtly easier. Recall rom the previous section that iven an extension 1 A E π G 1, a section σ o π is a set-theoretic map G E such that πσ() = or all G. We shall restrict our choice o maps to based sections only, meanin sections such that σ(1 G ) = 1 E. This choice is equivalent to takin the normalized bar resolution in our computation o H n (G; A), and while the ollowin proposition can be proved usin the unnormalized resolution, doin it this way simpliies the issue o the unit or the roup operation in E. Now or any, h G, σ(h) and σ()σ(h) both map to h under π, hence their dierence lies in A by exactness. We thus deine [, h] = σ()σ(h)(σ(h)) 1. This deines a unction [ ] : G G A dependin on E and σ called the actor set determined by E and σ. Proposition Let A be a G-module. A map o sets [ ] : G G A is a actor set i and only i it is a normalized 2-cocycle. Proo. A normalized 2-cycle is an element o Z 2 (G; A), meanin it is a set map [ ] : G G A satisyin that or all,, h G [,1] = [1,] = [,h] - [,h] + [,h] - [,] = Now suppose we have a actor set [ ]. This determines the roup operation in E, which is iven by (a, ) (b, h) = (a + b + [, h], h), where b denotes the G-module action o on b iven by conjuation in E. Now we have that (, ) ((, ) (, h)) = ([, h] + [, h], h) ((, ) (, )) (, h) = ([, ] + [, h], h) By associativity o E, we recover exactly the second condition in the deinition o a normalized 2-cocycle. The irst condition is obvious since σ(1) = 1.

16 16 RAPHAEL HO Now conversely, suppose we have a normalized 2-cocycle [ ]. Let E be the set A G, and deine an operation on E exactly as the one in the irst part o the proo. One easily checks (, 1) is the identity or this product. Associativity holds because o the second condition o a normalized 2-cycle. Given (a, ) E, we have that (a, ) ( 1 a 1 [, 1 ], 1 ) = (, 1). Thus E is a roup. The subroup A 1 is isomorphic to A and E/A 1 is G. We thus have that 1 A E G 1 is an extension, and the section G = G E ives us a actor set [ ] which is identical to our oriinal 2-cocycle. Lemma Let E be an extension o G by A with based section σ, and let [ ] be the actor set determined by σ. I E is an equivalent extension, then there exists a based section σ o E such that the actor set determined by σ is [ ]. Proo. Since E and E are equivalent, there exists an isomorphism β between them. I σ is a based section o E, then clearly σ = β σ is a based section o E. By deinition, [, h] = σ()σ(h)σ(h) 1, so we have β([, h]) =(βσ())(βσ(h))(βσ(h) 1 ) =(βσ())(βσ(h))(βσ(h)) 1 =σ ()σ (h)σ (h) 1. However, β restricts to the identity on A, thus β([, h]) = [, h], thereby ivin us the desired result. Lemma Given an extension A E π G, two dierent actor sets [ ] and [ ], correspondin to choices σ and σ o based sections respectively, dier by a 2-coboundary. Proo. Given two based section maps σ and σ, σ () lies in the same coset o A as σ(). That means that there exists an element α() A such that σ () = α()σ(). The correspondin actor set is [, h] =α()σ()α(h)σ(h)σ(h) 1 α(h) 1 =α() + σ()α(h)σ() 1 + σ()σ(h)σ(h) 1 α(h) =[, h] + α() α(h) + α(h). As we see, the dierence [, h] [, h] is precisely the coboundary dα(, h) = α() α(h) + α(h). The above three lemmas show that there is a well-deined map Φ rom the set o equivalence classes o extensions to H 2 (G; A). Lemma Two extensions o G by A with section maps σ i : G E i ivin rise to the same actor set are equivalent. Proo. As sets, we already have that E 1 = E2 = A G. Writin out the roup operation o an extension E with actor set [ ] under the bijection with A G, we see that the products (a, 1) (b, 1) = (a + b, 1), (a, 1) (, ) = (a, ) and (, ) (a, 1) = (a, a) are ixed. The roup structure is thus entirely determined by the product (1, ) (1, h) = ([, h], h). Thus i σ 1, σ 2 determine the same actor set, the bijections o E 1 and E 2 with A G ive a roup isomorphism E 1 = E2 and thus an equivalence o extensions.

17 CLASSIFICATION OF GROUP EXTENSIONS AND H 2 17 In particular, i [ ] =, it means that the set map σ is a roup homomorphism. In other words the extension is split, thus E is the amiliar semi-direct product A G rom roup theory. Lemma The map Φ is injective. Proo. Suppose [ ] and [ ] are two actor sets correspondin to based sections σ : G E and σ : G E respectively, that represent the same cohomoloy class in H 2 (G; A). That means that [, h] [, h] = α() α(h) + α(h), where α is a 2-coboundary. Let µ() = α() 1 σ(). Since α() lies in A, µ is also a based section o E. Its correspondin actor set is [, h] =α() 1 σ ()α(h) 1 σ (h)σ (h) 1 α(h) = α() σ ()α(h)σ () 1 + σ ()σ (h)σ (h) 1 + α(h) =[, h] α() + α(h).α(h). Thus [, h] [, h] =, so by Lemma 3.14, E and E lie in the same equivalence class. Theorem Equivalence classes o extensions are in 1-1 correspondence with the cohomoloy roup H 2 (G; A). Proo. We ve already shown that the map Φ was well-deined, and the previous lemma shows it is injective. By Lemma 3.11, every normalized 2-cocycle ives rise to a extension o G by A, thus Φ is also a surjection, thereby establishin the bijection. Acknowledments. I would like to thank my mentors, Rol Hoyer and Michael Smith, or the support and patience they showed durin the writin o this paper. Reerences [1] Weibel, C., An introduction to homoloical alebra, Cambride University Press, [2] Dummit, D. & Foote, R., Abstract alebra, John Wiley and Sons, Third Edition, 24. [3] May, J.P., Notes on Tor and Ext, may/misc/torext.pd