Solvable SSN Groups with ND

We are going to prove that if an SSN group in Theorem II.2.17(i) has the property ND, then its rational group algebra has only one Wedderburn component which is not a divi-sion ring. So within this subsection, we assume that G = P o Q where P is an elementary abelian p-group for a prime p, Q is a cyclic p⁰-group which acts faithfully on P, and every nontrivial subgroup of Q acts irreducibly on P. We first study some properties of such groups G. Denote Fpthe finite field of order p.

Lemma II.4.1. Let G = P o Q be as above. Then we have the following.

(i) P is an irreducible Fp[H]-module for every nontrivial subgroup H of Q. In partic-ular, N_G(P₁) = P for each nontrivial proper subgroup P₁of P.

(ii) G⁰= P.

(iii) P is the maximal abelian subgroup containing G⁰. (iv) If P is not cyclic, then |Q| is odd.

Proof. (i) Let 1 6= H ≤ Q. Because P is an elementary abelian p-group, one can regard P as an Fp[H]-module. Since H is a p⁰-group, Fp[H] is semisimple. If P is not an irreducible Fp[H]-module, then P = P₁× P₂where P₁and P₂ are nonzero Fp[H]-submodules. Thus, P₁ are P₂are H-stable. In other words, H acts reducibly on P, a contradiction. It follows that P is an irreducible Fp[H]-module. We can also conclude that every nontrivial proper subgroup of P is not H-stable. In particular, if P₁is a nontrivial proper subgroup of P and 1 6= h ∈ Q, then P₁^h6= P₁, for otherwise P1 is hhi-stable. Thus, N_G(P₁) = P because P is abelian.

(ii) Since G/P ' Q is abelian, we have G⁰⊆ P. Note that G⁰is normal in G and G⁰6= 1.

By (i), we have G⁰= P, for otherwise P = N_G(G⁰) = G.

(iii) Let A ≤ G be abelian and A ⊇ G⁰= P. Pick gx ∈ A for g ∈ P and x ∈ Q. Then h(gx) = (gx)h for each h ∈ P since A is abelian. Since P is abelian, we know that hg = gh and thus hx = xh for every h ∈ P. Because Q acts faithfully on P, we get that x = 1. Thus, A= P.

(iv) Now, assume that P is not cyclic. Suppose that |Q| is even. Let 1 6= g ∈ P and let x ∈ Q be of order 2. Then the element gg^xis fixed by x. In particular, x ∈ N_G(hgg^xi).

Then either hgg^xi = 1 or hgg^xi = P by (i). Thus gg^x= 1 as by assumption P is not cyclic.

Hence, g^x= g⁻¹ and we obtain x ∈ N_G(hgi). This again is impossible by (i). Hence, |Q|

is odd.

Lemma II.4.2. If P is cyclic, then Q[G] has only one matrix component Q[G](1 − eG⁰). In particular, G hasND.

Furthermore, the matrix component is isomorphic to M_|Q|(F) where F = Q(εp)^σ for some σ ∈ Gal(Q(εp)/Q) given by σ (εp) = εⁱ_p. Here, the integer i is given by g^h= gⁱ where P= hgi and Q = hhi. Thus,

Q[G] ' Q[Q] ⊕ M|Q|(F).

Proof. When P is cyclic, P is of order p. Since G is metabelian, we use Theorem II.1.5 to find primitive central idempotents e of Q[G] such that Q[G]e is non-commutative. Let (H, K) be a pair subgroups of G in Theorem II.1.5 such that K 6⊇ G⁰= P (Lemma II.1.4).

Then P ≤ H and H⁰ ≤ K ≤ H. It follows that H⁰ ≤ G⁰∩ K = P ∩ K = 1. So H is abelian and we obtain H = P by Lemma II.4.1(iii). Since P is of order p, we have K= 1. As a consequence, e(G, P, 1) is the only primitive central idempotent such that Q[G]e(G, P, 1) is non-commutative. Moreover, e(G, P, 1) = 1 − eP= 1 − eG⁰. Denote P = hgi and Q = hhi such that g^h = gⁱ for some i. Now, we can conclude from Propo-sition II.1.9 that Q[G]e(G, P, 1) ' Q(εp) ∗ Q ' (Q(εp)/F, σ , 1) ' M_|Q|(F) where Q ' hσ i ⊆ Aut(Q(εp)/Q) such that σ (εp) = εⁱ_p and F = Q(εp)^σ. Hence, Q[G] ' Q[Q] ⊕ M_|Q|(F).

When P is not cyclic, it is more complicated. First, we consider the Wedderburn decomposition of Q[G].

Lemma II.4.3. If P is non-cyclic, then

Q[G] ' Q[Q] ⊕ vM|Q|(Q(εp)) where

v= |P| − 1

|Q|(p − 1) ∈ N.

Proof. The following argument is similar to the proof of Lemma II.4.2. Since G is metabelian, we use Theorem II.1.5 to find non-commutative components Q[G]e for some primitive central idempotents e. Let (H, K) be a pair subgroups of G in Theorem II.1.5 such that K 6⊇ G⁰= P (Lemma II.1.4). Note that P ≤ H and H⁰≤ G⁰∩ K = P ∩ K < P.

Suppose H is nonabelian. We have H⁰6= 1 so P  H. There is an element gx ∈ H for g ∈ P and 1 6= x ∈ Q. Then x ∈ H. It follows that x ∈ N_G(H⁰). By Lemma II.4.1(i), this can not happen. Hence, H is abelian and, moreover, H = P by Lemma II.4.1(iii). Note that P/K is cyclic if and only if |P : K| = p, because P is elementary abelian. Consequently, we only

need to focus on primitive central idempotents e(G, P, K) with K such that |P : K| = p.

We remark here that (P, K) is a strong Shoda pair of G. Thus,

Q[G]e(G, P, K) ' M|Q|(Q(εp)) by Proposition II.1.10.

For convenience, we write down the explicit form of e(G, P, K). Let K ≤ P be such that

|P : K| = p. Then P/K ' C_p. It follows that ε(P, K) = eK−P. Clearly, P ⊆ Cene _G(ε(P, K)).

Let gx ∈ Cen_G(ε(P, K)) for g ∈ P and x ∈ Q. Then we obtain eK^x= eK. By Lemma II.4.1(i) again, x = 1 and Cen_G(ε(P, K)) = P. So Q is a right transversal of CenG(ε(P, K)) in G and we have

e_K= e(G, P, K) =

∑

x∈Q

( eK^x−P).e

Now, we determine how many distinct eK one defines this way. Note that Q also acts on the set S = {K | K ≤ P, |P : K| = p}. Every group in the orbit K^Q leads to the same idempotent e_K. If K₁^Q and K₂^Q are in distinct orbits, then K₁^x6= K₂^y for every x, y ∈ Q. Thus, fK₁^xfK₂^y= eP and then (fK₁^x−P)(fe K₂^y−P) = 0. It follows that ee _K1e_K2 = 0. In particular, we have e_K1 6= e_K2. Hence, the number of orbits K^Q is precisely the number of distinct e_K. The number |S| of subspaces of dimension n − 1 in P ' Fⁿ_p is the number of subspaces of dimension 1 and this is (pⁿ− 1)/(p − 1). Since each orbit K^Q has size

|Q|, we have v = |S|/|Q| = (|P| − 1)/((p − 1)|Q|) distinct orbits. Therefore, we have v distinct primitive central idempotents e₁, . . . , e_v such that Q[G]ei' M_|Q|(Q(εp)) for each i. Finally, ^L_iQ[G]ei has dimension v|Q|²(p − 1) = |G| − |Q|. Since |Q| = |G/G⁰|, it follows that these components Q[G]eifor i = 1, . . . , v are all non-commutative components of Q[G].

Proposition II.4.4. Assume that P is non-cyclic. If G has ND, then Q[G] has only one matrix component isomorphic to M_|Q|(Q(εp)). In this case, |P| − 1 = |Q|(p − 1) and |Q|

is a prime number.

Proof. Let G = P o Q where P and Q be as above. Assume P is non-cyclic and G has ND. Since Q is not normal in G, there are s ∈ G and y ∈ Q such that s⁻¹ys6∈ Q. Then α = (1 − y)s bQ= s(1 − y^s−1) bQis a nonzero nilpotent element of Z[G]. Write y^s−1 = g₀x for some 1 6= g₀∈ P and x ∈ Q. Then α = s(1 − g0) bQ. Let K be a subgroup of P such that P/K is cyclic and let e = e(G, P, K) be a primitive central idempotent of Q[G] in the proof of Lemma II.4.3. Because, by assumption, G has ND, eα ∈ Z[G]. It follows that es⁻¹α ∈ Z[G] which implies (1 − g0)e bQ∈ Z[G]. Now, supp((1 − g0)e) ⊆ P so (1 − g₀)e must have integer coefficients. Thus,

(1 − g₀)e = (1 − g₀)

∑

x∈Q

( eK^x−Pe) ∈ Z[P].

Since (1 − g₀) eP= 0, we have

(1 − g₀)

∑

x∈Q

Ke^x∈ Z[P].

If g₀∈ K^xfor some x, then (1 − g₀) eK^x= 0. Computing the coefficient of 1, it follows that

|{x ∈ Q | g₀6∈ K^x}|

|K| ∈ Z.

If |{x ∈ Q | g0 6∈ K^x}| = 0, then g₀ ∈ K^x for all x ∈ Q. However, ^T_x∈QK^x is a proper subgroup of P and it is normal in G. So g₀∈^T_x∈QK^x= 1 by Lemma II.4.1(i), a contra-diction. Thus, |{x ∈ Q | g₀6∈ K^x}| is a positive integer greater than or equal to |K|. In particular, |Q| ≥ |K|.

According to Lemma II.4.3, |P| − 1 = v|Q|(p − 1) for some v ∈ N. Now, we have

|P| − 1 ≥ v|K|(p − 1). Dividing by |K| on both sides, we get p −_|K|¹ ≥ v(p − 1). Since 0 < _|K|¹ < 1, it follows that p − 1 ≥ v(p − 1). Hence, v = 1. So, by Lemma II.4.3, Q[G]

has only one matrix component, as desired.

Finally, we claim that |Q| is prime. To see this, we first note that every nontrivial subgroup of Q is not normal in G. Otherwise, there will be an abelian subgroup of G properly containing P = G⁰which contradicts to Lemma II.4.1(iii). Thus, we can replace α by α = (1 − y)scQ₁ in the above proof where Q₁ is a subgroup of Q such that |Q₁| is the smallest prime divisor of |Q|. Then one also obtains |Q₁| ≥ |K|. If Q 6= Q₁, then

|Q| ≥ |Q₁|²≥ |K|²= |P|²/p²≥ |P| since |P| ≥ p². It follows that |Q| > |P| since p - |Q|.

However, this leads to |P| − 1 = |Q|(p − 1) > |P|(p − 1) ≥ |P|, a contradiction. Thus, Q= Q₁.

Combining Lemma II.4.2 and Proposition II.4.4, we can conclude the following result.

Corollary II.4.5. Let G be a group in Theorem II.2.17(i). If G has ND, then Q[G] has only one matrix component.

Before we end this subsection, we generalize SSN groups P o Q with non-cyclic P and ND.

Lemma II.4.6. If G = Cⁿ_po Cq is a group for some distinct primes p and q such that pⁿ− 1 = (p − 1)q and C_q acts nontrivially on Cⁿ_p, then G has SSN. In particular, Q[G]

has only one matrix component isomorphic to M_q(Q(εp)).

Proof. For convenience, we write P = Cⁿ_p and Q = C_q. We first note that this nontrivial action must be faithful because every nontrivial element of Q generates Q. Let N be a proper subgroup of P such that Q acts on N. We claim that N = 1. Note that P is an Fp [Q]-module and N is a sub[Q]-module. Since Fp[Q] is semisimple, there exists a complement

submodule H. Thus, P = N × H and Q acts on H. If N has an element n such that n^x6= n for some x ∈ Q, then n^x6= n for all 1 6= x ∈ Q and the orbit n^Q consists of q nontrivial elements. Thus, N \ {1} contains n^Q. But |N| − 1 ≤ pⁿ⁻¹− 1 < 1 + p + · · · + pⁿ⁻¹ = q, a contradiction. So Q acts trivially on N. If N 6= 1, then H is a proper subgroup of P.

By the same argument again, we can conclude that Q acts trivially on H. Thus, Q acts trivially on P, a contradiction. Hence, indeed we have that N = 1. As a consequence, Q acts irreducibly on P. Hence, G has SSN by Theorem II.2.17(i). Finally, Q[G] has only one matrix component isomorphic to Mq(Q(εp)) since v = (|P| − 1)/((p − 1)|Q|) = 1 by Lemma II.4.3.

Combining Lemma II.4.3, Proposition II.4.4 and Lemma II.4.6, we have the following conclusion.

Proposition II.4.7. Let G = P o Q be as in Theorem II.2.17(i) with non-cyclic P. Then G hasND if and only if G = Cⁿ_po Cq for some primes p, q such that pⁿ− 1 = (p − 1)q and C_qacts nontrivially on Cⁿ_p. In this case, Q[G] has only one matrix component isomorphic to M_q(Q(εp)).

RemarkII.4.8. For primes p, q and n ≥ 2 such that pⁿ− 1 = (p − 1)q, there always is a cyclic group C_q acting nontrivially on Cⁿ_p. To see this, we view Cⁿ_p as a n dimensional vector space over Fp. Then it is well-known that

|GL_n(Fp)| = (pⁿ− 1)(pⁿ− p) · · · (pⁿ− pⁿ⁻¹).

Since q is prime dividing pⁿ− 1, we can always find an element in GL_n(Fp) of order q, as desired. (Indeed, we only need the restriction q | pⁿ− 1.) Furthermore, we can also find an invertible linear operator of order q in SL_n(Fp) because q divides |GL_n(Fp)|/(p − 1) =

|SL_n(Fp)|. If p = 2, then q is a Mersenne prime. Thus, C₂²o C3' A₄, C³₂o C7, C₂⁵o C31, etc. are SSN groups. If p is odd, then so are q and n since q = 1 + p + · · · + pⁿ⁻¹6= 2 and q≡ n (mod 2). For instance, C³₃o C13, C₅³o C31 and C₇⁵o C2801are SSN groups.

Primes p and q satisfying pⁿ− 1 = (p − 1)q for some n are called repunit primes. For more information, one can refer to [Dub93].