# Solvable SSN Groups with ND

## II. Nilpotent Decomposition in Integral Group Rings

### II.4.1 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, NG(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 P1h6= P1, for otherwise P1 is hhi-stable. Thus, NG(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 = NG(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 ggxis fixed by x. In particular, x ∈ NG(hggxi).

Then either hggxi = 1 or hggxi = P by (i). Thus ggx= 1 as by assumption P is not cyclic.

Hence, gx= g−1 and we obtain x ∈ NG(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) = εip. Here, the integer i is given by gh= 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 gh = 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) = εip 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 ∈ NG(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 ' Cp. It follows that ε(P, K) = eK−P. Clearly, P ⊆ Cene G(ε(P, K)).

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

eK= e(G, P, K) =

## ∑

x∈Q

( eKx−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 KQ leads to the same idempotent eK. If K1Q and K2Q are in distinct orbits, then K1x6= K2y for every x, y ∈ Q. Thus, fK1xfK2y= eP and then (fK1x−P)(fe K2y−P) = 0. It follows that ee K1eK2 = 0. In particular, we have eK1 6= eK2. Hence, the number of orbits KQ is precisely the number of distinct eK. The number |S| of subspaces of dimension n − 1 in P ' Fnp is the number of subspaces of dimension 1 and this is (pn− 1)/(p − 1). Since each orbit KQ has size

|Q|, we have v = |S|/|Q| = (|P| − 1)/((p − 1)|Q|) distinct orbits. Therefore, we have v distinct primitive central idempotents e1, . . . , ev such that Q[G]ei' M|Q|(Q(εp)) for each i. Finally, LiQ[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 − ys−1) bQis a nonzero nilpotent element of Z[G]. Write ys−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

( eKx−Pe) ∈ Z[P].

Since (1 − g0) eP= 0, we have

(1 − g0)

## ∑

x∈Q

Kex∈ Z[P].

If g0∈ Kxfor some x, then (1 − g0) eKx= 0. Computing the coefficient of 1, it follows that

|{x ∈ Q | g06∈ Kx}|

|K| ∈ Z.

If |{x ∈ Q | g0 6∈ Kx}| = 0, then g0 ∈ Kx for all x ∈ Q. However, Tx∈QKx is a proper subgroup of P and it is normal in G. So g0Tx∈QKx= 1 by Lemma II.4.1(i), a contra-diction. Thus, |{x ∈ Q | g06∈ Kx}| 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 = Cnpo Cq is a group for some distinct primes p and q such that pn− 1 = (p − 1)q and Cq acts nontrivially on Cnp, then G has SSN. In particular, Q[G]

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

Proof. For convenience, we write P = Cnp and Q = Cq. 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 nx6= n for some x ∈ Q, then nx6= n for all 1 6= x ∈ Q and the orbit nQ consists of q nontrivial elements. Thus, N \ {1} contains nQ. 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 = Cnpo Cq for some primes p, q such that pn− 1 = (p − 1)q and Cqacts nontrivially on Cnp. In this case, Q[G] has only one matrix component isomorphic to Mq(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 Cq acting nontrivially on Cnp. To see this, we view Cnp as a n dimensional vector space over Fp. Then it is well-known that

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

Since q is prime dividing pn− 1, we can always find an element in GLn(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 SLn(Fp) because q divides |GLn(Fp)|/(p − 1) =

|SLn(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].