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 | x^p= y^qk = 1, yxy⁻¹= x^r0i

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 y^qk0 fixes x and y^ql does not fix x for every 0 ≤ l < k₀. It is easy to check that ord_p(r₀) = q^k0. Note that G⁰ = hxi

and M = hx, y^qk0i is the maximal abelian subgroup of G containing G⁰. Moreover, M = hxy^qk0i is cyclic of order pq^k−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(y^qk0)^j for some i, j with p - i, then K contains an element hxi = G⁰. Hence, by Theorem II.1.5, the elements

e_j= e(G, M, hy^qji),

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

e_k0= (1 −ex)gy^qk0 and e_j= (1 −x)(fe y^qj− gy^qj−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 G_k = G. Then G_k/hy^qji ' G_j for k0 ≤ j ≤ k. In particular, Q[Gk]fy^qj' Q[Gk/hy^qji] ' Q[Gj]. Denote

C_k0= Q[Gk]exgy^qk0 and C_j= Q[Gk]ex(fy^qj− gy^qj−1) for k₀< j ≤ k which are commutative algebras. Then

Q[Gk] ' Q[Gk−1] ⊕C_k⊕ Q[Gk]e_k and

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

Q[Gk−1] ' Q[Gk−2] ⊕C_k−1⊕ Q[Gk−1]e_k−1 so we have

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

C_k−1' C_k−1 and Q[Gk]e_k−1' Q[Gk−1]e_k−1.

Continuing this process, we can obtain that

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

Q[Gk] ' Q[Gk]fG⁰_k⊕

k−1

M

k0

Q[Gj]e_j⊕ Q[Gk]e_k.

Now, G_k0 is a group in Theorem II.2.17(i). By Lemma II.4.2,

Q[Gk0]e_k0 ' M_qk0(Q(εp)^σ)

for σ ∈ Aut(Q(εp)) given by σ (εp) = εp^r0. If all components Q[Gj]e_j, k₀< j < k, and Q[Gk]e_k are division rings, then Q[Gk] has only one matrix component, and vice versa.

In this case, G_k has ND. It suffices to consider only Q[Gk]e_k 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 = ord_m(r). Let

G_m,r= hA, B | A^m= 1, Bⁿ= A^t, BAB⁻¹= A^ri be a group and let

A_m,r= (Q(εm), σr, εs)

be a cyclic algebra where σ_r(εm) = ε_m^r. Then G_m,r can be embedded in a division ring D if and only if A_m,r is a division algebra; and then {∑ia_ig_i∈ D | a_i∈ Q, gi∈ G} ' Am,r

where the isomorphism is obtained by the correspondence: A ↔ ε_m, 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 = 2^αm⁰, s = 2s⁰, where α ≥ 2, m⁰, s⁰, n⁰ are odd numbers; gcd(n,t) = gcd(s,t) = 2, and r ≡ −1 (mod 2^α).

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

αp= v_p(m), n_p= ord_mp−αp(r), δp= ord_mp−αp(p)

and µ_p is the minimal integer satisfying r^µp ≡ p^µ0 (mod mp^−αp) for some integer µ⁰ where v_pis the p-adic valuation, in other words, v_p(m) is the largest nonnegative integer

such that p^vp(m) divides m. Let

δ⁰= µpδ_p/n_p.

Theorem II.4.10 ([Ami55, Theorem 4]). Let m, n, r, s,t be as in Theorem II.4.9. The cyclic algebra A_m,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, (p^δ0− 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 G_kto fit all notations in Theorem II.4.9. Since gcd(p, q^k−k0) = 1, choose p⁰such that pp⁰≡ 1 (mod q^k−k0). Let

A= xy^p0qk0, B= y, m= pq^k−k0 and n= q^k0. Then |A| = m. By Chinese Remainder Theorem, let r ∈ Z be such that

r≡ r₀ (mod p) and r≡ 1 (mod q^k−k0).

We have

BAB⁻¹= yxy^p0qk0y⁻¹= x^r0y^p0qk0 = x^ry^p0qk0r= A^r. Here, gcd(m, r) = 1 since p - r0. Let

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

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

s= q^k−k0 and t= p.

Note that r^qk0 ≡ r^q₀^k0 ≡ 1 (mod p) since y^qk0 fixes x. Thus, r^qk0 ≡ 1 (mod m) and n | q^k0. 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= q^k0. Therefore,

Bⁿ= y^qk0 = A^p= A^t.

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

be rewritten by A and B. It follows that G_k' H. In other words, G_k= hA, B | A^m= 1, Bⁿ= A^t, BAB⁻¹= A^ri.

In particular,

G⁰_k= hA^si and Z (Gk) = hA^ti.

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

Recall that e_k = e(G_k, M, 1) where M = hx, y^qk0i = hxy^qk0i is the maximal abelian subgroup of G_k containing G⁰_k.

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

Proof. We follow all notations as above. First of all, we note that M = hAi since |A| = m= |M|. Moreover, A^y= A^B= A^r. Observe that |G_k/M| = q^k0 = n. For 0 ≤ i < n, we fix preimages of yⁱ∈ G_k/M by yⁱ under the canonical homomorphism G → G/M. For 0 ≤ i, j < n, we have yⁱy^j= yⁿy^{i+ j−n}where 0 ≤ i + j − n < n. Moreover, yⁿ= Bⁿ= A^t= A^p. By Proposition II.1.10, we obtain that Q[Gk]e_k' (Q(εm), σr, εm^p). Now, εm^p= εs and the result follows.

As we mentioned, our group G_k 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 ord_p(r₀) = 2. Thus, we can assume r₀= p − 1 and r ≡ r₀ ≡ −1 (mod p). It follows that r ≡ −1 (mod m) since q^k−k0= 2 and r ≡ 1 ≡ −1 (mod q^k−k0). By Theorem II.4.10, Q[Gk]e_k= Q[G2]e₂is a division ring. In this case,

G₂' Q_4p = hx, y | x^p= 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 = q^k0, m = pq^k−k0. In particular, mhas two prime divisors p and q. For p, we have α_p= v_p(m) = 1 and n_p= ord_mp−αp(r) = ord_qk−k0(r) = 1; for q, we have αq= v_q(m) = k − k₀and n_q= ord_mq−αq(r) = ord_p(r) = q^k0. Note that p > 2 since G_k is nonabelian. Thus, the only possibility is the case (2)(a) since p6= q. Hence, we only have to check when

gcd(q, (p^δ0− 1)/s) = 1 happens without the situation n = s = 2.

We compute δ⁰. Since α_p= 1 and r ≡ 1 (mod q^k−k0), we have µp= 1. Moreover, n_p= 1. Then we get

δ⁰= δp= ord_qk−k0(p).

Recall the q-adic valuation v_q. It follows that gcd(q, (p^δ0− 1)/s) = 1 if and only if v_q((p^δ0− 1)/s) = 0 if and only if

v_q(p^{ordqk−k0(p)}− 1) = v_q(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]e_k is a division ring if and only if v_q(p^{ordqk−k0(p)}− 1) = k − k₀.

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

(i) If q 6= 2, then v_q(p − 1) = 1.

(ii) If q = 2, then either k = 2 and G2' Q_4p, or k> 2 and p ≡ 5 (mod 8).

Proof. Note that ord_p(r₀) = q^k0 and r₀^p−1≡ 1 (mod p). Thus, q^k0| p − 1. In other words, v_q(p − 1) ≥ k₀. By assumption, Q[Gj]e_jare division rings for k₀+ 1 ≤ j ≤ k. We first as-sume q 6= 2. Then, by Lemma II.4.12, we have v_q(p^δ0j−1) = j −k₀where δ⁰_j= ord_qj−k0(p) for each j. When j = k₀+ 1, we obtain δ⁰_j= 1 and v_q(p − 1) = 1 ≥ k₀. In particular, k₀= 1 and y^qfixes x.

Assume q = 2. Recall that n = 2^k0 and s = 2^j−k0 for the group G_j. When j = k₀+ 1, we obtain s = 2. If k₀ > 1, then n 6= s = 2 and we have v₂(p^δ0j− 1) = j − k₀ = 1 by Lemma II.4.12. However, we also have δ⁰_j= 1 and v₂(p^δ0j− 1) = v₂(p − 1) ≥ k₀> 1, a contradiction. Thus, k₀= 1 and n = 2. In this case, y^qfixes x.

If k = 2, then we can obtain that G2' Q_4p. Assume k ≥ 3. When j = k0+ 2, we have s= 2²6= n = 2. So v₂(p^δ0j− 1) = j − k₀= 2 by Lemma II.4.12 again. On the other hand, δ_j⁰= ord₂2(p) which divides 2. If δ⁰_j= 2, we have p ≡ 3 (mod 4). Then p²≡ 1 (mod 8) and v₂(p^δ0j− 1) = v₂(p²− 1) ≥ 3, a contradiction. Thus, δ⁰_j = 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 + q^ab and q - b. Then ordq^a+d(c) = q^d and v_q(c^qd− 1) = a + d for every d ∈ Z≥0.

Proof. Computing c^q, we have c^q= (1 + q^ab)^q= ∑^q_i=0 ^q_i
(q^ab)ⁱ. Note that v_q( ^q_i
) = 1 for i 6= 0, q. Thus, c^q= 1 + q^a+1b₁for some b₁∈ N and q - b1. Continuing this process, we can obtain c^qd = 1 + q^a+db_dfor some b_d∈ N and q - bd, for every d ∈ Z≥0 where b₀= b.

Thus, v_q(c^qd− 1) = a + d. Moreover, ord_qa+d(c) | q^d. If d = 0, then ord_qa+0(c) = 1.

Assume d > 0. Note that c^qd−1 = 1 + q^a+d−1b_d−1 6≡ 1 (mod q^a+d) since q - bd−1. It follows that ord_qa+d(c) = q^d, as desired.

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

(i) If q 6= 2, then v_q(p − 1) = 1.

(ii) If q = 2, then either k = 2 and G₂' Q_4p, or k> 2 and p ≡ 5 (mod 8).

In particular, the matrix component is isomorphic to M_q(Q(εp)^σ) where the automor-phism σ ∈ Aut(Q(εp)/Q) given by σ (εp) = ε^rp⁰.

Proof. One direction is actually Lemma II.4.13 and we prove the opposite direction. By assumption, k₀ = 1. Let q 6= 2 and v_q(p − 1) = 1. Then p = 1 + qb for some b ∈ N and q - b. We can conclude that δ_j⁰= ord_qj−k0(p) = q^j−k0−1 and v_q(p^δ0j− 1) = j − k₀for k₀+ 1 ≤ j ≤ k by Lemma II.4.14. Thus, by Lemma II.4.12, Q[Gj]e_j 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' Q_4p. In this case, Q[Q4p] has only one matrix component isomorphic to M₂(Q(εp+ ε_p⁻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 δ_j⁰= ord₂j−k0(p) = 2^j−k0−2and v2(p^δ0j− 1) = j − k₀for k0+ 2 ≤ j ≤ k by Lemma II.4.14.

So Q[Gj]e_jis 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 = C_poC_q^kbe 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]e_kis a division ring , then G_kis embedded in A_m,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 C₇oC3^k= hx, y | x⁷= y^3k = 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 C3^k= hx, y | x¹³ = y^3k= 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].