Solvable SSN Groups with ND, II

在文檔中 整群環的 Jordan 分解與冪零分解 (頁 88-95)

II. Nilpotent Decomposition in Integral Group Rings

II.4 Nilpotent Decomposition for Non-nilpotent SSN Groups

II.4.2 Solvable SSN Groups with ND, II

Now we focus on SSN groups G in Theorem II.2.17(ii). Write the presentation of G as follows:

G= hx, y | xp= yqk = 1, yxy−1= xr0i

for some distinct primes p, q, integer k ≥ 2, 2 ≤ r0≤ p − 1 and hyi acts non-faithfully on hxi. Then there is a positive integer 1 < k0< k such that yqk0 fixes x and yql does not fix x for every 0 ≤ l < k0. It is easy to check that ordp(r0) = qk0. Note that G0 = hxi

and M = hx, yqk0i is the maximal abelian subgroup of G containing G0. Moreover, M = hxyqk0i is cyclic of order pqk−k0. If K is a normal subgroup of M such that (M, K) is a strong Shoda pair of G, then Q[G]e(G, M, K) is commutative if and only if K ⊇ G0 by Lemma II.1.4. Moreover, if K contains xi(yqk0)j for some i, j with p - i, then K contains an element hxi = G0. Hence, by Theorem II.1.5, the elements

ej= e(G, M, hyqji),

for k0≤ j ≤ k, are the primitive central idempotents corresponding to non-commutative Wedderburn components. More precisely,

ek0= (1 −ex)gyqk0 and ej= (1 −x)(fe yqj− gyqj−1) for k0< j ≤ k and

Q[G] ' Q[G] eG0⊕ Q[G]ek0⊕ · · · ⊕ Q[G]ek.

Now, we fix p, q, r0 and denote Gk = G. Then Gk/hyqji ' Gj for k0 ≤ j ≤ k. In particular, Q[Gk]fyqj' Q[Gk/hyqji] ' Q[Gj]. Denote

Ck0= Q[Gk]exgyqk0 and Cj= Q[Gk]ex(fyqj− gyqj−1) for k0< j ≤ k which are commutative algebras. Then

Q[Gk] ' Q[Gk−1] ⊕Ck⊕ Q[Gk]ek and

Q[Gk] ' Q[Gk−2] ⊕Ck−1⊕Ck⊕ Q[Gk]ek−1⊕ Q[Gk]ek. If we denote x, y to be generators of Gjfor j < k, then

Q[Gk−1] ' Q[Gk−2] ⊕Ck−1⊕ Q[Gk−1]ek−1 so we have

Q[Gk] ' Q[Gk−2] ⊕Ck−1⊕ Q[Gk−1]ek−1⊕Ck⊕ Q[Gk]ek. By the uniqueness of Wedderburn decomposition, we have

Ck−1' Ck−1 and Q[Gk]ek−1' Q[Gk−1]ek−1.

Continuing this process, we can obtain that

Cj' Cj and Q[Gk]ej' Q[Gj]ej for k0≤ j < k. Thus,

Q[Gk] ' Q[Gk]fG0k

k−1

M

k0

Q[Gj]ej⊕ Q[Gk]ek.

Now, Gk0 is a group in Theorem II.2.17(i). By Lemma II.4.2,

Q[Gk0]ek0 ' Mqk0(Q(εp)σ)

for σ ∈ Aut(Q(εp)) given by σ (εp) = εpr0. If all components Q[Gj]ej, k0< j < k, and Q[Gk]ek are division rings, then Q[Gk] has only one matrix component, and vice versa.

In this case, Gk has ND. It suffices to consider only Q[Gk]ek by the reduction process above. To check that Q[G]ek is a division ring or not, we will use S.A. Amitsur’s result on [Ami55].

Theorem II.4.9 ([Ami55, Theorem 3]). Let m, r be two relatively prime integers and put s= gcd(r − 1, m), t = m/s and n = ordm(r). Let

Gm,r= hA, B | Am= 1, Bn= At, BAB−1= Ari be a group and let

Am,r= (Q(εm), σr, εs)

be a cyclic algebra where σrm) = εmr. Then Gm,r can be embedded in a division ring D if and only if Am,r is a division algebra; and then {∑iaigi∈ D | ai∈ Q, gi∈ G} ' Am,r

where the isomorphism is obtained by the correspondence: A ↔ εm, B↔ σr.

We list conditions [Ami55, (3C), (3D)] below for the next theorem where m, n, s,t are defined as in the previous theorem:

(3C) gcd(n,t) = gcd(s,t) = 1;

(3D) n = 2n0, m = 2αm0, s = 2s0, where α ≥ 2, m0, s0, n0 are odd numbers; gcd(n,t) = gcd(s,t) = 2, and r ≡ −1 (mod 2α).

We define the following notations. Let p be a fixed prime dividing m. We put:

αp= vp(m), np= ordmp−αp(r), δp= ordmp−αp(p)

and µp is the minimal integer satisfying rµp ≡ pµ0 (mod mp−αp) for some integer µ0 where vpis the p-adic valuation, in other words, vp(m) is the largest nonnegative integer

such that pvp(m) divides m. Let

δ0= µpδp/np.

Theorem II.4.10 ([Ami55, Theorem 4]). Let m, n, r, s,t be as in Theorem II.4.9. The cyclic algebra Am,r = (Q(εm), σr, εs) is a division algebra if and only if either (3C) or (3D) holds and one of the following holds:

(1) n = s = 2 and r ≡ −1 (mod m).

(2) For every q | n there exists a prime p | m such that q - npand that either (a) p 6= 2, and gcd(q, (pδ0− 1)/s) = 1 or,

(b) p = q = 2, (3D) holds, and m/4 ≡ δ0≡ 1 (mod 2).

To use Amitsur’s result, we have to rewrite the presentation of Gkto fit all notations in Theorem II.4.9. Since gcd(p, qk−k0) = 1, choose p0such that pp0≡ 1 (mod qk−k0). Let

A= xyp0qk0, B= y, m= pqk−k0 and n= qk0. Then |A| = m. By Chinese Remainder Theorem, let r ∈ Z be such that

r≡ r0 (mod p) and r≡ 1 (mod qk−k0).

We have

BAB−1= yxyp0qk0y−1= xr0yp0qk0 = xryp0qk0r= Ar. Here, gcd(m, r) = 1 since p - r0. Let

s= gcd(r − 1, m), t = m/s and n= ordm(r).

We compute these three numbers. Since p - r0− 1, we have p - r − 1. Moreover, qk−k0 divides r − 1. We obtain

s= qk−k0 and t= p.

Note that rqk0 ≡ rq0k0 ≡ 1 (mod p) since yqk0 fixes x. Thus, rqk0 ≡ 1 (mod m) and n | qk0. In other words, n is a power of q. On the other hand, rn0≡ rn≡ 1 (mod p). It follows that ynfixes x. By the definition of k0, we have

n= qk0. Therefore,

Bn= yqk0 = Ap= At.

Now, if H is a group generated by the above generators A, B and adding the stated rela-tions, then |H| = mn = pqk. There is an epimorphism from Gk into H since x and y can

be rewritten by A and B. It follows that Gk' H. In other words, Gk= hA, B | Am= 1, Bn= At, BAB−1= Ari.

In particular,

G0k= hAsi and Z (Gk) = hAti.

Observe that gcd(n,t) = gcd(s,t) = 1. So Gk satisfies the condition (3C).

Recall that ek = e(Gk, M, 1) where M = hx, yqk0i = hxyqk0i is the maximal abelian subgroup of Gk containing G0k.

Lemma II.4.11. Q[Gk]ek' (Q(εm), σr, εs) = Am,rfor k≥ 2.

Proof. We follow all notations as above. First of all, we note that M = hAi since |A| = m= |M|. Moreover, Ay= AB= Ar. Observe that |Gk/M| = qk0 = n. For 0 ≤ i < n, we fix preimages of yi∈ Gk/M by yi under the canonical homomorphism G → G/M. For 0 ≤ i, j < n, we have yiyj= ynyi+ j−nwhere 0 ≤ i + j − n < n. Moreover, yn= Bn= At= Ap. By Proposition II.1.10, we obtain that Q[Gk]ek' (Q(εm), σr, εmp). Now, εmp= εs and the result follows.

As we mentioned, our group Gk satisfies (3C). Now, we check the conditions in The-orem II.4.10(1). If n = s = 2, then q = 2, k0= 1 and k = 2. Furthermore, m = 2p and ordp(r0) = 2. Thus, we can assume r0= p − 1 and r ≡ r0 ≡ −1 (mod p). It follows that r ≡ −1 (mod m) since qk−k0= 2 and r ≡ 1 ≡ −1 (mod qk−k0). By Theorem II.4.10, Q[Gk]ek= Q[G2]e2is a division ring. In this case,

G2' Q4p = hx, y | xp= y4= 1, yxy−1= x−1i, the generalized quaternion group of order 4p for odd prime p.

For the case (2) in Theorem II.4.10, we observe that n = qk0, m = pqk−k0. In particular, mhas two prime divisors p and q. For p, we have αp= vp(m) = 1 and np= ordmp−αp(r) = ordqk−k0(r) = 1; for q, we have αq= vq(m) = k − k0and nq= ordmq−αq(r) = ordp(r) = qk0. Note that p > 2 since Gk is nonabelian. Thus, the only possibility is the case (2)(a) since p6= q. Hence, we only have to check when

gcd(q, (pδ0− 1)/s) = 1 happens without the situation n = s = 2.

We compute δ0. Since αp= 1 and r ≡ 1 (mod qk−k0), we have µp= 1. Moreover, np= 1. Then we get

δ0= δp= ordqk−k0(p).

Recall the q-adic valuation vq. It follows that gcd(q, (pδ0− 1)/s) = 1 if and only if vq((pδ0− 1)/s) = 0 if and only if

vq(pordqk−k0(p)− 1) = vq(s) = k − k0. Hence, we have the following

Lemma II.4.12. Assume the condition Theorem II.4.10(1) does not hold. Then Q[Gk]ek is a division ring if and only if vq(pordqk−k0(p)− 1) = k − k0.

Lemma II.4.13. Assume that Q[Gk] has only one matrix component for k ≥ 2. Then yq fixes x and the following holds.

(i) If q 6= 2, then vq(p − 1) = 1.

(ii) If q = 2, then either k = 2 and G2' Q4p, or k> 2 and p ≡ 5 (mod 8).

Proof. Note that ordp(r0) = qk0 and r0p−1≡ 1 (mod p). Thus, qk0| p − 1. In other words, vq(p − 1) ≥ k0. By assumption, Q[Gj]ejare division rings for k0+ 1 ≤ j ≤ k. We first as-sume q 6= 2. Then, by Lemma II.4.12, we have vq(pδ0j−1) = j −k0where δ0j= ordqj−k0(p) for each j. When j = k0+ 1, we obtain δ0j= 1 and vq(p − 1) = 1 ≥ k0. In particular, k0= 1 and yqfixes x.

Assume q = 2. Recall that n = 2k0 and s = 2j−k0 for the group Gj. When j = k0+ 1, we obtain s = 2. If k0 > 1, then n 6= s = 2 and we have v2(pδ0j− 1) = j − k0 = 1 by Lemma II.4.12. However, we also have δ0j= 1 and v2(pδ0j− 1) = v2(p − 1) ≥ k0> 1, a contradiction. Thus, k0= 1 and n = 2. In this case, yqfixes x.

If k = 2, then we can obtain that G2' Q4p. Assume k ≥ 3. When j = k0+ 2, we have s= 226= n = 2. So v2(pδ0j− 1) = j − k0= 2 by Lemma II.4.12 again. On the other hand, δj0= ord22(p) which divides 2. If δ0j= 2, we have p ≡ 3 (mod 4). Then p2≡ 1 (mod 8) and v2(pδ0j− 1) = v2(p2− 1) ≥ 3, a contradiction. Thus, δ0j = 1 and v2(p − 1) = 2. It follows that p ≡ 5 (mod 8).

We are going to prove that the converse of Lemma II.4.13 is also true. Before we prove it, we give an elementary computation.

Lemma II.4.14. Let a, b, c ∈ N be such that c = 1 + qab and q - b. Then ordqa+d(c) = qd and vq(cqd− 1) = a + d for every d ∈ Z≥0.

Proof. Computing cq, we have cq= (1 + qab)q= ∑qi=0 qi(qab)i. Note that vq( qi) = 1 for i 6= 0, q. Thus, cq= 1 + qa+1b1for some b1∈ N and q - b1. Continuing this process, we can obtain cqd = 1 + qa+dbdfor some bd∈ N and q - bd, for every d ∈ Z≥0 where b0= b.

Thus, vq(cqd− 1) = a + d. Moreover, ordqa+d(c) | qd. If d = 0, then ordqa+0(c) = 1.

Assume d > 0. Note that cqd−1 = 1 + qa+d−1bd−1 6≡ 1 (mod qa+d) since q - bd−1. It follows that ordqa+d(c) = qd, as desired.

Proposition II.4.15. Let k ≥ 2. Then Q[Gk] has only one matrix component if and only if yqfixes x and the following holds.

(i) If q 6= 2, then vq(p − 1) = 1.

(ii) If q = 2, then either k = 2 and G2' Q4p, or k> 2 and p ≡ 5 (mod 8).

In particular, the matrix component is isomorphic to Mq(Q(εp)σ) where the automor-phism σ ∈ Aut(Q(εp)/Q) given by σ (εp) = εrp0.

Proof. One direction is actually Lemma II.4.13 and we prove the opposite direction. By assumption, k0 = 1. Let q 6= 2 and vq(p − 1) = 1. Then p = 1 + qb for some b ∈ N and q - b. We can conclude that δj0= ordqj−k0(p) = qj−k0−1 and vq(pδ0j− 1) = j − k0for k0+ 1 ≤ j ≤ k by Lemma II.4.14. Thus, by Lemma II.4.12, Q[Gj]ej is a division ring for k0+ 1 ≤ j ≤ k and Q[Gk] has only one matrix component.

Assume q = 2. If k = 2, then G2' Q4p. In this case, Q[Q4p] has only one matrix component isomorphic to M2(Q(εp+ εp−1)) by [JdR16, Example 3.5.7]. Assume k ≥ 3 and p ≡ 5 (mod 8). Then p = 1 + 22bfor some odd integer b. Thus, we also have that δj0= ord2j−k0(p) = 2j−k0−2and v2(pδ0j− 1) = j − k0for k0+ 2 ≤ j ≤ k by Lemma II.4.14.

So Q[Gj]ejis a division ring for every k0+ 2 = 3 ≤ j ≤ k by Lemma II.4.12.

Finally, since k0= 1 and G1is the group in Theorem II.2.17(i), the matrix component comes from Q[G1]e1 which is isomorphic to the matrix ring described in Lemma II.4.2.

We complete the proof.

RemarkII.4.16. Let G = CpoCqkbe as a group in Theorem II.2.17(ii) with distinct primes p, q and k ≥ 2. If p ≡ 3 (mod 4), q = 2 and k ≥ 3, then Hales and Passi proved that G do not have ND by finding a tricky nilpotent element in [HP17, Theorem 4.1]. However, we still do not know in general whether G has ND or not if Q[G] has more than one matrix component.

Recall that if Q[Gk]ekis a division ring , then Gkis embedded in Am,rby Lemma II.4.11 and Theorem II.4.9. Note that v3(7 − 1) = 1 so we have the following infinite number of groups of odd order which can be embedded in division rings. See also [Ami55, Theo-rem 6].

Corollary II.4.17 ([LP13, Proposition 4.7]). The nonabelian group C7oC3k= hx, y | x7= y3k = 1, yxy−1= x2i for k ≥ 2 are all embeddable in division rings.

According to [Ami55, Theorem 6], the group C7o C9 above is the first group of odd order embedded in a division ring. Note also that v3(13 − 1) = 1. Then we have another infinite number of groups C13o C3k= hx, y | x13 = y3k= 1, yxy−1 = x3i of odd order for k≥ 2. When k = 2, the group C13o C9is the second group of odd order embedded in a division ring, see [Lam01a, (3.7) Theorem].

在文檔中 整群環的 Jordan 分解與冪零分解 (頁 88-95)