# p-Groups

## II. Nilpotent Decomposition in Integral Group Rings

### II.3.1 p-Groups

Recall that a group is called NCN if all its non-cyclic subgroups are normal. By Propo-sition II.2.15, a p-group has the SSN property if and only if it has the NCN property.

Note that the rational group algebras of Hamiltonian 2-groups have no matrix compo-nents. Moreover, non-Hamiltonian NCN p-groups had been classified by Z. Boˇzikov and Z. Janko [BJ09]. We state their result below.

Theorem II.3.1 ([BJ09, Theorem 1.1] or [Ber08, Theorem 16.2]). Let G be a finite non-abelian and non-Hamiltonian p-group all of whose non-cyclic subgroups are normal.

Then G is one of the following groups.

BJ1. G is metacyclic minimal nonabelian and G is not isomorphic to Q8.

BJ2. G = G0∗ Z, the central product of a nonabelian group G0of order p3with a cyclic group Z, where G0∩ Z =Z (G0) and if p = 2, then |Z| > 2.

BJ3. p = 2 and G = Q8× Z where Z is cyclic of order > 2.

BJ4. G is a group of order 34and maximal class with Ω1(G) = G0' C3×C3. BJ5. G = ha, b | a8= b8= 1, ab= a−1, a4= b4i, where |G| = 25.

BJ6. G = Q16. BJ7. G = D8∗ Q8.

BJ8. G = ha, b, c | a4= b4= [a, b] = 1, c2= a2, ac= ab2, bc= ba2i, where G is the min-imal non-metacyclic group of order25.

BJ9. G = ha, b, c, d | a4= b4= [a, b] = 1, c2= a2b2, ac= a−1, bc= a2b−1, d2= a2, ad= a−1b2, bd= b−1, [c, d] = 1i, where G is a special 2-group of order 26in which every maximal subgroup is isomorphic to the minimal non-metacyclic group of order 25 inBJ8.

Here a p-group is called minimal nonabelian (resp. minimal non-metacyclic) if it is nonabelian (resp. non-metacyclic) and all proper subgroups are abelian (resp. meta-cyclic). A p-group G is said to be maximal class if G has nilpotency class n where

|G| = pn+1. Moreover, the notation Ω1(G) is given by the following definition Ωi(G) =

hg ∈ G | gpi= 1i.

Now, we focus on each class of groups, separately.

BJ1 According to [BJ09, Proposition 1.3] or [BJ06, Lemma 3.1], groups in BJ1 have the following presentation (proved by L. R´edei)

G= ha, b | apm= bpn= 1, ab= a1+pm−1i = Cpmo Cpn

where m ≥ 2, n ≥ 1, |G| = pm+nand |G0| = p. It is easy to check in such a group apand bp are central. Moreover, we have G0 = hapm−1i since [a, b] = apm−1 and |G0| = p. We remark here that G has MJD (and so ND) when (p, m, n) = (2, 2, 1), (2, 2, 2), (2, 3, 1) and (3, 2, 1). Moreover, the proof of [Liu12, Lemma 2.8] shows that G does not have ND for p= 2 and (m, n) = (2, ≥ 4), (3, ≥ 2). Now we discuss when Q[G] has only one matrix component.

Lemma II.3.2. Let n = 1. Then Q[G] has only one matrix component for any p, m ≥ 2.

Moreover, this component is isomorphic to Mp(Q(εpm−1)).

Proof. Let A = hai. Then A is of index p in G. Thus, A is a maximal abelian subgroup of G containing G0= hapm−1i. Let (H, K) be a pair subgroups of G in Theorem II.1.5 such that G06⊆ K (Lemma II.1.4). Then (H, K) = (A, 1). In other words, e(G, A, 1) = ε (A, 1) = 1 − ]apm−1 = 1 − eG0is the only primitive central idempotent e such that Q[G]e is non-commutative.

Now we find out the structure of Q[G]e(G, hai, 1). Note that (hai, 1) is a strong Shoda pair of G. By Proposition II.1.10, Q[G]e(G, hai, 1) ' Q(ε|a|) ∗ (G/hai) ' Q(εpm) ∗ hσ i where σ ∈ Aut(Q(εpm)/Q) and σ (εpm) = ε1+ppm m−1. Note that σ fixes εpm−1 since εpm−1= εppm. It is easy to check that the fixed field Q(εpm)σ = Q(εpm−1). Note also that σpis an identity map because (1 + pm−1)p≡ 1 (mod pm). By Proposition II.1.9, we can conclude that Q(εpm) ∗ hσ i ' Mp(Q(εpm−1)).

Lemma II.3.3. Let n ≥ 2. Then Q[G] has only one matrix component if and only if p= m = n = 2. Moreover, this component is isomorphic to M2(Q).

Proof. First of all, if p = m = n = 2, then G = P in Theorem I.4.5 and Q[P] has exactly one matrix component isomorphic to M2(Q), see Section I.7. We claim that Q[G] has at least two matrix components for the remaining cases of p, m, n.

Denote G = Gp(m, n). Note that bpis central. If p = m = 2 and n ≥ 3, then we have G2(2, n)/hbp3i ' G2(2, 3). By [Gir06, Group Type 32/21 on p. 197],

Q[G2(2, 3)] ' 4Q ⊕ 2Q(ε4) ⊕ 2Q(ε8) ⊕ HQ⊕ M2(Q) ⊕ M2(Q(ε4)).

It follows that Q[G2(2, n)] has more than one matrix components. Now, assume either central idempotent different from e(G, A, K1).

We claim that M = Q[G]e(G, A, K) is a matrix ring if p is odd and m ≥ 2, or if p = 2

It follows that NE/Fp2) = ∏p−1j=0ε(1+p)

Lemma II.3.5. Let G be a group in BJ2. Then Q[G] has only one matrix component which is isomorphic to Mp(Q(ε|Z|)).

Proof. Let G = G0∗ Z, z0, a, b be as above and let A = haiZ. Note that ap∈Z (G0) = hz0i ⊆ Z,Z (G) = Z, a 6∈ Z and |G| = p2|Z|. If |a| = p, then hai ∩ Z = 1 and |A| = p|Z|;

if |a| = p2, then hai ∩ Z = hz0i and |A| = |hai||Z|/|hai ∩ Z| = p|Z|. In other words, A is a maximal abelian subgroup of G containing G0= hz0i. We will show that Q[G] has only one primitive central idempotent corresponding to a non-commutative Wedderburn component. Since ab= az0, A is normal in G. Moreover, |G/A| = p so G is metabelian.

Note that abi= azi0. Thus, primitive central idempotent such that e(G, A, K)Q[G] is non-commutative.

Now we assume |a| = p2. If |Z| = p and p > 2, then hapi = Z = hz0i. It follows that A= hai and K = 1. Then we have only one primitive central idempotent e(G, A, 1) = ε (A, 1) = 1 −ze0 corresponding to a non-commutative Wedderburn component. Assume

|Z| > p. The remaining proof is similar to the case |a| = p. Observe that if tk= ap∈ hz0i

1 −ze0is a primitive central idempotent, as desired.

Finally, note that (A, K) is a strong Shoda pair of G. When |a| = p, K = haz0ji and (az0j)b6∈ K; when |a| = p2, K = hatji and (atj)b6∈ K. Thus, K is not normal in G. For both situations, A = CenG(K) ⊆ NG(K) and A is of index p in G. We have NG(K) = A. Since |A| = p|Z|, we have [A : K] = |Z|. By Proposition II.1.10, Q[G]e(G, A, K) ' Mp(Q(ε|Z|)).

BJ3 Note that Q8× C4 has MJD and we know that their rational group algebras has only one matrix component (see Section I.7). The following result imitates from [Liu12, Lemma 2.13] that Q8× C8 does not have MJD. For convenience, we denote ≡n to be the equivalence relation on Z[G] modulo n for n ∈ N. In other words, α ≡nβ if α − β ∈ nZ[G].

Lemma II.3.6. G = Q8×C2n does not haveND for n ≥ 3.

Proof. Let Q8= ha, b | a4= 1, b2= a2, ab= a−1and C2n = hxi. Write z = a2= b2. Then ba= abz. Note that z,t are central. Denote t = x2n−3. Let

r= (a + bt)(1 − t2)(1 + t4)(1 − z) and s= (a + bt2)(1 − t4)(1 − z).

Clearly, rs = sr = 0 since t8 = 1. Observe that (a + bt)2 = z(1 + t2) + abt(1 + z) and (a + bt2)2= z(1 + t4) + abt2(1 + z). So we have r2= s2= 0 due to t8= 1 and z2= 1. Let

α = 1

2(r(1 − t) + s(1 − t)3).

Then α2= 0. (We do not have to check α 6= 0 when we have eα 6∈ Z[G] below.) We claim that α ∈ Z[G]. Note that 1 − t ≡21 + t and (1 + t2k) ≡2(1 + t)2k for every k ∈ N. Thus, r(1 − t) ≡2(a + bt)(1 + t)7(1 − z) and s(1 − t)32 (a + bt2)(1 + t)7(1 − z). Moreover, (a + bt) + (a + bt2) ≡2bt(1 + t). Thus, 2α ≡2bt(1 + t)8(1 − z) ≡2bt(1 + t8)(1 − z) ≡20.

In other words, α ∈ Z[G].

Let e = et4 which is a central idempotent. We have er = r and es = 0. Thus, eα = r(1 − t)/2. Observe that r(1 − t) ≡2(a + bt)(1 + t7)(1 − z) 6≡20 since the coefficient of a is odd. Hence, eα 6∈ Z[G], as desired.

RemarkII.3.7. The above proof gives us the following idea. Let G be a finite group and H ≤ G. Assume that H does not have ND. So there are a nilpotent element α ∈ Z[H]

and a central idempotent e ∈ Q[H] such that αe 6∈ Z[H]. Note that α is still a nilpotent element of Z[G]. If e is also central in Q[G], then we have αe 6∈ Z[G] and G does not have ND. We will use this idea for the group in BJ9.

BJ4 Let G be a group in BJ4. Then the proof of [LP13, Lemma 3.8] shows that G' hx, y, z | x9= y3= 1, xy = yx, xz= xy, yz= x−3y, z3= x3i.

By the proof of [LP10, Lemma 2.1], it shows that G does not have ND. For convenience, we present the proof below.

Lemma II.3.8 ([LP10, Lemma 2.1]). The group above does not have ND.

Proof. Let α = (1 − x)(1 − y) ∈ Z[G] and e = ex3 ∈ Q[G]. Since x3 is central, e is a central idempotent. Clearly, eα 6∈ Z[G] via the coefficient of 1. Thus, e(αz) 6∈ Z[G]. We claim that (αz)3= 0. Since zαz−1= αz, we have (αz)3= ααzαz

2z3. It suffices to show that ααzαz

2 = 0. Note that xz2 = x−2y2 and yz2 = x−6y= x3y. Since xy = yx, we have α αzαz

2 = (1 − x)(1 − x)z(1 − x)z2(1 − y)(1 − y)z(1 − y)z2. Computing directly, we have

(1 − x)(1 − x)z(1 − x)z2 = (x−1− x)yb+ (x3− 1)(x−1y+ x−2y2) and

(1 − y)(1 − y)z(1 − y)z2 = bx3(−y + y2).

Becauseby(−y + y2) = (x3− 1) bx3= 0, we get ααzαz

2 = 0, as desired.

Similarly, proofs of [Liu12, Lemma 2.15 and 2.16] show that groups in BJ5 and BJ8 do not have ND. Moreover, groups in BJ6 and BJ7 have MJD and we know that their rational group algebras has only one matrix component. For convenience to read, we present the proofs for BJ5 and BJ8 below.

BJ5

Lemma II.3.9 ([Liu12, Lemma 2.15]). Let G = ha, b | a8= b8= 1, a4= b4, ab= a−1i be the group inBJ5. Then G does not have ND.

Proof. Clearly, a4and b2are central elements. Let e = (1 − a4)/2 be a central idempotent of Q[G]. Let

α = b(1 − ab)(a2b(1 − b2) + (1 − a2))(1 + b2) ∈ Z[G].

Since b4= a4, we have (1 − b2)(1 + b2)(2e) = 4e ∈ 2Z[G]. Thus, we obtain b−1α (2e) ≡2 (1 − ab)(1 − a2)(1 + b2)2e. Moreover, 1 ∈ supp((1 − ab)(1 − a2)(1 + b2)2e) with the coefficient 1. We have b−1α (2e) 6∈ 2Z[G] and then αe 6∈ Z[G].

We claim that α2= 0. Let βi= a2ib(1 − b2) + (1 − a2i). Then

α2= b2(1 − a−1b)β−1(1 − ab)β1(1 + b2)2

since bab−1= a−1 and bβ1b−1= β−1. Let A = 2e = 1 − a4= 1 − b4. Because aβ−1a= a−2b(1 − b2) + (a2− 1) and a−2A= −a2A, we have

−1a(1 + b2) = a−2bA+ (a2− 1)(1 + b2) = −a2bA− (1 − a2)(1 + b2) = −β1(1 + b2).

Thus, −β−1ab(1 + b2) = −a−1−1ab(1 + b2) = a−1β1b(1 + b2) = a−1−1(1 + b2). So

α2= b2(1 − a−1b)(1 + a−1b)β−1β1(1 + b2)2= b2−1β1(1 + b2) since (1 − a−1b)(1 + a−1b)(1 + b2) = A. Moreover,

−1β1= A(−a2b(1 − b2) + (1 + a2))(a2b(1 − b2) + (1 − a2)) by a−2A= −a2Aagain. Then

−1β1(1 + b2) = A(−b2(1 − b2)A + (a4+ 1)bA + A(1 + b2)) = 0 and hence α2= 0.

BJ8

Lemma II.3.10 ([Liu12, Lemma 2.16]). Let G = ha, b, c | a4 = b4 = [a, b] = 1, c2 = a2, ac= ab2, bc= ba2i be the group in BJ8. Then G does not have ND.

Proof. Let α = (1 − a2b2)(1 + a)(1 + b)c ∈ Z[G] and let e = ea2 be a central idempotent in Q[G]. We claim that eα 6∈ Z[G]. It is equivalent to e(αc−1) 6∈ Z[G]. Now, ba2(αc−1) = ab2(1 − b2)(1 + a)(1 + b) ≡2 abbb6≡2 0. We have e(αc−1) 6∈ Z[G], as claimed. Now we show that α2= 0. Since a2, b2 are central and (1 − a2b2) = 2(1 − a2b2), we have α2= 2(1 − a2b2)(1 + a)(1 + b)(1 + ab2)(1 + a2b)c2. Moreover, (1 − a2b2)(1 + a)(1 + ab2) = (1 − a2b2)(a + ab2) and (1 − a2b2)(1 + b)(1 + a2b) = (1 − a2b2)(b + a2b) since (a2b2)2= 1. Note that (1 + a2)(1 + b2) = (1 + a2b2)(1 + b2). Hence, α2= 0.

BJ9 We use the idea in BJ8 to prove the following result.

Lemma II.3.11. The group G = ha, b, c, d | a4= b4= [a, b] = 1, c2= a2b2, ac= a−1, bc= a2b−1, d2= a2, ad = a−1b2, bd= b−1, [c, d] = 1i in BJ9 does not have ND.

Proof. Let A = b, B = a and C = cd. Then A4 = B4 = [A, B] = 1 and C2 = c2d2= (a2b2)a2= A2. Moreover, AC= bcd = (b−1)c= (b3)c= (bc)3= a2b−3= AB2and BC= acd= (a−1b2)c= a(a2b−1)2= A2B. Let H = hA, B,Ci and G0be the group in BJ8 with generators a0, b0, c0. Then there is a group epimorphism G0→ H given by a07→ A, b07→ B and c07→ C. By the proof of Lemma II.3.10, (1 − a20b20)(1 + a0)(1 + b0)c0is a nilpotent element of Z[G0]. It follows that α = (1 − A2B2)(1 + A)(1 + B)C is a nilpotent element of Z[H] ⊆ Z[G]. Note that A2= b2= a2c2= d2c2which commutes with c, d and, of course, a. Thus, e = fA2is a central idempotent in Q[G]. We claim that eα 6∈ Z[G]. It is equivalent to e(αC−1) 6∈ Z[G]. Now, cA2(αC−1) = cA2(1 − B2)(1 + A)(1 + B) ≡2A bbB= bba. Sinceb bbba= [ha, bi 6∈ 2Z[G], it follows that e(αC−1) 6∈ Z[G], as claimed.