The 2-groups of rank 2

JOURNAL

OF ALGEBRA

149, 1-31 (1992)

The 2-Groups

of Rank 2

DAVID J. RUSIN Department of Mathematical Sciences, Northern Illinois University, De Kalb, Illinois 60115 Communicated by George Glauberman Received January 17, 1986

The goal of this paper is to describe the (finite) groups of the title. Here, a group is said to have p-rank I if its largest elementary-abelian p-subgroup has order pr. The study of p-groups with given p-rank is an old one. The classification of p-groups of rank 1 is classical (see Lemma 1). For p odd, Blackburn [2] gave a classification of p-groups of rank 2. (In fact, he uses only the weaker hypothesis that all normal elementary-abelian subgroups have order at most p’.) For p = 2, the problem is still open. Konvisser [S] considered the slightly smaller family of 2-groups with a unique 4-group (elementaryabelian of order 2’). Some related questions are discussedin [ 1; 3, Sect.5.4; 4, Sect. 111.121. Our own interest in the problem begins with a paper of Quillen [7], in which he shows that the cohomology ring H*(G, Z,) has Krull dimension equal to the p-rank of G. As the Krull dimension measures in a sense the “complexity” of this ring, we would find “simple” cohomology rings by finding those of groups with low p-rank. This has been done for 2-groups of rank 1 and for certain classesof rank-2 2 groups [8]. Blackburn’s results for odd p suggest we consider the classification this way: it is easy to see that in any extension A -+ G + B, the rank of G is at most rank(A) + rank(B). So we should try to show conversely that all rank-2 G have a presentation of this form with A and B of rank 1. For 2-groups, this is a fairly large class of groups, since A and B would then be either cyclic or generalized quaternion. Unfortunately, there are many examples of rank-2 groups G that have no such presentation. Nonetheless, certain facts which would follow from such a classification turn out to be true. These are our principal results. First, we know (Lemma 1) that G almost always contains a normal 4-group F; if this is not central in G, it is certainly central in a subgroup of index 2. 1 0021-8693/92 $5.00 CopyrIght G 1992 by Academic Press, Inc. All rights of reproductmn m any form reserved.

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DAVID .I. RUSIN

THEOREM A. If G contains a central 4-group F, then G/F is a split extension A . B with A and B each cyclic or dihedral.

This is expected, up to a point: quaternion groups have dihedral quotients. However, it is not obvious that the extension should always be split. THEOREM

B. A 2-group of rank 2 contains a split metacyclic group of

index at most 2’.

(This bound is sharp, as, as is shown already by the wreathed quaternion groups.) Finally, we observe that the distinct G can then be obtained as extensions F -+ G -+ A B corresponding to cohomology classesin H*( A . B, F). The calculations of all these cohomology groups, and of the classesin them that lead to groups of rank 2, is discussed in Section 4. The resulting groups G are too numerous to permit succint description; apparently there are O(n3) distinct 2-groups of rank 2 with order 2”. We leave unsettled the problem of describing those 2-groups of rank 2 without a cental 4-group. An indication of the results is given in Section 4, but the rather lengthy details will be deferred to a later paper, as will the complete calculation of the rings H*(G, ZP). Our method continues the work [S]; the techniques used here do not exhaust the possibilities of this analysis and might extend to groups of higher rank. The idea is to find a series of normal subgroups of G, 1 < F, < F, < . . < F, < G and to describe each G/F, as an extension

to see that all the G/F,, and in particular G/F itself, have the desired form. As may be expected, these extensions are described cohomologically. In the process, we use a number of cohomological techniques: ring structure, Steenrod-algebra structure, Serre and Eilenberg-Moore spectral sequences, StiefelLWhitney classes, and the corestriction. I suspect that many of the steps can be accomplished without this machinery, but it does have side benefits (in addition to seeming quite straightforward to the author). We have arranged the paper as follows. First, Section 1 contains the statements of the lemmas which together constitute the proof of the main theorems. In Section 2 we have gathered background material from group theory and cohomology that will be used throughout. Then Section 3 contains the proofs of the stated and auxiliary lemmas. Finally, in Section 4

we present additronal destructions of the 2-groups of rank 2 and their cohomoiogy rings. We encourage the reader to consider throughout the paper some exnmples of 2-groups of rank 2. Here are a few of the most noteworthy: (a) All metacyclic groups; (b) wreathed groups Z, { Z, and Q, { Z,: (c) the groups SE Syl,(U,(4)) of order 2’j; (d) products D,oQ,, with amalgamated center Zz~

We present here the lemmas needed to prove Theorems A and B. The proofs of these, and of additional technical lemmas, Lemmas 7, 10, 11, 12, 15’, and 16, appear in Section 3. While the proofs are often cohomological, the statements below are not. So we assume G is a (finite) 2-group not containing Z:. Let us first recall LEMMA 1. G contains a normal 4-gruup F unless G is cyclic, quaternion, dihedral, or semidihedral.

See,e.g., f3, p. f99]= Note that these exceptionaf casesare all metacyclic, so Theorem B holds. As ~-AI&(F) 2 Z,, we see [G : C&F)] 6 2. Thus, to prove Theorem 3, it suRces to show G has a spfit metacyclic subgroup of index 4 when G contains a cent& 4-group F. This hypothesis holds for the rest of Section 1, as we prove Theorem A and Theorem B. (We discuss the case C, Ff G iti Section 4.) We designate by R the image in G/F of a subgroup Wg G. LEMMA 2. (7 contains a normal 4-group l7, except perhaps if l? is cyclic or dihedraI (so Theorem A holds); in the exceptional cases cl has a metacyclic strbgroup of index at most 2 (so Theorem B holds). Let N< G be the complete inverse image (“pullback”) of i7 in G. Note H (I G. ff we are willing to repface r7 by another normal 4-group in G, we will see LEMMA 3. We may assume H il: 2: I except perhaps z$ G is dihedral of order 4 or 8. In the exceptional cases, Tlzearenzs A and B ho/d. The key to the rest of the proof is the subgroup K = C,(H) Q G; it will provide a chain of subgroups on which we can work inductively. As noted in [ 11, K is metacyclic, as will then be its homomorphic images. However, we showed in KS} that if G is a metacyclic group with central 4-group F, then G/F is too, unless G/F is cyclic or dihedral.

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DAVID J. RUSIN

So let us set F, = F and then inductively define F, 4 K by letting F, + , IF, be the (unique) central 4-group of K/F,. We choose v minimally so that K/F, is cyclic (possibly trivial) or dihedral of order at least 8. We then set F Y+ 1 = K.

(A similar construction was performed in [S], but with a slightly different notion of length 1. This old definition will reappear after Lemma 6.) LEMMA 4.

(a) Each F,+I/F,+,=Zi

(b) (c)

v(K) b 2. KJF,, is cyclic (possibly trivial).

(d)

K/F,-,-Z,xZ,for

(,forO
some N>2.

Actually, K is split metacyclic (Lemma 15), and Fj + , /Fi ~ I 6 Z( K/F, ~, ) (Lemma 5(d)). Note that each F, is normal in G, as it is characteristic in K 4 G. Our goal is to prove that successively larger quotients G/F, have the general form we wish. We begin with a few odds and ends. LEMMA 5.

(a) G/F,, + , is elementary-ahelian.

(b) Fi, l/F;< Z(GIFi). (c) If an involution t E G/F, is the image of one in G/F,- , , then it is the image of one in G/F, (i d v).

(d)

Each F, + , /FiP 1 is isomorphic as a G-module -(i 6 v - 1).

Let us say that a group has property Pl if it is the product of an elementary-abelian group and a cyclic or dihedral group. Then LEMMA 6. G/F,, has property Pl, and K/F,> is contained in the cyclic or dihedral factor.

Now, it is possible that G/F,+, also has property Pl (e.g., G = Z4 x Z,), However, G/F,,+ 2 contains FY/FYPZ N Zi and so cannot have this property. Thus we choose 1 so that G/F, has property Pl but not G/F,- 1 (I= v or v - 1). This agrees with the definition of length 1 in [S]. While we are assuming v Z 2, it could be that I= 1. LEMMA 8.

Theorem A holds if I= 1.

(Of course in this case G is Z, x Z; or D, x ZS,if not elementary abelian; we are simply claiming s < 2-observe that Z: N D, !) The cases G= D, x Z; and G= Z; follow from a theorem of Anne MacWilliams [6] on 2-groups with no normal abelian subgroups of rank 3.

THE%GROUPSOFRANK 2

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So we assume I> 2 and move to describe G/E;-, . Let us say that a group has property P3 if it is a split extension A. B with A and B each cyclic or dihedrai; let us also say that a quotient of G has (the weaker) property P2 if it is a product Xx Y of an elementary-abelian Y (possibly trivial) and a subgroup X which has property P3 and contains the image of K. We first prove LEMMA9. G/FjPl has property P2. (We will, more precisely, describe the extensions X in F,/F,_ 1+ X-t LT.) On the other hand, we show X = G/F,_ , : LEMMA 13. Ff d @G, and (so) G/F[-, has property P3.

The longest lemma is then LEMMA 14. If G/F,+, has property P3 (1< i < I - I ), then so does G/r’,.

These few lemmas have similar proofs and follow this pattern. We work with the cocycle q describing the extension Fi+ , jFi + G/Fi -+ G/F,, 1. As a further extension FiiF~_, -+ G/Fi- , -+ G/Fi is possible, we find a strong constraint on q (“Steenrod invariance”) which results from spectral sequence arguments (Lemma 19). On the other hand, as G/F, contains F,, JF, N .Z& there is a strong non-triviality result for q. As G/F,,, has property P3, we can describe its cohomology ring fairly precisely (there are several cases to consider), and so it is possible to find all q meeting these two conditions. In fact, if A is suitably rechosen (within A . (F,+,/Fi+ ,)-see Lemma 18), this q turns out to be exactly the cocycle of an extension Z: -+ A”. 3 + A . B with 2 and B again cyclic or dihedral (Lemma 12). The conclusion of Lemma 14, of course, is that C = G/F, has property P3, which is Theorem A. If we pass from 6?= A . B to subgroups of index at most 4, we may assume that first B, then A, is cyclic (i.e., G is split metacyclic). Theorem B then follows from LEMMA 15. If G is split ~~tacyc~jc, so is G.

(This proves Theorem B, even for the groups of length 1.) 2. BACKGROUNDMATERIAL Before the lemmas, it will be necessaryto fix our notation and to provide some preparation for the unaccustomed reader. Certain groups will occur frequently: the cyclic, dihedral, and quaternion

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DAVID J. RUSIN

groups of order IZ will be written Z,, D,, and QII, respectively. We will sometimes write 0: for the rotation subgroup of D,, so 0,’ N ZniZ. Certain subgroups of a group are particularly important, including its center Z(G), the Frattini subgroup Q(G) = G’G’, and the subgroup Q(G) generated by the involutions (elements of order 2) in G. We write G#=G-11). If G has a normal subgroup A and G/A 4 B, we will express this fact by writing A -+ G -+ B. If there is a splitting (i.e., complement to A) B --+G, we write G = A . B. If A is abelian, this defines a cohomology class, or “cocycle,” in H2( B, A ). Most commonly, A will for USbe a central elementary-abelian subgroup of G. If we choose a basis for this A, we can write H*(B, A)=H*(B,

Z,)@ .‘. @H*(B, Z,),

and so q = (q/, .... qm) with each qj~H*(B, Z,). Of course, a different choice of basis for A will result in different components for q (e.g., it might repla= qt by qi +q2). Recall that the maps A -+ G and G + B induce maps (“backwards”) in cohomology, denoted by Resz and infg (or infg,,). We also use, briefly, the corestriction Gory associated to an inclusion A + G. If A -=IG1then this map has the property that

We will often need to consider more than one group extension at a time. A useful result is this. LEMMA 17. Supposewe have two extensions X -+ Y + Z and R + 5’ + Y, the latter with R abelian and corresponding to a cocycle CL E H2( Y, R).

(a) If Res$(cc)= 0, then S contains a subgroup S, isomorphic to R. X (the semidirect product), which maps onto X < Y under the projection S -+ Y. (b) Suppose X acts trivia& on R, so we have a map inf: H*(Z, R) -+ H*( Y, R). IJ‘cr = inf(r’) for some LY’,then S contains a subgroup S, N R x X, with both ,factors normal in S. Moreover, the extension R -+ S/X -+ Z correspond.~to the cocycfe cx’E H2(Z, R). (c) If in (b) Mjeassume Y= X.Z, then S= X. U. ProoJ Standard theory. We view S = {(r, JJ)E R x Y} with multiplication (r, y)(r’, JS’)= (r(r’)-‘y( y,v’), yv’); the map S -+ Y is just the projection onto the second component. Here, fe Z*( Y, R) represents the cohomology class ~1.Then, the statement Res(a) = 0 meansfl xx x = 6g for some function g: X-+ R. Then S, = R x X is the semidirect product of the subgroups Rx (1) and X’= {(g(x),x)lx~X~.

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In (b), we take f to be inflated from a cocycle f’ E Z2(Z, R) representing cc’; then f( y, y’) = f( x, x’) if y~x and y’- x’ mod X. In this case, Res(rr)= 0 and S, = R x X, with g identically zero. Naturatly, R -a S by construction, but also XQ S: (1, x)(‘,~) = (1, y-‘xy) follows from the cocycle relation and the dependence of f’(x, y) only on xX and yX. Here we compute that S/X= (R x Y)/(l x X) may be viewed as the set of pairs (v, z) E R x Z with group law (v, z)(r’, 2‘) = (r(f)’ .f’(?, 2’)). Here, 5 and z’ are inverse images of z and 2’ in Y, so f(Z, 3) =f’(z, 2’) Then this is just the product prescribed by the cocycle f’, so that indeed S/X corresponds to CC. (c) Let U= R x Z d R x Y. As f‘ is inflated from f' E Z2(Z, R), U is a subgroup of S. It meets X= 1 x X trivially as Xn Z= (11 in Y. As Y= YZ, S = XU. Finally, as Xa S (from part (b)), we are justified in writing s=x- u. Note that we are assuming X acts trivially on R. If Y= Xx 2, then all of U fixes X (pointwise), so S= Xx U. 1 Note that the copy of X normal in S (in part (b)) is not unique, so S/X is not, either. For example, take R = X= Z = Z2, Y = Z,, and c(= 0, so that S = Z, x Z, with R = Z, x 1. We can write r = inf(cr’) with either a’ = 0 or CI’# 0 E H2(Z2, Z,). In the one case,X= 1 x Z, and S/X- Zi; in the other, X is the diagonal subgroup of QS, and S/X 2?Z4. The cohomological description of extension theory requires a knowledge of HZ. We will more generally be interested in the cohomoiogy ring H*(G) = H*(G, Z,). We use the Kunneth formula H*(Xx Y) ilz H*(X)@ H*(Y). So for example, since H*(Z,) = Z,[X], the cohomology ring H*(G) is a polynomial ring if G is elementary-abelian. We remark, too, that if M6 G is maximal, then the kernel of Resg is the principal ideal generated by the L E H’(G) = Hom(G, Z,) with kernel M. A few remarks about H’(G) = Hom(G, Z,) are also in order. The elementsf~ Hom(G, Z,) are in one-to-one correspondence with the maximal subgroups M= Ker(f) of G; we will say M corresponds to f~ H’G. Note that f restricts to zero on M and then on @p(G);in particular, the ideal spanned by H’(G) restricts to zero on @i(G). Also well known is the cohomology of the dihedral group (see,e.g., [S]). If n 2 8 we may write

ff*P,)=

z,cz, -? .?Jl/(Y2= VI,

where x, y E H’ and z E HZ. If we view Z, as Dz,, then the restriction map Resg: is onto, and its kernel is the principal ideal (-x), so that H*(Z,) = Zz[z, yl/(.? = 0)

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DAVID J. RUSIN

if n 3 4. (We sometimes write H*(Z,) = Z,[z, x, JJ] with the understanding that x=0.) A few properties of this ring R = H*(L),) are of interest. R is a reduced ring, that is, it has no non-zero nilpotent elements. Our x E H’(D,) is not a zero divisor, but the ideals (y) and (x + y) are each other’s annihilators. (Of course, similar remarks apply to the ring H*(D,> x Zs).) In the given presentation of H*(L),), the ~EH~ is not unique, but for any choice of the possible z we have Res,,,,Jz) ~0. It is easy to check from the relations given that H2(D,)= {clz-i-L2]a=0 or 1, LEH’(D,,)). Also, L2 =xL for LE H’(L),) (or LEH’(Z,)). There is a nice way to specify z E H’(D,) canonically, as can be checked from the description above: z is the cocycle of the extension Z, + D,, -+ D,. Upon restriction to Dz, we see that ZE N*(Z,) is the cocycle of Z2 -+ Z,, + Z,. Some care must be taken with the smallest groups. If G = Z1 = D,, then H’(G) =O, as is the cocycle of the extension Z, + (Z2 = 02) + G. If G=Z?= D, then H*(G)=Z,[x], so H2(G)= (0, x2). This x2 is the cocycle of Zz + Z, + Z,, and 0 is the cocycle of Z, + D, -+ D2. Finally, if G = D,, then the cocycle of Z, -+ D, -+ D, is (in a suitable basis of G) x . y E Z,[x,

y] = H*(G).

It follows that the cocycle of the extension Z; -+ Zz + Zf may be written (x2, y2) E Hz E H2(Z:)2, if basesin the two copies of Zi are chosen suitably. Another way we will describe low-dimensional cohomology classes is with Stiefel-Whitney classes. A representation p: G -+ GL,( R) determines cohomology classes w,(p) E H’(G) for i < n. If n = 1, the correspondence p t-) w’(p) is the isomorphism Hom(G, Z,) 2 H’(G) mentioned before. For rt > 1, wl(p) = w,(det p). If n = 2, the image of p is cyclic or dihedral. The inclusions: p(G) -+ GL,R is unique up to conjugacy, and so wi(i) E H’(pG) is well-defined. Then w,(p) = infFG(wi(j)). The relevance of these classesis that when G is cyclic or dihedral and p is faithful, then w&) is exactly the cocycle z defined above (corresponding to larger cyclic or dihedral extensions). This even holds for small groups: w2(p) = 0 if G = (1); if G 2: Z,, then there are two inclusions p: G + GLzR {a rotation and a reflection) with, respectively, w,(p) = x2 and w*(p) = 0. If G = Da, then an injection p: G --) G2R must send two of the three involutions to a reflection; write the maximal subgroups they generate as Ker(x) and Ker(y); then w2(p) = x. y is again the cocycle needed. We find w,(p) = x for the representation above (when G contains elements of order 4; otherwise W,(P) = 0 unless G = D,, where wi( p) = x+.Y)If G is not cyclic or dihedral, one can create more representations p: G + GL, R by choosing QE Hom(G, Z,) and forming p 0 e. The formulas

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w,(pOa) = Wz(P)+ W’(P)u”(O)+ MJf2 are relevant for this next lemma, LEMMA 18. Suppose X = A 13 is a split extension with A and B each cyclic or dihedral. Assume B has exponent > 2, and that QZ(B) acts trivially on A. Suppose given a cocycle q E H2(X) of the form q = z + L(x + L) with L E H’(X) and with x and z the Stiefel- WhitneJf classes of the faithful representation of XJA N B. Then q = w*(p’), where p’ is a representation of X with kernel A’ N A. In fact, A’ and A are maximal subgroups of A + = A ’ fJZ( B), and X = A’ . B (thus, ifX=AxB then X=A’xBas well).

broom If L is inflated from H’(B), then one may compute q = z, so p’ = p, A’ = A, and there is nothing to prove. So we may view L as the Stiefel-Whitney class pi of some representation cf: X -+ Zz with A not in Ker(a). We noted that q= VV&@O), so let p’= p@cr. What is A’= ker(p’)? If XE ker(p’), then either a(x) = 1 (and p(x)=Z) or a(x) = -1 (and p(x)= -I). In the first case, x~Ker(p)= A (and in ker(o) as well). In the second case, we see p(x) E QZ(pX), so as p is faithful on X/A 2 B, we obtain x E A . SZZ(B) = A+. Combining the two, we see that A’ = ker(o) Q A + (of course, A d A + is also maximal, as A+ z A x Z,). Let a E A\A’, and let bE QZ(B)#. Then our hypotheses make bE@(B)<@(X), so o(b)= 1; thus @@o)(b)= --I, and so abE ker(p’). Therefore we have A = (A n A’, a) and A’E (A n A’, ab). Now, we have assumed b acts trivially on A, so a and ab act equally on A n A’ and have equal squares in A n A’. This shows A and A’ are isomorphic extensions of AnA’ by Z,. A’= ker(p’) is clearly normalized by B. It meets B trivially, since if (~@a)(x)=Z, then p(x)= I1, so XESZZ(B)<@(B), so p(x)= (p 0 c)(x) = I, so x = 1. It is complemented by B, as A’. B includes both B and A d A + = A’ . OZ(B). Thus, we are justified in writing X = A’ . B. 1 We will use this lemma frequently to replace a cocycle z + L(x + L) by the simpler cocycle z, at the expense of replacing A by A’. One other property of the cohomology ring H*(G) we use heavily is that it is an algebra over the mod-2 Steenrod algebra. This means that there are certain “natural” additive maps Sq’: H”(G) + H”+‘(G) with a few properties, most notably (a) if a E H”(G), then Sq’fa) = a, Sq”(a) = a’, Sq’(a) = 0 for i > n; (b) Sq’(a -b) =x.j+k=i S&(a) .Sqk(b) (the Cartan formula).

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For example, these axioms suffice to calculate the action of Sq’ on II*( as soon as we give Sq’(z). It follows from Wu’s formula that Sq’(z) = x2 (and so Sq’(z) -0 in H*(Z,)). We will usually need to consider only Sq’, and so when convenient we write Sq for Sq’. Note from (b) that Sq’ is a derivation of H*(G). Finally, there is the matter of spectral sequences.We shall use primarily the Serre spectral sequence of the group extension A -+ G -+ 3. This is a first-quadrant spectral sequence of rings with EgY cxFp(H”fy(G))/ Fpf ‘(HPfq(G)), where each H”(G) has been “filtered,” so that H”(G)=F”aF’>

>F”>F”“=O.

If XE H”(G) lies in some Fp(H”(G)) but not in Fpf ‘, we will say x represents the element .x + Fpf ’ E EgY. We have E2.O=H*(B), E2* N inf~(~*(~)), and dually E>,* 2: Res~(~*(~)) c H*(A). The differentials d, : E/‘+”-+ Ee + r,q~ ‘+ i are all derivations, so that E, + , inherits a ring structure. The filtered subgroups FP(H*(G)) are all ideals, and the isomorphism E$ * r, Fp/Fp+ ’ is one of rings. Very useful for us is the calculation of some differentials. If A 2: Z’J is central, we have seen we may write H*(A) 2 Z,[x,, .... x,] and 4 = (4, >.*.3qm) (both depending on the basis of A). Then d2(xi) =qi, Moreover, the transgressions u’,: EF’-’ + E;’ commute with the Steenrod squares, so that in particular &(xf) = d,(Sq’(x,)) = Sq’d,(x,) is just SY’(cri). It will be necessry to know which elements in H*(G) have a property we will call “Steenrod invariance.” Let us say that a collection of elements {qi) c H*(G) is Steenrod-invariant if the ideal Z=C qiH*(G) is closed under the action of the Steenrod algebra. We will be interested in the case in which each q, is in H’(G); from the axioms of the Steenrod algebra, we see that one need only show each Sq’(q,) E I. We will write Sq’(qi) = C L,q, with L, E H’(G), or simply, or Sq(q) = L . q. LEMMA 19. Suppost Sz-+ X -+ Y is a central extension with 521: Z;I, corre~p~nd~~gto the cocycle q E H2( Y, St). If there is u further centrffl extension 52’+ Z + X with Q’ 2i Z;, in which each element ofst” rifts to an element qf order 4 in Z, then q is Stee~rod-invariant.

Proof: Choose a basis of D so that q = (ql, .... qn)E H2( Y)m with 4’ = (4, , ‘..>q,) linearly independent and q,+ , = . . . = qm= 0. We must show q’ is Steenrod-invariant. If {x i, .... x,) is the basis of N’(B) dual to the chosen basis of Q, then we may compute the cohomology of X from a spectral sequence with ET* = H*( Y)[x,, ...) x,] and with d,(x,)= qi. Note in particular that x I + 1, .... X, will survive to E, .

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Let Q be the cocycle of the second extension, and Q = Res,(Q). Once a basis of !2’ is chosen, we may write Q = (Q,, .... Q,) with Q, E H2(X). Now, this Qi represents the element Qic E22 of the previous spectral sequence; in particular, d2(Qj) = d3(Qj) = 0. Write this Qi as a quadratic form in H*(SZ)=Z[x,, .... x,], say ni=ck<,“k/XkX/.

Then

d’@,)

=Ck

(C&k

%kq,+C,,k

ak,q,)

xk.

As

-%

is the free H*( Y)-module on the xk’s, this shows c ICk

alkql

+

c I>k

ak/q/=o

for all k. As qi = 0 if i > r and the other qi are linearly independent, this says ak, = 0 if k d r (unless k = 1). Thus, we have

Q;=c akkx:+1 ak[xkxI. k,cr

r
akkSq’(qk). Since this is to be 0 in E,, we must have Then d3(Qi)=Ck
at&+(qk)=~

Li,4j.

(*I

Now, these computations are valid irrespective of the basis chosen for 52’. Let us choose a basis to diagonalize (A&) = (uhk) as follows. Write s2=sZ,xsZ,, where 52,=ker(x,+,)n ... nker(x,) and S2,= ker(x,) n ... n ker(x,). Then we have shown Qi restricts on Sz, to a cocycle (C a,,~~)~; in particular, the pull-back of 52, is abelian, so from the hypothesis of the lemma, it has the form Zi x ZzPr. Choose a partial basis of 1;2’consisting of the squares of the chosen basis in 52,; complete the basis arbitrarily. In this basis, we will have a,& = 0 unless k = i, and a,, = 1. Thus Eq. (*) shows Sq’(q,) =cL,q,, as needed. a We will in our applications usually have m = n = r = 2. Frequently, but not always, we will be able to reline Lemma 19 to say that L is diagonal. For example, we show LEMMA 20. Suppose G is elementary-abelian, and suppose q E H’(G)“’ is Steenrod-invariant, say Sq(q) = L. q. Then L is diagonalizable, that is, Sq(Mq) = D . (Mq) for an invertible (scalar) matrix A4 and a diagonal matrix D. Equivalently, there is a basis for the ideal generated by q which consists of elements which are (individually) Steenrod-invariant.

Proof: It is, of course, not true in general that D = MLM- ‘, since L is not unique if the elements of q are not linearly independent in H2(G).

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DAVID J. RUSIN

Thus, we must proceed as in Lemma 19: if we reorder the q,, we may assume q’= (q,, .... qr) is linearly independent and q” = (ql+, , .... qm) lies in the span of q’. Then if we replace q by Mq (where M is the appropriate matrix in GLJZ,)), we may assume q” =O. The matrix L may now be written (“,’ x), where Sq(q’) = L’ . q’. This L’ is now uniquely determined by q’. It suffices to show L’ is diagonalizable. So we will assume that q consists of linearly independent classes.Then the equations 0=Sq’(Sq’(q))=Sq’(L~q)=Sq’(L)‘q+

LY!?q’(q)= (Sq’(L)+LZ)q,

which follow as Sq’ is a derivation, force Sq’(L) = L2. If (x,, .... x,) is a basis of H’(G), then we may write L = C x,L, for some matrices Lie M,(Z,). Comparing coefficients of xix1 in the equation Sq’(L) = L2 (in M,(Z,)[x,, .... .x,1), we conclude L:=Li

and

LiLi= LiLi

for all i and j; that is, the Li are all commuting idempotents. In particular, it is easy to seethat they may be simultaneously diagonalized, that is, each Di = ML,M-’ is diagonal, for some single matrix M. But then MLM-’ = I: xiDi is also diagonal. Since Sq(Mq) = MSq(q) = MLq = MLM - ’ . Mq = L?. Mq, we are done. 1 In a few easy cases,we can describe Steenrod-invariant elements. LEMMA 21. If G is elementary-abelian and Q E H’(G) has Sql(Q) = 0, then Q = L2 for some L E H’(G).

Proof Indeed, write Q=xisja,x,x,; then Sq’(Q)=&aO(xfxj+xj.$), so all these coefficients (with i < j) are zero. j LEMMA 22. If G is elementary and Q E H’(G) has Sq’(Q) = L . Q, then Q= L’(L+ L’) for some L’E H”(G).

ProoJ: Restrict these equations to kerfl); by the previous lemma, Res(Q)= (L’)’ for some L’E H’(ker(L))
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Wemayassume IBIa8. TherrH*(G)=H*(A)@Z,[z,~,y]/y~= xy. All the one-dimensional generators restrict trivially to SZZ(B) 6 G(G), but the Stiefel-Whitney class z does not. Thus, Q lies in the subring H*(A)@Z,[x, y]/y(x+ y) inflated from G/@(G), say Q =inf(Q). This Q is unique to within ker inf= (y(x + y)). Note similarly that Sq(&) + Lo is a multiple of y(x + y). So write Sqg = LQ + L’y(x + y). As Sq’ is a derivation, we conclude 0 = L’(L’ f L + x) . y(x + y) in this polynomial ring. Hence either L’ = 0 or L’ = L + x. In the latter case, we may replace & by Q + JJ(X+ y) (which also inflates to Q) and conclude Sq(Q) = LQ, which is also true in the former case. Then the preceding lemma shows Q = L’(L + L’). 1 Proof:

3. PROOFS OF THE LEMMAS Proof of Lemma 2. If indeed G contains no normal 4-group R, then as in Lemma 1, we conclude G is cyclic, quaternion, dihedral, or semidihedral. We must rule out the quaternion and semidihedral cases. In each of these cases, the extension F -+ G -+ G would correspond to a cocycle qc H2(C)OH2(G). Now, it is clear from [8] that H’(G) is spanned by products of one-dimensional classesand so restricts to zero on Q(G). But then q restricts to zero on the central involution r of G, and so G contains the extension F + Zs -+ (t), violating the assumption that G has rank 2. Thus indeed G is cyclic or dihedral (of order > 16). If G is cyclic, then G is a cyclic extension of a central subgroup F, hence abelian. Since G has rank 2 and G rank 1, we see Gg Z2 x Z,,, for some N. This is already metacyclic. If G is dihedral, let G+ 6 G be the inverse image of G’ -CG. Then [G : G+] = 2 and G+ is metacyclic from the preceding paragraph, so Theorem B is true in this case. 1 Proof of Lemma 3. Suppose G contains no normal copy of 2:. Let

Hd G be the group of order 16 mapping to the normal 4-group R_a G. It is easy to see that H = 2, . Z,, the split non-abelian metacyclic group, or H=Q8xZ2.

We observe that Z(H) = F; if not, then H would be a cyclic extension of a central subgroup (of order 8), hence abelian. Thus we can even show K = C,(H) must equal F. If not, then Ra G would be non-trivial and so would meet ZG non-trivially; choose i, E $2(Rn ZG)#. Note ii 4 n, since C,(H) = F. So choose i2 E 52(R n ZG) #, and we will have ( il, i2) a central (hence normal) 4-group in G. Its inverse image in G is abelian, as [t,, f2] E [K, H] = 1. As noted in the first paragraph, this inverse image would then be Z,j.

14

DAVID J. RUSIN

So we may assume C,(H) = F, giving an injection G/F + Aut(H). Now there is an exact sequence 0 + Hom(H, F) -+ Aut(H) + Aut(l;i) x Aut(F). Since G acts trivially on F, and since the 2-part of Aut(R) is Z,, it follows that X= C,(R) maps to the subgroup Hom(R, F) of Aut(H). In particular, X/F is elementary-abelian. I claim that in fact X/F is of rank 2. First suppose X= G, let x, y E H be elements of order 4 with [x, y] = .Y’, and pick z E X- H. Consider the three elements [x, z], [y, z], and [xy, 21 = [x, z][ y, z]. If, say, [M;,z] = 1, then H* = (M’, z) is isomorphic to Z$, and normal in X as it contains X’ d F. So if G = X, the existence of this H* contradicts our assumption on G. Similar constructions work if w =x, u’= y, or w = .xy. Otherwise, all three commutators are distinct, and so one of them equals [x, y]. If, say, [x, 2]= [x, y] , then take H* = (x, yz); if [y, z] = [x, y] then H* = (xz, y) works; and if [xy, z] = [x, y] = ]xy, y], take H* = (xy, yz). In any case, we obtain a contradiction as in the previous paragraph. So if G = X, we conclude G = X is of rank 2, so G = D, (Theorem A) and G = H is metacyclic (Theorem B). If G > X, we still claim X is of rank 2. Indeed, let g E G - X and view X as a (g)-module; it will be the sum of some number a of copies of the trivial module and b > 1 copies of the faithful indecomposable module ( NR). We need to show a = 0 and b = 1. If not, we can find a non-zero tiEim(l-g)=G’<@(G), and also a non-zero ijEker(l-g)=ZGdistinct from U. As V is central in G = G/F, its action on G itself is r-’ = 2:. p(x) for some p E Hom(G, F), so in particular u’ = u as ii E Q(G). Then H* = (u, u) 2 Z4 is (g)-invariant, hence normal in G, contrary to hypothesis. So, X= Z?, and G= (Q, g) = D, (Theorem A), which contains the metacyclic H of index 2 (Theorem B). 1 Proof of Lemma 4. Let A 4 K be a cyclic subgroup with B = K/A also cyclic. Then conjugation induces a map B + Aut(A) N Z, x Z, for some N (generators of these two cyclic factors are known to be the automorphisms a-+a-’ and a + us of A). The second factor is precisely the centralizer in Aut(A) of Q,(A) 2 Z,. Since we are assuming K= C,(H) and H N Zi, it follows that Q,(A) d H 6 Z(K), so B maps onto this second factor of Aut(A). Choose b E K to generate B so that b acts on A as a + aif2”-’ for some t (0 < t Q IZ- 2, where /A\ = 2”). With a suitable generator a of A, we then write

K= (a, bla2”= 1, b2m=a2’,ah=a’+2n-‘).

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If we write K(n, m, t, i) for the group on the right, then we seeimmediately that RN K(n - 1, m - 1, t - 1, i - 1) (with negative parameters replaced by zeros). We note that m - t > 2 if 2: < Z(K). So now (a) is easy by induction: (m - 1) - (t - 1) >, 2, so K has a central Zi, which can only be FX;,,as F3,iFz? F3/F2 contains all involutions of &XlZ~ E K/F,. Part (b) is a matter of definitions. We know K contains a central Zz, so we can certainly find F, and F2; however, one does need to check that K/F, cannot already be cyclic, or dihedral of order greater than 8. (Note that D, does contain a normal 4-group, just not a central one.) By definition, K/F, is to be cyclic or dihedral, so for (c), it is necessary only to note that K(n, m, t, i) is never dihedral. It is easiestjust to observe that Q,(A) N Z, is central. Then, finally, K/F, _ , is a cyclic extension of the central subgroup F,/F, _ 1N Zi, hence abelian; this will prove (d). fl so as HaG we have a Proof of Lemma 5. (a) F, + 1= K=C,(H), and H=Zz, G acts conjugation map G/F” + , qAut(H). As F
Let t E G map to this involution in G/F,_ , , and let X= (F;, t ) < G. This is of rank 2 and contains F in its center. If w also had rank 2, then as in [l], X would be metacyclic. But then, so would X/F,_ r, which contains the elementary-abelian group (F,/F,_ ,) x ( tFj_ , ) of order 8, a contradiction. So instead x has rank at least 3. Let u E W--c be an involution, As t maps to an involution in G/F,, t* E F,, so 8-- Fi= fc. In particular, u= t mod F,, so u maps to the given one in G/F,. (d) As F,, , is metacyclic, write Fi+JF/- f = (x, 4’). Note that Fi+,/F;-, = (x2, y2) as Fi+,/Fiz Zi. Thus, by Lemma4(a), x and ~1 commute with x2 and y’. Now, fix g E G and write XR= x”yb, YR= x’jld Then b and c are even by Lemma 5(b), so y’, ,xL’ are central, and ix’)” = (I’)“*, ( JJ’)~= (.x*)‘( F’)~. This shows that the map uFi + u*F,- , is an isomorphism of G-modules, F,+ JFi -+ F, + I/Fj_, . 1 Proof of Lemma 6. We need to study the extension C = F,,+ ,lF, -+ G/F, -+ GiFv + 1. This C is cyciic (Lemma 4(c)), and the quotient is elementary-abelian (Lemma 5(a)). Of course, if F, + 1= F,,, we are done, so

we may assume C # { 1}. It is helpful to put most of the proof in the following lemma. Note that the proof follows the framework mentioned in Section 1. 481,149x?

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DAVID J. RUSIN

LEMMA 7. Suppose C + X -+ Z; is an extension with C # { 1 } cyclic, and suppose there is a further extension Z, -+ Z -+ X in which C pulls back to a cyclic group. Then X is the product qj’ a cyclic or dihedral group containing C and an elementary-abelian group.

Proqfi We will view X as an extension sZC-+ X-+ X. By Lemma 19 the cocycle qE H*(R) of this extension is Steenrod-invariant. We now proceed by induction on 1C/. If /Cl = 2, then 8= Zz, and so y is the product of two linear forms in H’(R) (Lemma 22). With a suitable basis for H’(X) chosen, we have q = 0, .x2, or my. In the first case, Xr C’ x z. In the second case, X- Z, x Zz- I, and in the third, X2: D, x Z,s ‘. In either case, C = @(X) lies in the first factor. So suppose now that /Cl > 2; let C= C/G?C< X. Then the extension c -3 x--f Z; meets the hypotheses of the lemma (the necessary Z is our original X). By induction, we may write x= A x B with A elementaryabelian, and C < B = cyclic or dihedral. If B 2: Z,, we are done as in the preceding paragraph (it may be necessary to alter the cyclic or dihedral factor to ensure that it contains C N Z,, but this is no problem as C. Z(X) contains all elements of order 4 in X). Thus, we assume B contains elements of order 4, so that

H*(8)=H*(A)@Z,[z,x,

yyy*=xy

with the added relation x = 0 if B is cyclic. Now, q is to be Steenrod-invariant and to restrict non-trivially to Szc. As 0 6 G(z), all generators of H*(X) restrict to zero there except z, and so q=z+xL,+yL,+Q

for some I,,, Lz, and Q E H*(A). We compute in this case Sq(q) = x(z + XL, + yL2) + XL: + yL: + Sq(Q)

so Steenrod invariance here means Sq(q) = xq (compare coefficients of z). This forces the relation yL: + x( L: + Q) + Sq( Q) = 0.

Comparing coeffkients of y and the constant term show reprectively L: = 0 (so L2 =0) and Sq(Q)=O (so Q= L2 for some L). (If B is dihedral, x#O, so we in fact have Q = Lf .) In any event, this means q=z+xL+Lf

with L E H’(z).

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2

By Lemma 18, q = wz(p) for a representation p: x+ M, R with kernel A’ N A, where x= A’ x B. By Lemma 17(c), XE A’ x Y, where Z2 + U -+ B corresponds to w*(p) E H’(B), that is, U is again cyclic or dihedral. In fact, the construction there shows U/BC= B, and so contains c: this means U contains C. This completes the induction. 1 We now complete the proof of Lemma 6. Since the metacyclic K/F,-, of rank 2, the extension F,/F,,-,

-K/F,,-,

+ F,,+,/F,,=

is

C

is non-trivial. Thus for some subgroup Y/F,, , < F,/F,.- I of order 2, we must have Z,rF,,/Y-*K/Y+C

also non-trivial. So applying Lemma 7 with X= G/F,, and Z = G/Y, we conclude that GJF, has the desired form. 1 Proof F-G-G.

of

Lemma 8. Let

q= (ql, q2) E H*(G)*

be the cocycle of

(a) Suppose G z Z;; we wish to show s < 4, so it suffices to derive a contradiction when s = 5. Here, each qi is a quadratic form in s variables. If F is to have rank 2, then q1 must restrict, on any subspace V< ZS,isotropic with respect to q2, to a quadratic form without isotropic vectors. Now, any quadratic space (V, q) may be decomposed as 1/ = V’ I Y” I V”’ (and q = q’ I q” I q”‘), where we may choose bases of the subspaces as follows. First, q’ = 0, xl, or xy $- .x2+ y2 with l=dim V’=O, 1, or 2, respectively. Next, q” =x1 y, + . . + x,y,, and dim Y” = 2m. Also, q”’ = 0; let n = dim Y”‘. In this notation, V has isotropic subspaces of dimension m + n. In particular, if q is to have no isotropic vectors, V = v’ (of dimension < 2). Thus we conclude our q2 can only have isotropy subspacesof dimension at most 2. The only possibilities for (/, m, n) are (2, 1, 1) or (1, 2,0), so that in a suitable basis of G we have q2 of form either (1) q2=(x2+xy+y2)+(zw)+0 or (2) q2 = (x2) + (yz + WV). In both cases,there are isotropic subspaces V of dimension 2, on which q1 will be anisotropic, and hence “quaternionic”: q, = a* + ab + b* in any basis {a, b} of V. Case 1. The possible V include the planes x = y = z = 0 and x = y = w = 0; on either one, (u, z + w} is a basis, so that the polynomial ql(x, Y, z, w, 0) must be q1=02+u(z+w)+(x+w)~

mod (x7 Y, z) n (x, Y, w) = (x, Y, zw).

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DAVID J. RUSIN

Now restrict q, and q2 to the hyperplane z = w. There, we have q, = 1” mod (x, y, z’), say q1 = II2 + u(ax + fly) + Q(x, y, z)

with Q a polynomial in x, y, and z. Note that q2 = (x + y + z)’ + xy on this hyperplane. (We will not leave this hyperplane in C.) On the plane x= y= -I, q2 =O, so q, must be quaternionic: with basis I), 1 X), we see Q+ fi = 1 (and Q(.x, X, X) =x2). But similarly, the planes x =O, y=z and y = 0, x = z show that j?= 1 and OL = 1, respectively, a contradiction. Case 2. Here, q2 =x2 + J’Z+ MC is zero on the four planes given by x=0 and either y=“=O, y=r=O, z=i~=O, or z=u=O. Thus, q,=(I’+Z)2+(W+U)2$(y+z)(W+U) mod(x,r,u)n(x,y,w)n(x,,-,~)n(x,z,w)=(x,yz,M?u).

This time, we restrict to the hyperplane M:= u, on which q2 = x2 + JU and q, E (y + z)* mod (x, yz, 1~~).On the plane .x= y=O, q2 is zero, and so q1 E z2 mod +v~is to be quaternionic, which is impossible for either qr = z* or q,=z* or q,=z2+w2. (b) SupposeG N Z; x Z, with n > 2. Then H*(G) = Z,[z, y, x, , .... x,1/ ( y2 = 0). If a E SZ@G#is the involution in Q(c), and b is any involution in Z,;, then any cocycle q=z + Q with QEZ~[Y, x1, .... x,] is zero on either b or ah. Now, q= (ql, q2) must be non-zero on a, so say Res(,,(q,) = 0 # Res(,,(q,). Then q= (z + Q, Q’) with Q (and Q’) as above. Since z + Q is zero on one of the involutions, Q’ is non-zero there, and hence non-zero on both. We conclude Q’ is non-zero on all of Z;. It follows from the preceding paragraph that s d 2. (c) This follows from either (a) or (b), as DZnx Z; contains both Z;+‘and Z,,xZ;; so sd2. 1 Proof of Lemma 9. Let q be the cocycle of the central extension F,/F,-,

-+ G/F,.. 1 --f G/F,.

Since F,/I;;-, 1: Zi (Lemma 12(c)), we may apply Lemma 19 to conclude q is Steenrod-invariant. By definition of 1, G/F, has property Pl. If in fact it is elementary abelian, then we may choose a basis of F,/F,_ i so that q= (L, LZ, L3L4) (Lemma 20) with L,E H’(G/F,). We aim to find such a simple form for q in the other two cases,G/F/g A x B with A elementary-a~Iian and B cyclic or dihedral, of exponent at least 4. Note that by Lemma 4(a) or (d), the

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central involution of B lifts to an element of order 4 in G/F,_ 1, so one of the qi must restrict non-trivially there; we will suppose henceforth that q, # q2 = 0 when restricted to QZB. We will write H*(G/F,) = H*(A) 0 Z,[z, x, y]/y2 = xy as in Lemma 7, with z, x, and y inflated from (G/F,)/A and the other generators inflated from (G/F,)/B. Then q = (z + XL, + yL, + Q,, XL, + yL4 + Q,) with Lj and Q;eH*(A). It seems easiest to separate the two cases: x = 0 (B cyclic) and x # 0 (B dihedral). LEMMA 10. If q E H2(A x B)’ is Steenrod-invariant, with B cyclic, then (replacing q, by q, + q2 if necessary) q is either

q+ =((z+xA,+A~),A,A,)

or

q ~ =((z+xA,+n:,)+yA1,A:)

for some A, E H ‘( A x B). (Of course, x = 0 when B is cyclic.) Proof: Write Sq(q) = M. q; comparing coefficients of z shows M has the form (z g;). As Sq’ is a derivation, we have (Sq’(M)+M*)q=O as in Lemma 20; we calculate this as (M, + M,) M, q2 = 0. Since H*(A x B)/(y) is a polynomial ring, one of these three factors lies in ( y). We will write q as we did before Lemma 10 and write Mj= yi y + N, (N, E H’(A)). If we compare coefficients of y in the entries of ,Sq(q)= M. q, obtain in four relations,

(1) %(Q,)=N,Qz

(3) fMQ2) = N,Q,

(2)

(4) L~=y,Qz+N,L,.

L:=Y,Q~+N,L

Suppose first that q2E (y), i.e., Q2 = 0. Then (2) forces L: = N1L,, making L, a multiple of L4. If we replace q, by q1 + q2, we replace L2 by L, + L,, so we may assume L, = 0. Also, Q2 = 0 and (1) force Q, to be a square (Lemma 21) say Q1 = L& Then q = (z + L& yL,) is of type q+. So we may suppose q2$ ( y). If also M1 $ (y), then M, + M, E ( y). Since the substitution of q1 + q2 for q, substitutes M, + Mz for M,, we may assume M, E (y) after all. So we have N, = 0. Then (1) makes Q, = Li again, and (3) makes Qz = L,(L, + N,) for some L5~ H’(A). Then (4) shows L,(L, + N,) = y2L,(L, + N,), so that L4 = y2L5 or N2 + y2L,. This at least makes q2=(L5+y)(Ls+Nz+(l+y,)y)=~,~,. Now, if Lz=O, then q=q+ and we are done. Otherwise, (2) shows y1 = 1 and L2= L,(L,+ N2), so that L5= L2=L5+N2 (so N,=O). If y2 = 0, then q = (z + Li + yL,, (L, + y)*) = q-. Otherwise, y2 = 1, and (41+42,42)=(z+(Lo+Ls)2,~,~2)=s+. I

20 LEMMA

DAVID J. RUSIN

11. The conclusion

qf Lemma 10 also

holds if B is dihedral.

Proqf: Again, we compare coefficients of Sq(q) = Mq to conclude Sq(q,)= xq, + M,q, and Sq(q.J = M,q,. By Lemma 23 we write qz = A, A, for some ni E H’(G). We need only worry about the nature

ofq,. The preceding lemma shows that if we restrict q to the abelian maximal subgroup H = ker(x), we may assume q = q+ or q - (if we replace q, by q1 + q2 when necessary). As ker(Res) = (x), we may write q,=z-tLi+xL,

or

q,=z-tL;:

+yA,+xL3

for some L,EH’(G). Then the Steenrod invariance Sq(q,)-xq, E (q2) shows x( L, + LX)’ E (qzf or x(L,, + L,)’ + y/l: E ( q2), respectively. We wili write L, for Lo + L,. We observe that if L E H’(G), we may read any equation mod (L) by restricting that equation to the subgroup ker(L) of index 2. As G is a product of a dihedral group and an elementary one, so is ker(L), unless L = (x). Thus, H*(G)/(L) is a ring without nilpotent elements, in which (the image of) x is not a zero divisor. Consequently, XL: = ML in H*(G) forces L, E (L) as long as L #x. (If L = x, we have L: = M.) Let us begin with the case Res,(q,) =O. Then whether Res(q) = q’ or q-1 we may write q=(z+L:,+xL,,xA,).

Then Steenrod invariance forces X’ L:E (xn,), as shown above-say, XL: = (Mx) n 2. Then as in the preceding pargraph, either L4 E (A,) or A2 =x. In the first case, write L, = MA, and replace q, with q1 + ctq2 to obtain q=(z+xL,+L&xA2)=qf.

In the latter case, q = ((z + xLO + Li) + xL4, x2) = q-. So we may assume Res,(q,) # 0, so neither .4r nor A, = x. Suppose first that Res(q) = q+; then we saw above that Steenrod invariance meant XL: is a multiple of q2. Restricting to H shows that the multiple restricts to zero, so either XL: =0 or XL: ==xq2. Divide by x to obtain L4 = 0 or q2 = Lz. In the former case, LO=L, so q, =z +xL, + Lg, as needed. In the latter case, replace q, by q1 + q2 = (z + Li + xL3) + (Lo + L3)’ = ir7 + XL, + L$. In either case, q = q+. Finally, suppose Res,(q) = q- (and Res,(q,) # 0), so that q, = z + Li + XL, + y/i, and XL: + y/f: E (A, A,), As in the preceding paragraph, we restrict this last equation to H to see

If we restrict (*) to ker( y), we obtain XL: 1 ax(ii, ,4,), and so Li = a,4i A, (both mod (y)). Similarly, Li E (1 + c()Ai,42 mod (x+ y). As c(= 0 or 1, one of these reads L: z 0, so that L, E (0, y, x + y}. If /1,=/i,, then (*) reads x(L,+aA,)2=0, so L,=aA,, and then (qt+aq,,q,)=((zfxL,+L:)fyA,,A:)=q’”‘. So we may assume As nz#xx, A, #A,. If then Lq=O, then (*f reads y,1f=Omod,4,. H*(ker AZ) is a reduced ring, so yl+ I = O+and thus A, =X + y mod A,. + A, = x+y+A,. Then, (ql -t-q2,qJ) = ((z-+-x(L,+A,) As n,#nz, (L,+n,)‘&f,n,)=q”.

So we may assume L, # 0. Reduce f*) mod ,4, to obtain XL: z 0 mod A,, so as in the preceding paragraph, conclude L, = 0 mod n r, so L4=A,. Then (*) reads O=/i,((x+y)n,t-(y~ax)/iz). Now, we noted earlier that L, E {0, y, x + y}, so 0 # L, = AL shows A, = y + bx, so then Ann(/i,)=(y-t-(1 -tfl)x): (,~+Y)(y+Bx)-tfytax)j12~0

mod (.v+ (f +B)x).

SimpIify* and cancel an x, to obtain ( i t a + $f if, z ax mod f y + ( I+ jI)x), As Res,(q2)#0, A,$0 and n2fx, so l-i-a+fi=O, so @=O, i.e., L,=A,=y. Then q,=z+xL3+L:. 1 We remark that the substitution q1 + q1 I- q2 is effected by choosing a new basis for F,jF,- 1. We also observe that Sq(q) =Mq with M=( 6 .,i!.,) or G 3 (Iand so M is not necessarily diagonizable). We summarize the situation with the help of Lemma 18: we may write G/F,= A x B with A elementary-abelian, B cyclic or dihedral, so that with an appropriate choice of basis in El/r’;- l, we have either q=q+ =(aL,L,) or q=q- = (z+ yt,, L:) for some Lifr H’(G/F!) and with z = w2(ps) inflated from ~~((G/F~)/A). (This interpretation is now vaiid even if B = Zz or D,.) Let Y=ker(L,)nker(L,)>,A. If A= YxA,, then G/F,rXx Y, where 2 = A, x B and the cocycle q is inflated from (G/F,)/Y. Then from Lemma 17(c), we conclude G/F,_, E Xx Y, where X is the me-image of 8 in G/F,-, . So, to prove G/F,- 1 has property P2, it will suffice to prove X has property P3. This is not hard, given our description of the cocycie q: let 4 = Resa(g), the cocycle of the extension

We need only this characterization of extensions with property P3. LEMMA 12. Let R= A . B be a group with property P3, and let Z: + X-+ z be a central extension corrmponding to the cocy& (qE,q2)EH2(X)‘. Zj’q2=inf(zB) and Res,(q,)=z,*, then X is again an

22

DAVID

J. RUSIN

extension A + X -+ B with A” and B cyclic or dihedral. In our present circumstances, this extension splits.

We remark that the Stiefel-Whitney classeszA and zs must be properly interpreted if A or B is elementary-abelian, but that the lemma is valid in these casesas well. ProoJ:

Write this central Z: as Z,e, @Z,e,. The extension

corresponds to q, by Lemma 17(b), so the pre-image A” of A u B in X/(e,) is the central extension defined by Res,(q,) = z,.,, that is, 2 is again cyclic or dihedral (as A is). Of course, (X/(e,))/A” E X/A r B is still cyclic or dihedral. Now, the extension (ez) -+ X+ X/(e,) corresponds to q2 = inf(z,). Again by Lemma 17(b), there is a normal copy of 2 in X and B = X/J is again cyclic or dihedral. We would also like to say the extensions splits; however, it is not clear that in general a copy of B may be found in X/(ez). Certainly there is no problem if B is cyclic: If 6 E R generatesB, then any inverse image b E X has the property that (b) Zz N B x Zz is the whole inverse image of B, so in particular b has order exp(B), and so (b) < X maps isomorphically to fi. (If B= {l}, then n=Z, and we need only take b#l.) If 3 is dihedral, we need to use Lemma 5(c), so assume X= G/F, , for some i and X= G/F,. If b and C are generating involutions of B B X, then they map to involutions in G/F,+, to which Lemma 5(c) may be applied. We conclude that there are involutions b and c in X which map to 6u and Cv in X, for some u and v in F, + , IF,. As F,, , /F, 6 Z(G/F,), 6u Cv has the same order as 1;(;and maps onto B under the map 8-+ x/A N B. Thus we may replace B = (6, C) with (bu, Ftl) and assume B is generated by images of involutions in X. Let B = (b, c) d X. This is clearly a dihedral group mapping onto B under X + x. Since (by induction on i) the central involution in (h, T) lies in F,, 1/Fi, it is the image of an element of order 4 in F, + ,/F, , (Lemma 4(a)), and so B has order 2 1BI. Moreover, the central involution of 8 is e, E Zz, while A”n Zz = (e, ), so A n B = { 1}. Thus, X = 2 . B. (There is some diffkulty if 1BI < 8; fortunately, this never occurs, as can be seen by induction (I El = 2 IBJ) except perhaps if i = v. But in this case, the existence of the splitting B< G/F,-, can be seen by inspection below.) 1 We complete the proof of Lemma 9. We wish to show that in the extension F,/‘F,p, -+X+x

THE Z-GROUPS OF RANK

2

23

corresponding to q+ or q-, X has property P3. We use Lemma 12. If 4=4+, rewrite q = (L, L,, z), and take A0 and B for the A and B of the lemma; then all conditions are met. (Note A0 N ZS, with s 6 2, so 2 = Z,, (n = 1,2, or 4) or D, (n =4 or 8).) If instead q = q’- = (z + JJL,, L:), then either A, = { 11, or A, N Z, and A*
usual. It suffices to show that in the extension

the normal subgroup is spanned by squares and commutators. This is easy to check from Table I. Now let 8~ G/F,... 2 be the inverse image of Y. As [G/F,- , , Y] = 1, we ,
We conclude Y acts trivially on Ft/Fl _2, and thus by Lemma Sd on H= FJF, as well. This makes Y lie in the image of K= C,(H), and so in X (this is part of the definition of property P2). Thus, G/F,-, = X. 1 Proof of Lemma 14. We first make the following useful observation: LEMMA

15’. If G/F, is of rank 2 and 0 < i < 1, then G is split metacyclic.

24

DAVID J. RUSIN TABLE I

I. Xelementary tq= (LA W‘IJ) H*(x) = Z,[x. ?; 2, ..‘f

X=2?, X=2:

x=zj II.

B cyclic H*(x) = Z,[z, J’, 1,, ..]/,$ = 0

A=Z,

B dihedral H’(R)=Z,[z,

R=A.B=Z: Z4.G D4.G Z.SXD, 24-Q Z,.D, D,.D, Dd.4 D, x 4

(2, .P) (z+.?y,.P) (z, yi) (z, il.??)

Z,” x z4 zz,. z, D, I Z,, z,, x 4

A=jl]

(z, 2) (2 + .xy, x2) t:* x2)

A=.&

(z, 22) (z, .i* + 4.v) (z. 2 + iv)

z;z, z,.z, Dn.Z, 4. x Z, Z,. D,, Z,.D2,=Z,.D, =(z,xz,),z? D,, Z, D,.Dln=Zn.Ds =(Z,xD,).22 f), D,, &. x D,

A=Z;

III.

q = (X2, p) (2 + XL-,y’) (.xy, .x2) (2, yz) (xy, z(z +x)) (xv, z(z +x + J)) (JCL’, z(x + y)) (XL’,5’) (.ry, zw)

x, y, a,...$y=xy

A=ZZ

(i, y.2) (z, i,i*)

ProofI As G/Fi is of rank 2 and contains Fi+ 1/Fi 2: Zg in its center, it follows that all involutions of G/F, lie in Fi+ I/Fi, and thus by Lemma 4a, lift to elements of order 4 in G/F,_ I. Thus, the only involutions of G/F, ~.1 are those in Fi/Fie 1: G/F,- I is again of rank 2. Then by induction one shows similarly that G/F, is of rank 2 as well. But then as in Lemma 5(c), we see G is metacyclic. Let us show G is split. This is clear if G is abelian; otherwise choose i minimal with F,> G’. Write G= (a, b) with G’d (a). Then the image subgroup (5) < G/Fi has a complement (a> as G/F; is abelian. Clearly the only involutions here are Q(ci) x Q(g> on the one hand and Fj+ $/Ff on the other. Since F,, ,/F,_ , N Z$, it follows in particular that the preimages b’ E G/F,+, of 8 are of order 2.0(8), and that the involution in (h’) is distinct from that in (a’) (with a’ any pullback of ti). This shows G/Fi- 1= (a’) . (6’). Similar arguments then show our conclusion holds for each i, including i = 1: G = (a) . (V’). 1 We remark that the cocycle of the extension

THE2-GROUPSOFRANK

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25

must necessarily have one component (“q2,” in a suitable basis of Fi+ ,/E;) inflated from the Stiefel-Whitney class of B = ( G/Fi+ , )/Ai. The proof is as in Lemma 17(b). Lemma 15’ is more than enough to prove Lemma 14 in the special case where A -+ G/F,+ 1 -+ B with A and B both cyclic. We need to consider the possibility that at least one is dihedral. We will assume first that both contain elements of order 4, so H*(A) and H*(B) have the usual “generic” forms. During this lemma, set G = G/F,+ 1 and let q = (q,, q2) be the (Steenrod-invariant) cocycle of the extension

Step I. We first suppose A is cyclic, so by Lemma 15’ we suppose B is dihedral. LEMMA

16. We may assume q2 = infGIA(zB)mod (inf(x,)).

Proof: There is a peculiar difficulty here: suppose in fact that G = A x B. Then it seemsjust as natural to view G as an extension G = B. A; we will prove in Step II that in fact G/F, has a dihedral normal subgroup B. However, it need not have a normal cyclic A” mapping to A. Thus, we carry out a little song-and-dance to exclude this phenomenon. Write B+
Resg+(q,)=inf$:(z,+).

(*I

Let us first study the possible actions of B on A. The dihedral B is generated by two (non-central) involutions in G. Note in particular that they cannot lie in Fi+ z/Fi+ 1~ Z(G), and thus not in K/F, + 1, SO as in Lemma 12, we may assume they are images of involutions in G/F,. So we can find two involutions r, s E G/F, mapping into these generators. Note that B= (r, s) < G/F, is also dihedral; it maps onto B, has a kernel contained in Fi+ 1/Fi Q Z( G/F,), and does not map injectively to B (the pre-images of the central involution in ZB < F,, 2/Fi+ 1 are of order 4 in F,+,/F,). So in fact, 1171 = 2 IBI. It is also easy to find a cyclic A”< G/F, of order 2 JAI with X= A(F,+ ,/Fi) 4 G/F,. Unfortunately, A” need not itself be normal in G/F,. Write X r 2 x Z,; then the action of each b E B on X can be described with a matrix (j :). As B acts trivially on Fi+ , /Fi < Z(G), we need k = 0. If b2=1, we compute i2+jk=i2z1 modJA[, so that i- +I modIA”l/2.

26

DAVID

J. RUSIN

In particular, r and s act as f 1 on A = A/.&?. Our assumption of nontriviality then requires at least one of r or s to act as - 1 on A. If r acts trivially, but not s, then G’ = A . B+, where the generator 7%of B+ acts as - 1 on A. Observe that this makes (G+)‘=@(A), and @(G’)=@(A) @(B+). By Lemma 15’, we can write c’ = U. V with U and V cyclic, and with ResG+(q2) = infG+;.,,(z, ), so it will suffice to show U=A. Clearly, @(G’)=@(U)@(I/), and CJ~(C+)‘=@(A), so @(U)=@(A). Then if U#A, we would have A=(a) and U=(cr.rs); this would yield both (a ‘t-s)* E Q(U) = @(A)
= infg: ,4 Resi+(zB) = Resg+ inf$,(z,)

so that q2 = infG,,Bzg mod ker Resgi = (xe).

1

This gives q2 approximately. Can we obtain it exactly? Well, H*(G) may be computed from the Serre spectral sequence of A -+ G -+ B. As each H’(A) E Z,, we have trivial action and E, - H*(A) 0 H*(B) is generated as a ring by E,*.‘g H*(B), E?’ = Z2. yA, and EF* = Z, .zA. This last element must persist to E, , as it is represented by the cocycle q, (we know Res,(q)#O and Res,(q,)=O, so Res,(q,)#O). Thus, the only possible non-trivial differential is d2(xA) E H*B; but as the extension is split, inf g is an injection, so Ez,’ u ET.‘. Thus, E, = E, and with an additional relation of H*(G) = Z,C y.4, z.4 ; xs, Y,, -,lh; = XBYE,> the form yt = y,(crx, + /IJ~). Note that G+ = ker(x,), and as noted we may take zA = ql. So we have q = (zA, z-B+ x,L) for some L E H l(c). Since q is Steenrodinvariant, we may compare the coefficients of zA and zs in Sq’(q,) = aq, + bq, to conclude Sq’(q,) = x,q,, and so x,L: = 0. Yet it is already clear from the spectral sequence that xs is not a zero divisor (this also follows since Resg+ is onto by inspection). Thus, L: =O. Writing L = axg + by8 + cy,.,, we would obtain (cya )* = (ax, + by,)* = (axB + byB) x8 smce y,2 = xs y,, . hence q2 = zs + -~B(~Y.yA) + (CYAJ2. We then use Lemma 18 to write G = A’ f B with q, = infe,,.(z,).

Clearly

THE i&GROUPS

OF RANK

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27

Res,(q,) #O as above, so Res,(q,)=z,,. Then Lemma 12 completes the argument. We point out that the conclusion of the argument had this form: Res,(q,) was easily zero, but we needed to replace A by A’ in order to obtain precisely q2 = inf(z,). In the next step, we will need to alter A even to obtain Res,(q,) = 0; but then the conclusion q2 = inf(z,) will be forced. Step II. We prove Lemma 14 when c = A . B with A dihedral of order at least 8. By Lemma 12, it will suffice to show that we may arrange to have qz = inf(z,) and Res,(q,) = zA. First, note that q does not restrict to zero on SZZ(A)= A n (F,+,/F,+ *) N Z2. By choosing a suitable basis of F, + ,/Fi, we may then assume q2, but not ql, restricts to zero on l2Z(A). Let us first suppose B acts trivially on H*(A). Then the Serre spectral sequenceof the extension A + G -+ B has E2 term H*(A) 0 H*(B), which is generated by &?.’ N H*(B), ,!?(?)I2N’(A), and Z~E ET2. I claim this sequencecollapses. Certainly E2.O persists to E, for dimension reasons,As the extension G= A . B is split, we even have ET%*N E:,‘, so that d2 I E-f.’ =O; hence E’$’ persists as well. FinalIy, we saw q, restricted non-trivially to &Z(A), so Res,(q,) =z,~ + L2 for some LE H’(A). Thus, q, E H’(G) represents the element za + L’ E E:‘; as we have seenL2 E E:-‘, it follows that zA persists to E, as well. Thus, when necessary,we may write H*(G) =

z2[x,,

I’B, ZB, -VA?.YA,-7,f]/,$S=-~&,

with an additional relation y: -t-xA JI~~E (Z?x, + Z, yA). (Z,x, + Z2 yB). (( y: -t-xA yA) cannot involve zs E E$,O as zs restricts non-trivially to 51ZB < G, but none of the other terms do,) As Res,,,(q,) = 0, Res,(q,) = L2 for some L E H’(A). As above, this L is restricted from some L E H’(G). Thus, given the preceding description of H*(G), we may write qz=L2+Q+OLZB with Q~inf H’(B) u (Z,.u, + Z, y,,,), and a = 0 or 1. Of course, as q2 cannot restrict to 0 on F,, ,/F. , + I 4 Q(G), we must have c(= 1. Thus, if we let Q’=Q-txsL, then q2 = (z,r,+ .xBL + L2) + Q’. As in Lemma 18, we may then write q2 = & + Q’, and G = A’ +B, with & = infcia, (z& Note that Q’ is a sum of products of two linear forms, one of which is zero on A (and of course on @p(B),< @G), which may thus be viewed as inflated from C/A’ as well (as A’ d A + = A . @B). So, if we replace A by A’, we may assume L=O, that is, Res,(q,) =O.

28

DAVID J. RUSIN

Then the Steenrod invariance of q forces that of Res,(q), and hence of Res,(q,). As this restricts non-trivially to LIZA, Res,(q,)= z + Ly; invariance makes L, = 0. That is, Res,(q, ) = zA, as we desire. (Thus, in the presentation of H*(G) given before, we may take q, for zA.) Invariance also makes Sq(q2)=L,q, + L,q, for some L,eH’(c). As q, = zA and q2 = z,+ Q’, we compare the coefficients of z,, and zg in the presented ring H*(G) to conclude Sq(q,) =xsq2, and so Sq(Q’) =x,Q’. Writing Q’ =xsM, + yBA4, (Mic H’(G) are zero or else restrict nontrivially to A-that is, Mie E$‘), we are forced to have x,44: = y,M:. But this is not even possible in the associated graded ring E,: if B is cyclic, xs= 0 and we would need Mi = 0, and so M, = 0 as EF* is a reduced ring; if B is dihedral, E$ * is the free H*(A)-module on xs and y,, so M: = M: = 0, and again M, = M, = 0. Either way, Q’ = 0, and so q2 = 0. As noted in the first paragraph of Step II, this is sufficient. Now suppose B acts non-trivially on H*(A). Certainly B leaves fixed the Stiefel-Whitney classes x and z of the faitful respresentation of A, so the only possible action is to exchange ~1 and x + y. Let B’= C,(H*(A)), so [B : B’] = 2. Note that we may write G’ = A B’= A’. B’ with Res,(q,) = 0, by the first part of Step II. (One needs to modify the proof if B’ is elementary-abelian.) As in the first part of Step II, we see Res,(q,) = L*. Note that L is then B-invariant. Since A+ = A . fiZB is certainly normalized by B, and A’ d A + is ker(L), it follows that A’ is normalized by B, too, so G= A’. B. As before, we conclude from Res,(q,) = 0 that Res,.(q,) = zA.. We replace A’ by A, and try to show Res,(q,) = 0 forces q2 = inf,(z,). We have already seen that Resc(q2) =inf,(z,.), so that q2 = inf(z,.) + iL for some L E H’(G), where G’ = ker(i) (that is, 1,= inf(l,*).) I claim that 3&Llies in the image of inf,,G that is, IL represents an element of Ey. As 1.~ EO;‘, the only other possibility is that LE EL0 and ~L#OE EL’. Indeed, choose L’ E H’(G’) with Res,(L’) = y; thus L’ corestricts to an element L” in H’(G) which then restricts on G’ to L’ + L’“, and then on A to y + +v”= x. Hence L” represents the non-zero element of Et’, so that L = L” mod Eke = im(inf). But AL” = I Cor(L’) = Cor(Res(;l) L’) = 0, as Res(J.)= 0. Thus, q2 =inf(z,+ A,L,) for some LB~ H’(B). A glance at a spectral sequence then shows that the Steenrod invariance makes Sq(q,) = Lq, for some L E E$‘. Restricting now to B< G shows Sq(z,+ 1,L,) = L . (zs + I,L,); comparing coefficients of zg gives R, Li = 0. If Bis cyclic, then A,L,EH~(B)uZ-Z’(B)=O, and q2=0. Ifinstead Bis dihedral, then since we saw E.,L, is nilpotent, ABLB= 0 again, so q2 = 0 still. In any case, q again has the desired form, and so by Lemma 12 we are done.

THE'i?-GROUPSOFRANK

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29

Step III. We assumed in the first two steps that A and B had exponent at least 4. Now in general, G 2 Fj+ 3/Fj+ , N Zi, and so it is not difficult to show that both A < G and B g c/A contain elements of order 4. The only difficulty can be if G = G/F,_, (that is, i = I- 2). We described these groups in Table I and noted that in every case, B had exponent at least 4; A did as well, except for the five cases with A 2: D,, and A non-central. Let B’ = C,(A) as in Step II, and I?’ = A B’ = A x B’. (Note G’ contains F,F,- I 6 Z(G).) Thus, G’ is a group with property Pl. Its inverse image G’
4. OTHER DIRECTIONS There are several ways we would seek to continue this work; we shall discuss each of them briefly here. They are (1) (2) (3) (4) (5)

the calculation of G from G; a description of the groups with normal, non-central 4-group F, the calculation of H*(G) for the 2-groups of rank2; an explanation of the situation for odd primes; expectations for groups of higher rank or non-p-groups.

(1) We have calculated the groups G fairly precisely; the author has improved the description by giving all possible actions of one cyclic or dihedral group on another. In particular, it is possible to identify all involutions in G. The calculation of H*(G) has been given approximately during Lemma 14. More accurate results have been obtained by following the procedure of [S], using the Eilenberg-Morre spectral sequence to find H*(G/Fi) for each i. This has the advantage that it identifies most generators as inflated from G/F,-, , so that their action on the various involutions may be calculated. The remaining generators turn out to be two-dimensional and can be viewed as the components (ql, q2) of the extension cocycle that describes G as a quotient of a group G* with

30

DAVID J. RUSIN

property P3. (In fact, G is actually a Baer sum of G* with one of a fairly small number of groups.) This enables us to calculate all of H*(G) and the restriction maps H2(G) -+ N2(Z2) to every involution in G. Since G will have rank 2 if and only if each such restriction carries either 4, or q2 to a non-zero class, we see that the possible pairs 4 may be computed readily. However, we hesitate to do this at the present time, since it is difhcult to decide when two such constructed groups G are isomorphic. Moreover, we Iack a succinct description of these groups. (2) It is not yet clear how to describe these groups. One may show that they contain a unique normal 4-group unless G = D,. Z, or D, . Q,. However, even if we could strengthen this show G contained, in most cases, a unique 4-group, it is not clear how to use this to our advantage; see [S]. Taking our cue from the metacyclic groups [8] and the amalgamated products D, . Qm, we find it not unreasonable to conjecture. Conjecture 1. G N X/F,, where X is a group of rank 2 with a central 4-group F, and Z, = F,, # F. This would imply Conjecture 2. Gjl2ZG has property P3.

It appears the two are actually equivalent, as we can specify the cocycles q E N*(G) needed to create X from G (and F,, ‘v Z,).

(3) We mentioned the calculation of H*(G) in (I). If G has a central 4-group, we can calculate H*(G) from the Serre spectral sequence fairly easily: higher differentials are computed knowing H*(G) as an algebra over the mod-2 Steenrod algebra. It would be preferable to compute N*(G) as such an algebra as well. Unfortunately~ the technique of [8] leaves a few ambiguities in the structure constants (all moduio inf N*(G/F, _ ,)). In each individual case, arguments can probably be made to resolve the di~culties, but no general techniques have evolved. More interesting is the case when G does not contain a central 4-group. We might conjecture that H*(G) has depth 1 unless G is dihedral. In addition, when G has a non-central 4-group, the variety by H*(G) need not be an affine plane (equivalently, a line in P’). Among the possibilities found is the union of live lines in P’ meeting at a point. (4) If p > 3, the p-groups of rank 2 are mostly metacyclic; see [2]. The groups with a central $-group are indeed metacyclic, which is the only option available in Theorem A. It would be of some interest to adapt the present proof to the case of odd primes and thus to show, for example, that these groups are split metacyclic. In this case, H*(G) is known; cf. [8].

THE 2-GROUPSOF RANK 2

31

Those groups without a center of rank 2 are also mostly metacyclic; however, Blackburn’s classification allows for other groups as well. Here, our conjectures for p = 2 are not correct; G/QZG 2: Zz x Z,, which for odd p is a form not allowed by Theorem A. (5) It is of some interest to describe the groups G of p-rank 2 which are not p-groups. Thus one would be able to find all groups with H*(G) of Krull dimension 2, and to find all possibilities for this subvariety of P’. Of course, the enumeration of these groups in general is probably hopeless--even the calculation of the simple groups of 2-rank 2 is difficult. We do observe that the Sylow 2-subgroups P of G are on our list of 2-groups; thus N,(P)/C,(P) is effectively computable. In most cases, in fact, we see N,(P) = C,(P), that is, Aut(P) is a 2-group. This is due to an exact sequence 0 + Hom(P/F,, F,/F,_ ,) -+ Aut(P/F,

,) -+ Aut(P/F,) x Aut(Fi/Fj_ ,)

and induction. One needs to check that there is a characteristic subgroup of order 2 in P/FimI, so that the image of Aut(P/F,_,) in Aut(F,/F,_,) is of order at most 2. (Usually, if P/F, = A ’ L?,we can show that the maximal cyclic subgroup of A is characteristic in P/Fl.) In another direction, one might hope for a description of p-groups of larger p-rank. The analysis of the present paper seemsfairly effective in this problem as well. The missing ingredient is a description of those groups without a normal abelian subgroup of maximal rank (cf. Lemma 1). Still, it is not implausible to suggest Conjecture 3. If G is a p-group with the same rank n as its center, then G/S2ZG-((A,.A,).A,).... .A, witheachAjcyclicordihedral.

REFERENCES 1. J. ALPERIN, Centralizersof abeliannormal subgroups,J. Algebra 1 (19&l), 110-113. 2. N. BLACKBURN. Generalizations of certain elementary theorems on p-groups, Proc. London Mah. Sot. (3) 11 (1961),l-22. 3. D. GORENSTEIN,“Finite Groups,” Harper & Row, New York, 1968. 4. B. HUPPERT, “Endliche Gruppe,” Springer-Verlag, New York, 1967. 5. M. KONVISSER, Z-groups which contain exactly three involutions, burl. 2. 130 (1973k 19-30. 6. A. R. WILLIAMS, On 2-groups with no normal abelian subgroups of rank 3, and their occurrence as Sylow 2-subgroups of finite simple groups, Trans. Amer. Math. Sot. 150 (1970), 345-408. 7. D. QUILLEN, The spectrum of an equivariant cohomology ring, I, II, ,4nn. qf Math. 94 (1971). 549-602. 8. D. RESIN, The mod-2 cohomoiogy of metacyclic 2-groups J. Pure A&. Algebra 44 (1987), 315-327.

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