CALCULATING THE GENUS OF A DIRECT PRODUCT OF CERTAIN NILPOTENT GROUPS

The Mislin genus G(N) of a finitely generated nilpotent group N with finite commutator subgroup admits an abelian group structure. If N satisfies some additional conditions —we say that N belongs to N1— we know exactly the structure of G(N). Considering a direct product N1 × · · · × Nk of groups in N1 takes us virtually always out of N1. We here calculate the Mislin genus of such a direct product.


Introduction.
By N 0 we denote the class of finitely generated infinite nilpotent groups N with 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 N is a nilpotent group and we consider the short exact sequence 0 → T N → N → F N → 0, where T N is the torsion subgroup of N and F N the torsionfree quotient, then N ∈ N 0 if and only if T N is finite and F N is free abelian of finite rank.If additionally (1) T N is abelian; (2) 0 → T N → N → F N → 0 splits on the right, so that N is the semidirect product for an action ω : F N → Aut(T N) of F N on T N; (3) the action ω satisfies ω(F N) ⊆ Z Aut(T N), where Z denotes the centre, then we say that N ∈ N 1 ⊂ N 0 .Note that (finite) direct products of members of N 1 inherit properties (1) and (2) above, but not, in general, property (3).
Recall from [3] that, given (1), ( 3) is equivalent to requiring that for each ξ ∈ F N, there exists u ∈ Z, prime to exp T N, such that ξ • a = ua for all a ∈ T N (T N is here written additively).Now if t is the height of ker ω in F N (meaning that t is the largest positive integer m such that ker ω ⊆ mF N ), then we know from [3] 1.1.Theorem.G(N ) ∼ = (Z/t) * /{±1}, for N ∈ N 1 .
In [5] the authors calculate G(N k ) for N ∈ N 1 with F N cyclic and k ≥ 2, obtaining the following theorem.

Theorem.
For N ∈ N 1 with F N cyclic and for any k ≥ 2, we obtain G(N k ) from G(N ) by factoring out those residues m mod t such that m ≡ i mod p λi 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(N 1 × • • • × N k ), where N 1 , . . ., N k ∈ N 1 .
In the third section we will show that, if the direct product involves a group N j ∈ N 1 with a non-cyclic torsionfree quotient F N j , 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 N 1 ∈ N 1 with F N 1 not cyclic and N 2 ∈ N 0 , we have G(N 1 × N 2 ) = 0.
In the case where the direct product only involves groups N 1 , N 2 , . . ., N k ∈ N 1 , all with a cyclic torsionfree quotient F N i , 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 a in some given group.
Let 2 ≤ x ≤ min(s 1 , s 2 ) + 1, and suppose that Then we say that a x (1) obstructs an isomorphism between (T N 1 ) p and Similarly we speak of a x(2) obstructing an isomorphism.We call the order of obstruction (of (T N 1 ) p , (T N 2 ) p ) the maximum of the orders of all generators of (T N 1 ) p , (T N 2 ) p obstructing an isomorphism.Of course, the order of obstruction is independent of the choice of direct sum decomposition of (T N 1 ) p , (T N 2 ) p .
In the course of the fourth section we will prove our main theorem, namely, a for a ∈ T N i .Define P to be the set of prime divisors p of t such that there are distinct r, v ∈ {1, . . ., k} for which the following conditions hold: (1) exp(T N r ) p = exp(T N v ) p ; (2) u v ∈ u r , u r ∈ u v , where u r , u v are viewed as elements of (Z/ exp(T N v ) p ) * ; (3) On those generators of (T N r ) p and (T N v ) p that obstruct an isomorphism between these two torsion groups, the actions of ξ v , ξ r are trivial.This means that u v ≡ u r ≡ 1 modulo the order of obstruction.
Then we obtain G(N It is also interesting to note that the condition (2), namely, However if p = 2 and m ≥ 3, the group (Z/2 m ) * is not cyclic, and in this case we cannot replace the given condition by the weaker condition |u r | = |u v |, 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 N 1 , N 2 such that exp(T N 1 ) = exp(T N 2 ), since we insist in Definition 1.5 that x ≥ 2.

Some preliminary results.
Recall from [2], [3] the following exact sequence (where Here T = T (N ) is the set of prime divisors of n = exp T N, QN = N/F ZN , F ZN being the free center of N , e = exp QN ab , and T -Aut N is the semigroup of self T -equivalences of N .Recall also how θ acts.For any T -automorphism ϕ, θ(ϕ) is the residue class modulo ±1 of det ϕ, ϕ being restricted to F ZN. (In [1], [2] it is shown that a T -automorphism sends F ZN to itself).Moreover in [5] the authors show the following.

Lemma. An endomorphism ϕ of N induces a commutative diagram
and only if α is an automorphism and ψ is a T -automorphism.

Lemma.
(i) For all ξ ∈ F N and for all a ∈ T N, we have α(ξ , for all ξ ∈ F N and for all a ∈ T N. Then we may find ϕ : N → N making 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 N 1 with a non-cyclic torsionfree quotient.

Proof of Theorem 1.4:
we have the following exact sequence: where e = lcm(e 1 , e 2 ).We show that we can realize the residue class of any m, prime to e, by some T -automorphism φ of N 1 × N 2 .In other words we show that for any m prime to e, there exists a commutative diagram where α is an automorphism and φ, ψ are T -automorphisms, such that det ψ = m.
Choose a basis for F N 1 such that and (in additive notation) where l remains to be determined Then we have only to verify the compatibility condition (Lemma 2.3) for .
We now have one of the following three possibilities: (1) If e is even and t 1 is even, then m is odd, m − 1 is even and thus Moreover, by the same argument as in Theorem 1.1 of [4], we can show that in any case u 2  1 ∈ u 2 .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.

The genus of a direct product of
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

Proposition. Consider the following commutative diagram:
where α is an automorphism and ϕ, ψ are T -automorphisms.Let p | t.
If m = v, then This means that, for all m, v, r ∈ {1, . . ., k}, for all i ∈ {1, . . ., s v }, and for all ∈ {1, . . ., s r }: We now assert that ∀ j ∈ {1, . . ., k} ∃!f (j) ∈ {1, . . ., k} such that p α 1(j) q(f (j)) for some q.Indeed, since p does not divide the determinant of α p , there certainly exists such a f (j).And if we suppose that there exist v, v such that p α where (t j ) p stands for the p-part of t j .Hence it would follow that det ψ = det(β ij ) ≡ 0 mod (t) p .However, this is impossible, since p det ψ (ψ being a T -automorphism).The assertion assures us that the matrix of α p , reduced mod p, looks like Note that the above also implies that exp(T N j ) p ≤ exp(T N f (j) ) p .Thus we have set up a map f : {1, . . ., k} −→ {1, . . ., k} : j −→ f (j).We claim that this map f is a bijection.Indeed, if we suppose that f (j) = f (j ) = v, meaning that p α 1(j) q(v) for some q ∈ {1, . . ., s v } and that p α 1(j ) q (v) for some q ∈ {1, . . ., s v }, then we would get from (4.2) that and thus If j = j , this would again imply that det ψ ≡ 0 mod (t) p , yielding a contradiction.So we conclude that f is a bijection.The above means that the following situation is impossible for the matrix of α p (each column being reduced mod p): q(f (j)) for some q implies (using (4.1)) that and by (4.2) that and thus Moreover, restating (4.1), (4.2), we get that, for all r ∈ {1, . . ., k}, for all ∈ {1, . . ., s r }, and for all q ∈ {1, . . ., s f (j) }, If we take r = j in (4.5), then we know that β f (j)r ≡ 0 mod (t r ) p (see (4.4)) and thus, again using (4.5), If we take r = j and m = f (r) in (4.6) (note that we then still have m = f (j), since r = j), then we get But of course u since, due to (4.3), this congruence is true mod exp(T N r ) p .Thus we get Let us reformulate what we have already proved.We have a bijection q(f (j)) for some q ∈ {1, . . ., s f (j) }, and f is uniquely determined by this property.Indeed, q(f (j)) a (r) |; ∀ r = j, ∀ q ∈ {1, . . ., s f (j) }, ∀ ∈ {1, . . ., s r } (4.12) So we have already established (3) in the statement of the proposition, which is simply (4.11) and (4.12).
We now distinguish two cases.
We then see, from (4.9) (4.10), that, for all j, Thus the matrix of ψ, reduced mod (t) p = p λ , looks like the identity matrix, and thus det ψ ≡ 1 mod p λ .
Case 2: f (j) = j for some j ∈ {1, . . ., k}.Suppose that Note that, since f is a bijection, we are certainly able to form a "closed chain" for f as above.Note also that the above implies that Thus all exponents are equal, meaning that all p-torsion groups appearing in a "closed chain" have the same p-exponent.This already establishes (1) in the statement of the proposition.Moreover, in this chain, we also have From this it follows that (4.14) β jys β ysys−1 . . .β y1j ≡ 1 mod (t j ) p , while all the other β's in the columns j, y 1 , y 2 , . . ., y s are congruent to 0, modulo (t) p .Moreover, from the above, it is also clear that u f (j) = u y1 ∈ u j and that u j ∈ u f (j) = u y1 if we view these as elements of (Z/ exp(T N j ) p ) * .This establishes (2) in the statement of the proposition.Repeating this process of constructing "closed chains" for f , until we have exhausted the whole of {1, . . ., k}, we obtain a number of congruences of the form (4.14), while all the other β's are congruent to zero modulo (t) p .
Combining all this together, we get It only remains to verify that Of course, since it is sufficient to prove that u f (j) ≡ 1 modulo the order of obstruction of (T N j ) p , (T N f (j) ) p .
We now have two cases.Either the order of obstruction is the order of some a x(j) or else it is the order of some a x(f (j)) .Suppose that the order of obstruction is equal to |a x(j) |.Then it is sufficient to prove that there exists v = f (j) such that p α x(j) i(v) , for some i ∈ {1, . . ., s v }.Indeed, this would imply that |α x(j) i(v) a x(j) | = |a x(j) |, which in turn would imply, by (4.12), that u f (j) ≡ 1 mod |a x(j) |, as required.Reduce the matrix of α p modulo p, and consider the columns of a 1(j) , a 2(j) , . . ., a x−1(j) .We may suppose that , for all v = f (j) and all i ∈{1, . . ., s v }.
Indeed, otherwise we would obtain that |α for some < x, which would lead by (4.12) to u f (j) ≡ 1 mod |a (j) |, and thus mod |a x(j) |.But of course, p det α p .So we can find q 1 , q 2 , . . ., q x−1 such that q 1 = q 2 = • • • = q x−1 and such that p α Moreover, we then know that Of course, due to the supposition on the orders and the order of obstruction, we are now unable to find for a x(j) a generator a qx(f (j)) , such that p α x(j) qx(f (j)) with q x = q 1 , q 2 , . . ., q x−1 , implying that |a qx(f (j)) | ≥ |a x(j) |.This means that, since p det α p , there exists v = f (j) such that p α In other words, in the column of a x(j) in the matrix of α p , reduced modulo p, we must have a non-zero number on a row outside α(T N f (j) ) p , which is what we had to prove.
Suppose on the other hand that the order of obstruction is equal to |a x(f (j)) |.It is then sufficient to prove that there exists r = j such that Indeed, by (4.11), it would then follow that u f (j) ≡ 1 mod |a x(f (j)) |.We will use the following notations: (T N j ) p = the direct product of all (T N i ) p , except (T N j ) p pr j = the projection onto the (T N j ) p -component pr = the projection onto the rest, that is, onto (T N j ) p .
We may suppose that the second components in these sums have smaller order than |a 1(f (j)) |, . . ., |a x−1(f (j)) | in (T N j ) p , respectively.Indeed, otherwise there would exist r = j such that for some q ∈ {1, . . ., x − 1} and some , which would imply by (4.11) that u f (j) ≡ 1 mod |a q(f (j)) | and thus also mod |a x(f (j)) |, which is what we wished to prove.But then we need that the first components of these sums have orders ) So H q (for q = 1, . . ., x − 1) is a subgroup of (T N j ) p of order |a q(f (j)) |.Moreover, we claim that Then there exist λ 1 , λ 2 (where we may assume that either p λ 1 or p λ 2 ) such that But then it is easy to see that this yields a contradiction with the injectivity of α p .We now have We see that in the quotient we have to factor out cyclic subgroups of orders |a 1(f (j This quotient thus has exponent < |a x(f (j)) |.It is then easily seen that, since the order of the first component in the above sum in (T N j ) p is < |a x(f (j)) |.This means that the second component of the sum has order equal to |a x(f (j)) | in (T N j ) p , so that there exists r = j such that |α (r) which is what we wished to prove.This concludes the proof of Proposition 4.1.
Thus we know that any m in im θ must fulfil the conditions given in Theorem 1.6.We now proceed to the converse; that is, we show that any such m is realizable.

Proposition.
Let m ∈ (Z/t) * with m ≡ i mod p λi i , for all i ∈ {1, . . ., k} where i = 1 or −1.Additionally if i = −1, then suppose that there exist r, v ∈ {1, . . ., k} such that r = v and (1) exp(T , which satisfy the compatibility condition of Lemma 2.3, such that det ψ ≡ m mod t.It will follow that any endomorphism ϕ of (N 1 ×• • •×N k ), compatible with α and ψ, will be a T -automorphism realizing m.We will determine α completely, but we will only determine the matrix of ψ mod t.
Fix a particular p among the prime divisors of t and let The idea is the following.If m ≡ 1 mod p λ , we will construct α p as the identity on (T N 1 × • • • × T N k ) p and the matrix of ψ, reduced mod p λ , should look like the identity matrix.
If m ≡ −1 mod p λ , then α p should map (T N r ) p to (T N v ) p and vice-versa as much as possible.This means that we map the respective generators with the same order (for example a 1(r) and a 1(v) ) on each other.On the generators of (T N r ) p obstructing an isomorphism, and on later generators, we define α p to be the identity, and likewise for (T N v ) p .On all other p-torsion subgroups (T N j ) p for j = r, v, we also define α p to be the identity.The matrix of ψ, reduced mod p λ , will look like the identity matrix outside the r th and v th columns.These two columns contain β rv and β vr such that u v ≡ u βvr r , u r ≡ u βrv v mod exp(T N v ) p .Then, as we will show, det ψ will be congruent to −1 mod p λ .
Case 2: m ≡ −1 mod p λ .We then know that there exist r, v ∈ {1, . . ., k}, r = v such that α p = Id for the generators of (T N r ) p and (T N v ) p obstructing an isomorphism and for later generators; α p (a j(r) ) = a j(v) and α p (a j(v) ) = a j(r) for the other generators of (T N r × T N v ) p ; and let β rr ≡ 0 mod p r β rv and β vr be chosen such that (which is always possible, by hypothesis) Remark that in both cases we can solve all the congruences (by the Chinese Remainder Theorem) and that β ij will be determined mod t j , so that the entries of the matrix of ψ will be determined mod gcd(t 1 , . . ., t k ) = t.We will now check that α and ψ, as constructed above, satisfy the compatibility condition (Lemma 2.3).
Case 1: m ≡ 1 mod p λ α(ξ s • a q ) = ψ(ξ s ) • α(a q ) (a q ∈ T N q ) (q = s) ⇐⇒ a q = u βsq q a q , and the latter holds since ⇐⇒ u s a s = u βss s a s , and the latter holds since β ss ≡ 1 mod p s .
Case 2: m ≡ −1 mod p λ If {q, s} = {v, r}, we get similar equations to those above.If {q, s} = {v, r}, we have for generators a j(r) , a j(v) that are mapped under α on each other: and the latter relations all hold.
If {q, s} = {v, r}, we have for generators a j(r) , a j(v) on which α is defined as the identity (that is, generators obstructing an isomorphism or later generators): and the latter relations all hold, by (3) above.
Finally we look at det ψ.For each p|t, we have However in the second case we know From this it follows that β vr β rv ≡ 1 mod p r = p v , so in either case we have det ψ ≡ m mod p λ .
Thus det ψ ≡ m mod 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 N 1 .

Example
Of course, we can explicitly describe the groups in the genus of N 1 × N 2 , using the descriptions of the groups in G(N ), for N ∈ N 1 , given in [3].
Of course, Corollary 1.7 gives us the simple formula for G(N 1 × N 2 ) in this case, since only the prime 2 is involved; and the value of G(N 1 × N 2 ) is unaffected by whether we can find ψ ∈ T -Aut(N 1 × N 2 ) 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 p as 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.
Note that this is indeed a generalization of Theorem 1.3.For ifN 1 = N 2 = • • • = N k ,then P would consist of all primes p i dividing t, so that G(N k ) would be obtained from (Z/t) * by dividing out those residues m mod t which are congruent to 1 or −1 mod p λi i for all p i .