Calculating the genus of a direct product of certain nilpotent groups

Publicacions Matem`atiques, Vol 39 (1995), 241–261. CALCULATING THE GENUS OF A DIRECT PRODUCT OF CERTAIN NILPOTENT GROUPS Peter Hilton and Dirk Scevenels Abstract The Mislin genus G(N) of a finitely generated nilpotent group N with finite commutator subgroup admits an abelian group structure. If Nsatisfies some additional conditions —we say that N belongs to N1— we know exactly the structure of G(N). Considering a direct product N1×···×Nkof groups in N1takes us virtually always out of N1. We here calculate the Mislin genus of such a direct product. 1. Introduction. By N0we denote the class of finitely generated infinite nilpotent groups Nwith finite commutator subgroup [N,N]. From [1], [2] we know that the (Mislin) genus G(N), for N∈N 0, may then be given the structure of a finite abelian group. Moreover, if Nis a nilpotent group and we consider the short exact sequence 0 →TN →N→FN →0, where TN is the torsion subgroup of Nand FN the torsionfree quotient, then N∈N 0if and only if TN is finite and FN is free abelian of finite rank. If additionally (1) TN is abelian; (2) 0 →TN →N→FN →0 splits on the right, so that Nis the semidirect product for an action ω:FN →Aut(TN) of FN on TN; (3) the action ωsatisfies ω(FN)⊆ZAut(TN), where Zdenotes the centre, then we say that N∈N 1⊂N 0. Note that (finite) direct products of members of N1inherit properties (1) and (2) above, but not, in general, property (3). 242 P. Hilton, D. Scevenels Recall from [3] that, given (1), (3) is equivalent to requiring that for each ξ∈FN, there exists u∈Z, prime to exp TN, such that ξ·a=ua for all a∈TN (TN is here written additively). Now if tis the height of ker ωin FN (meaning that tis the largest positive integer msuch that ker ω⊆mF N), then we know from [3] 1.1. Theorem. G(N)∼ =(Z/t)∗/{±1}, for N∈N 1. Moreover it is proved in [4] that 1.2. Theorem. For N∈N 1with FN not cyclic, G(Nk)=0for any k≥1, where Nkis the kth direct power of N. Now, if exp TN =n=pm1 1...p ms s,p1<p 2<··· <p s,mi≥1, we know that tmust have the form t=pλ1 1pλ2 2...p λs s, with 0 ≤λi<m i (i=1,2,... ,s); we write pλi it,i=1,2,... ,s. We also write T(N) for the collection of primes (p1,p 2,... ,p s). In [5] the authors calculate G(Nk) for N∈N 1with FN cyclic and k≥2, obtaining the following theorem. 1.3. Theorem. For N∈N 1with FN cyclic and for any k≥2,we obtain G(Nk)from G(N)by factoring out those residues mmod tsuch that m≡imod pλi i, i=±1,i=1,2,... ,s. In this paper we will generalize these calculations to obtain a result for the genus of a direct product, G(N1×···×Nk), where N1,... ,N k∈N 1. In the third section we will show that, if the direct product involves a group Nj∈N 1with a non-cyclic torsionfree quotient FNj, then the genus of the direct product is trivial. Note that this is a generalization of Theorem 1.2. In fact, we prove 1.4. Theorem. For N1∈N 1with FN1not cyclic and N2∈N 0,we have G(N1×N2)=0. In the case where the direct product only involves groups N1,N 2,... ,N k∈N 1, all with a cyclic torsionfree quotient FNi,an important role is played by the so-called generators that obstruct an isomorphism. In the definition below, we write |a|for the order of the element ain some given group. The genus of a direct product of nilpotent groups 243 1.5. Definition. Let N1,N 2∈N 1and p∈T=T(N1×N2). Suppose that (TN1)p=a1(1)⊕a2(1)⊕···⊕as1(1), where exp(TN1)p=|a1(1)|≥|a2(1)|≥···≥|as1(1)|; (TN2)p=a1(2)⊕a2(2)⊕···⊕as2(2), where exp(TN2)p=|a1(2)|≥|a2(2)|≥···≥|as2(2)|. Let 2 ≤x≤min(s1,s 2) + 1, and suppose that |a1(1)|=|a1(2)|,|a2(1)|=|a2(2)|,... ,|ax−1(1)|=|ax−1(2)|. Then we say that ax(1) obstructs an isomorphism between (TN1)pand (TN2)pif either |ax(1)|=|ax(2)|or x=s2+1 ≤s1. Similarly we speak of ax(2) obstructing an isomorphism. We call the order of obstruction (of (TN1)p,(TN2)p) the maximum of the orders of all generators of (TN1)p,(TN2)pobstructing an isomorphism. Of course, the order of obstruction is independent of the choice of direct sum decomposition of (TN1)p,(TN2)p. In the course of the fourth section we will prove our main theorem, namely, 1.6. Theorem. Let N1,N 2,... ,N k∈N1with G(Ni)∼ =(Z/ti)∗/{±1}. Set t= gcd(t1,... ,t k)=pλ1 1...p λs s. Let FNi=ξiwith ξi·a=uiafor a∈TNi. Define Pto be the set of prime divisors pof tsuch that there are distinct r,v∈{1,... ,k}for which the following conditions hold: (1) exp(TNr)p= exp(TNv)p; (2) uv∈ur,ur∈uv, where ur,uvare viewed as elements of (Z/exp(TNv)p)∗; (3) On those generators of (TNr)pand (TNv)pthat obstruct an isomorphism between these two torsion groups, the actions of ξv,ξr are trivial. This means that uv≡ur≡1modulo the order of obstruction. Then we obtain G(N1×···×Nk)from (Z/t)∗by factoring out the residue class of −1and those residues mmod tsuch that m≡1modpλi ifor all pi/∈P m≡1or −1modpλi ifor all pi∈P. 244 P. Hilton, D. Scevenels Note that this is indeed a generalization of Theorem 1.3. For if N1= N2=···=Nk, then Pwould consist of all primes pidividing t, so that G(Nk) would be obtained from (Z/t)∗by dividing out those residues mmod twhich are congruent to 1 or −1modpλi ifor all pi. 1.7. Corollary. Assume further that t=pλ. Then, with no further hypothesis, G(N1×···×Nk)∼ =(Z/t)∗/{±1}. It is also interesting to note that the condition (2), namely, ur∈uv, uv∈uris in fact equivalent to |ur|=|uv|, if the group (Z/exp(TNv)p)∗ is cyclic. This group is indeed cyclic if p= 2. However if p= 2 and m≥3, the group (Z/2m)∗is not cyclic, and in this case we cannot replace the given condition by the weaker condition |ur|=|uv|, as Example 4.4 will show. We anticipate that the notions of generators obstructing an isomorphism and the order of obstruction to an isomorphism may prove to be of interest beyond the scope of this paper. Notice that we only apply these notions to groups N1,N2such that exp(TN1) = exp(TN2), since we insist in Definition 1.5 that x≥2. 2. Some preliminary results. Recall from [2], [3] the following exact sequence (where N∈N 0) TAut Nθ −→ (Z/e)∗/{±1}→G(N)→0. Here T=T(N) is the set of prime divisors of n= exp TN,QN = N/FZN,FZN being the free center of N,e= exp QNab, and T-Aut N is the semigroup of self T-equivalences of N. Recall also how θacts. For any T-automorphism ϕ,θ(ϕ) is the residue class modulo ±1 of det ϕ,ϕ being restricted to FZN. (In [1], [2] it is shown that a T-automorphism sends FZN to itself). Moreover in [5] the authors show the following. 2.1. Lemma. Let ϕ:N→Nbe an endomorphism. Then ϕinduces ψ:FN →FN.Ifϕ(FZN)⊆FZN, then det(ϕ|FZN) = det ψ. So, for a T-automorphism ϕof N,θ(ϕ) is in fact the residue class of det ψ.ForN∈N 0satisfying conditions (1) and (2) of N1, we also have the following ([5], [6]). The genus of a direct product of nilpotent groups 245 2.2. Lemma. An endomorphism ϕof Ninduces a commutative diagram 0−−−−→TN −−−−→N−−−−→FN −−−−→0   α   ϕ  ψ 0−−−−→TN −−−−→N−−−−→FN −−−−→0 and ϕis a T-automorphism if and only if αis an automorphism and ψ is a T-automorphism. 2.3. Lemma. (i) For all ξ∈FN and for all a∈TN, we have α(ξ·a)=ψ(ξ)·α(a). (ii) Suppose that a diagram 0−−−−→TN −−−−→N−−−−→FN −−−−→0   α  ψ 0−−−−→TN −−−−→N−−−−→FN −−−−→0 is given, such that α(ξ·a)=ψ(ξ)·α(a), for all ξ∈FN and for all a∈TN. Then we may find ϕ:N→Nmaking a commutative diagram as in the previous lemma. We call (i) above the compatibility condition. 3. The genus of a direct product, involving a group in N1 with a non-cyclic torsionfree quotient. Proof of Theorem 1.4: Set T=T(N1)∪T(N2). Since N1×N2∈N 0, we have the following exact sequence: TAut(N1×N2)θ −−−−→(Z/e)∗/{±1}−−−−→G(N1×N2)−−−−→0 where e= lcm(e1,e 2). We show that we can realize the residue class of any m, prime to e, by some T-automorphism φof N1×N2. In other words we show that for any mprime to e, there exists a commutative diagram 0−−−−→TN1×TN2−−−−→N1×N2−−−−→FN1×FN2−−−−→0 α    φ  ψ 0−−−−→TN1×TN2−−−−→N1×N2−−−−→FN1×FN2−−−−→0 where αis an automorphism and φ,ψare T-automorphisms, such that det ψ=m. 246 P. Hilton, D. Scevenels Choose a basis for FN1such that FN1=ξ1,ξ 2,... ,ξ r,ker ω1=t1ξ1,t 2ξ2,... ,t rξr where t=t1|t2|... |trand ω1is the action of FN1on TN1. Let ξi·a=uiafor a∈TN1. Remark that the order of uimodulo exp(TN1) is then ti. Now set α=Id TN 1×TN 2 and (in additive notation)      ψ(ξ1)=mξ1+lξ2,where lremains to be determined ψ(ξj)=ξj(j=1) ψ|FN2=Id FN 2. Then we have only to verify the compatibility condition (Lemma 2.3) for ξ1.Now α(ξ1·a)=ψ(ξ1)·α(a) for all a∈TN1×TN2 if and only if u1a=um 1ul 2afor all a∈TN1, which is equivalent to (3.1) um−1 1ul 2≡1 mod exp(TN1). We now have one of the following three possibilities: (1) If eis even and t1is even, then mis odd, m−1 is even and thus um−1 1∈u2 1; (2) If eis even and t1is odd, then u1∈u2 1, since t1is odd; (3) If eis odd, then t1is odd (because t1|e1|e) and u1∈u2 1, since t1is odd. Moreover, by the same argument as in Theorem 1.1 of [4], we can show that in any case u2 1∈u2. Thus in each of the three cases it is clear that we can always solve (3.1) for l. Moreover det ψ=m, which completes our proof. 4. The genus of a direct product of N1,... ,N kin N1, each Ni having a cyclic torsionfree quotient FNi. Let N1,N 2,... ,N k∈N 1with G(Ni)∼ =(Z/ti)∗/{±1},tibeing defined as in Section 1. Set t= gcd(t1,... ,t k)=pλ1 1...p λs sand set T=T(N1× N2×···×Nk). Suppose that, for p|tand for i=1,... ,k, (TNi)p=a1(i)⊕a2(i)⊕···⊕asi(i) The genus of a direct product of nilpotent groups 247 with exp(TNi)p=|a1(i)|≥|a2(i)|≥···≥|asi(i)|. Let FNi=ξiwith ξi·a=uiafor a∈TNi. To calculate G(N1×···×Nk) for N1,... ,N k∈N 1, we will use the exact sequence TAut(N1×···×Nk)θ −→ (Z/e)∗/{±1}→G(N1×···×Nk)→0, which is valid, since clearly N1×···×Nk∈N 0. In the following two propositions we will give a description of im θ. From these we can then conclude how to use Theorem 1.6 to obtain G(N1×···×Nk). 4.1. Proposition. Consider the following commutative diagram: 0−→ TN1×···×TNk−→ N1×···×Nk−→ FN1×···×FNk−→ 0   α   ϕ  ψ 0−→ TN1×···×TNk−→ N1×···×Nk−→ FN1×···×FNk−→ 0 where αis an automorphism and ϕ,ψare T-automorphisms. Let p|t. Let ψ(ξm)=k j=1 βmjξjfor m=1,... ,k, and let αp(ai(v))= s1  =1 α(1) i(v)a(1) + s2  =1 α(2) i(v)a(2) +···+ sk  =1 α(k) i(v)a(k) for v∈{1,... ,k}and i∈{1,... ,s v}. Then there exists a bijection f:{1,... ,k}−→{1,... ,k}:j−→ f(j), where f(j)is the unique index such that pα1(j) q(f(j)) for some q∈{1,... ,s f(j)}. Moreover, we also have (1) exp(TNj)p= exp(TNf(j))p; (2) uj∈uf(j),uf(j)∈uj, with uj,uf(j)viewed as elements of (Z/exp(TNj)p)∗; (Note that, of course, (1) and (2) become trivial if j=f(j).) (3) uf(j)≡uf(r)≡1mod|α(r) q(f(j))a(r)|for all r=j, for all &∈ {1,... ,s r}and for all q∈{1,... ,s f(j)}. Further, if f(j)=jfor all j∈{1,... ,k}, then det ψ≡1modpλ; and if there exists j∈{1,... ,k}such that f(j)=j, then det ψ≡±1modpλ. In the latter case we need, moreover, the condition uj≡uf(j)≡1modulo the order of obstruction of (TNj)p,(TNf(j))p. 248 P. Hilton, D. Scevenels Proof: The compatibility condition (Lemma 2.3) tells us that ψ(ξm)·α(ai(v))=α(ξm·ai(v)) for m∈{1,... ,k},v∈{1,... ,k},i∈{1,... ,s v}. This yields the following. If m=v, then   uβv1 1α(1) i(v)a(1) +···+  uβvk kα(k) i(v)a(k) =  uvα(1) i(v)a(1) +···+  uvα(k) i(v)a(k). If m=v, then   uβm1 1α(1) i(v)a(1) +···+  uβmk kα(k) i(v)a(k) =  α(1) i(v)a(1) +···+  α(k) i(v)a(k). This means that, for all m,v,r∈{1,... ,k}, for all i∈{1,... ,s v}, and for all &∈{1,... ,s r}: If m=v, then uβvr rα(r) i(v)a(r)=uvα(r) i(v)a(r); If m=v, then uβmr rα(r) i(v)a(r)=α(r) i(v)a(r). Thus If m=v, then uβvr r≡uvmod |α(r) i(v)a(r)|;(4.1) If m=v, then uβmr r≡1mod|α(r) i(v)a(r)|.(4.2) We now assert that ∀j∈{1,... ,k}∃!f(j)∈{1,... ,k}such that pα1(j) q(f(j)) for some q. Indeed, since pdoes not divide the determinant of αp, there certainly exists such a f(j). And if we suppose that there exist v,vsuch that pα1(j) i(v)for some i∈{1,... ,s v} and pα1(j) i(v)for some i∈{1,... ,s v}, The genus of a direct product of nilpotent groups 249 then it follows from (4.2) that ∀m=vu βmj j≡1mod|a1(j)|= exp(TNj)p and ∀m=vuβmj j≡1mod|a1(j)|= exp(TNj)p. If v=v, this would imply that ∀mβ mj ≡0mod(tj)p, where (tj)pstands for the p-part of tj. Hence it would follow that det ψ= det(βij)≡0mod(t)p. However, this is impossible, since pdet ψ(ψ being a T-automorphism). The assertion assures us that the matrix of αp, reduced mod p, looks like a1(j)α(TNf(j))p                        0 0 0 0 ... ... . . .... ... ... * * ... ... . . .... ... ... 0 0 0 . . . 0                        Note that the above also implies that exp(TNj)p≤exp(TNf(j))p. Thus we have set up a map f:{1,... ,k}−→{1,... ,k}:j−→ f(j). We claim that this map fis a bijection. Indeed, if we suppose that f(j)=f(j)=v, meaning that pα1(j) q(v)for some q∈{1,... ,s v}and that pα1(j) q(v)for some q∈{1,... ,s v}, then we would get from (4.2) that ∀m=vu βmj j≡1mod|a1(j)| ∀m=vu βmj j≡1mod|a1(j)|, and thus ∀m=vβ mj ≡0mod(tj)p ∀m=vβ mj≡0mod(tj)p. 256 P. Hilton, D. Scevenels 4.2. Proposition. Let m∈(Z/t)∗with m≡imod pλi i, for all i∈{1,... ,k}where i=1or −1. Additionally if i=−1, then suppose that there exist r,v∈{1,... ,k}such that r=vand (1) exp(TNr)p= exp(TNv)p(that is |a1(r)|=|a1(v)|); (2) uv∈ur,ur∈uv, where ur,u vare viewed as elements of (Z/exp(TNv)p)∗; (3) uv≡ur≡1modulo the order of obstruction of (TNr)p,(TNv)p. Then we can realize m, that is, [m]∈im θ. Proof: We will construct an automorphism α∈Aut(TN1×···×TNk) and a T-automorphism ψ∈TAut(FN1×···×FNk), which satisfy the compatibility condition of Lemma 2.3, such that detψ≡mmod t.It will follow that any endomorphism ϕof (N1×···×Nk), compatible with αand ψ, will be a T-automorphism realizing m. We will determine α completely, but we will only determine the matrix of ψmod t. Fix a particular pamong the prime divisors of tand let p1t1,p 2t2,... ,p ktk,p λt. Set      ψ(ξ1)=β11ξ1+···+β1kξk ... ψ(ξk)=βk1ξ1+···+βkkξk. The idea is the following. If m≡1modpλ, we will construct αp as the identity on (TN1×···×TNk)pand the matrix of ψ, reduced mod pλ, should look like the identity matrix. If m≡−1modpλ, then αpshould map (TNr)pto (TNv)pand vice-versa as much as possible. This means that we map the respective generators with the same order (for example a1(r)and a1(v)) on each other. On the generators of (TNr)pobstructing an isomorphism, and on later generators, we define αpto be the identity, and likewise for (TNv)p. On all other p-torsion subgroups (TNj)pfor j=r,v, we also define αpto be the identity. The matrix of ψ, reduced mod pλ, will look like the identity matrix outside the rth and vth columns. These two columns contain βrv and βvr such that uv≡uβvr r,ur≡uβrv vmod exp(TNv)p. Then, as we will show, det ψwill be congruent to −1modpλ. Case 1: m≡1modpλ. Define αp=Id:(TN1×···×TNk)p→(TN1×···×TNk)pand let βii ≡1modpifor all i∈{1,... ,k} βij ≡0modpjif j=i. The genus of a direct product of nilpotent groups 257 Case 2: m≡−1modpλ. We then know that there exist r, v ∈{1,... ,k},r=vsuch that (1) exp(TNr)p= exp(TNv)p(that is |a1(r)|=|a1(v)|); (2) uv∈ur,ur∈uvviewed as elements of (Z/exp(TNv)p)∗; (3) uv≡ur≡1 modulo the order of obstruction of (TNr)p,(TNv)p. Define αp:(TN1×···×TNk)p→(TN1×···×TNk)pas follows              αp= Id outside (TNr×TNv)p; αp= Id for the generators of (TNr)pand (TNv)p obstructing an isomorphism and for later generators; αp(aj(r))=aj(v)and αp(aj(v))=aj(r) for the other generators of (TNr×TNv)p; and let                        βii ≡1modpifor i=r, v βvv ≡0modpv βrr ≡0modpr βrv and βvr be chosen such that uv≡uβvr r,u r≡uβrv vmod exp(TNv)p (which is always possible, by hypothesis) βij ≡0modpjotherwise . Remark that in both cases we can solve all the congruences (by the Chinese Remainder Theorem) and that βij will be determined mod tj, so that the entries of the matrix of ψwill be determined mod gcd(t1,... ,t k)=t. We will now check that αand ψ,as constructed above, satisfy the compatibility condition (Lemma 2.3). Case 1: m≡1modpλ α(ξs·aq)=ψ(ξs)·α(aq)(aq∈TNq)(q=s) ⇐⇒ aq=uβsq qaq,and the latter holds since βsq ≡0modpq α(ξs·as)=ψ(ξs)·α(as) ⇐⇒ usas=uβss sas,and the latter holds since βss ≡1modps. Case 2: m≡−1modpλ If {q,s}={v,r}, we get similar equations to those above. If {q,s}= {v,r}, we have for generators aj(r),aj(v)that are mapped under αon 258 P. Hilton, D. Scevenels each other: α(ξv·aj(r))=ψ(ξv)·α(aj(r))⇐⇒ aj(v)=aj(v) α(ξv·aj(v))=ψ(ξv)·α(aj(v))⇐⇒ uvaj(r)=uβvr raj(r) α(ξr·aj(r))=ψ(ξr)·α(aj(r))⇐⇒ uraj(v)=uβrv vaj(v) α(ξr·aj(v))=ψ(ξr)·α(aj(v))⇐⇒ aj(r)=aj(r), and the latter relations all hold. If {q,s}={v,r}, we have for generators aj(r),aj(v)on which αis defined as the identity (that is, generators obstructing an isomorphism or later generators): α(ξv·aj(r))=ψ(ξv)·α(aj(r))⇐⇒ aj(r)=uβvr raj(r) α(ξv·aj(v))=ψ(ξv)·α(aj(v))⇐⇒ uvaj(v)=aj(v) α(ξr·aj(r))=ψ(ξr)·α(aj(r))⇐⇒ uraj(r)=aj(r) α(ξr·aj(v))=ψ(ξr)·α(aj(v))⇐⇒ aj(v)=uβrv vaj(v), and the latter relations all hold, by (3) above. Finally we look at det ψ. For each p|t,wehave det ψ= det(βij)≡1modpλif m≡1modpλ −βrvβvr mod pλif m≡−1modpλ. However in the second case we know ur≡uβrv v≡(uβvr r)βrv mod exp(TNr)p. From this it follows that βvrβrv ≡1modpr=pv, so in either case we have det ψ≡mmod pλ. Thus det ψ≡mmod t, which concludes the proof of Proposition 4.2. With these two propositions our main result, Theorem 1.6, is established. We now give an example of how one can use Theorem 1.6 to calculate the genus of a direct product of groups in N1. 4.3. Example. Let N1∈N 1with TN1=Z/9⊕Z/49, FN1=ξ1and ξ1·a=22a for all a∈TN1.Sou1=1+3·7 and t1=3·7 = 21. Let N2∈N 1 The genus of a direct product of nilpotent groups 259 with TN2=Z/9⊕Z/3⊕Z/343 and FN2=ξ2where ξ2·b= 148bfor all b∈TN2.Sou2=1+3·72and t2=3·7 = 21. Thus t= 21. Then P={3}(see Theorem 1.6). Note that 22 ≡148 ≡1 mod 3 (3 being the order of obstruction of (TN1)3,(TN2)3), but plainly 7 ∈ P.Thuswe have to factor out of (Z/21)∗the residue classes of 1,−1 and of those m such that m≡1mod7,m≡±1 mod 3. This means factoring out the group Hgenerated by {−1,m}, where m≡1mod7,m≡−1mod3. Thus H=−1, 8.Thus G(N1×N2)∼ =(Z/21)∗/H ∼ =Z/3. Of course, we can explicitly describe the groups in the genus of N1×N2, using the descriptions of the groups in G(N), for N∈N 1, given in [3]. Finally we give the promised example to show that the condition ur∈ uv,uv∈urcannot be replaced by the weaker condition |ur|=|uv| in Theorem 1.6. That is, we will give an example where 2 ∈ P, although the prime 2 satisfies (1), (3) and the weaker form of (2); and where the compatibility condition of Lemma 2.3 excludes the condition det ψ≡ −1mod2 λ. 4.4. Example. Let N1=x, y |x16 =1,yxy−1=x3 N2=x, y |x16 =1,yxy−1=x5. Then N1∈N 1,with TN1=Z/16 = a1,FN 1=Z=ξ1and ξ1·a1=3a1 N2∈N 1,with TN2=Z/16 = a2,FN 2=Z=ξ2and ξ2·a2=5a2. Moreover t1=4 and t2=4, so t=gcd(t1,t 2) = 4. Let ψ∈TAut(FN1× FN2) be given by ψ(ξi)=βi1ξ1+βi2ξ2,for i=1,2, and let α∈Aut(TN1×TN2) be given by α(aj)=αj1a1+αj2a2,for j=1,2. 260 P. Hilton, D. Scevenels Expressing the condition α(ξi·aj)=ψ(ξi)·α(aj) for i,j=1,2 yields the following equations : 3α11a1+3α12a2=α113β11 a1+α125β12 a2 (1) α21a1+α22a2=α213β11 a1+α225β12 a2 (2) α11a1+α12a2=α113β21 a1+α125β22 a2 (3) 5α21a1+5α22a2=α213β21 a1+α225β22 a2 (4) Now, we have at least one of two cases; either 2 α11 or 2 α21.If 2α11, then we need 2|α21 (otherwise (1) and (2) contradict) so that 2α22 (because 2 det α) and so 2|α12 (otherwise (3) and (4) contradict). However, in this case we obtain          β11 ≡1mod4 β12 ≡0mod4 β21 ≡0mod4 β22 ≡1mod4 So that det ψ≡1mod4. If 2 α21, then analogously 2|α11,2α12 and 2|α22. Here we need          β11 ≡0mod4 β22 ≡0mod4 3≡5β12 mod 16 5≡3β21 mod 16 The two last congruences however have no solution. We thus can conclude that for each ψ∈TAut(FN1×FN2) we have det ψ≡1mod4; and det ψ≡−1 mod 4 is impossible. Of course, Corollary 1.7 gives us the simple formula for G(N1×N2) in this case, since only the prime 2 is involved; and the value of G(N1×N2) is unaffected by whether we can find ψ∈T-Aut(N1×N2) with det ψ≡ −1 mod 4. To obtain a counterexample to the statement of Theorem 1.6 with the weaker version of condition (2), we need to complicate our Example 4.4 by involving another prime pas a factor of t, in addition to the prime 2, and arranging that p/∈P. We would thereby obtain an example in which all the hypotheses of Theorem 1.6 were verified, except that condition (2) is replaced by the weaker version, but the conclusion of the theorem is false. The genus of a direct product of nilpotent groups 261 References 1. G. Mislin,“Nilpotent groups with finite commutator subgroups,” Lecture Notes in Math. 418, Springer-Verlag, 1974, pp. 103–120. 2. P. Hilton and G. Mislin, On the genus of a nilpotent group with finite commutator subgroup, Math. Z. 146 (1976), 201–211. 3. C. Casacuberta and P. Hilton, Calculating the Mislin genus for a certain family of nilpotent groups, Comm. in Alg. 19(7) (1991), 2051–2069. 4. P. Hilton and C. Schuck, On the structure of nilpotent groups of a certain type, Topological Methods in Nonlinear Analysis, Journal of the Juliusz Schauder Center 1(1993), 323–327. 5. P. Hilton and C. Schuck, Calculating the genus of certain nilpotent groups, Bull. Mex. Math. Soc. 37 (1992), 263–269. 6. P. Hilton, Non-cancellation properties for certain finitely presented groups, Quaestiones Math. 9(1986), 281–292. Peter Hilton: Department of Mathematical Sciences State University of New York Binghamton New York 13902-6000 U.S.A. Dirk Scevenels: K.U. Leuven Fakulteit Wetenschappen Departement Wiskunde Celestijnenlaan 200B B-3001 Heverlee BELGIUM Rebut el 5 de Setembre de 1994