**I. 4.2 3-Groups and {2, 3}-Groups**

**I.7 Remaining Problems and Wedderburn Components**

We list the finite groups which are known to have MJD in the following table where each pis an odd prime.

We next present the Wedderburn decomposition of Q[G] for above groups G such that Q[G] has some nonzero nilpotent elements (see Theorem I.1.3). We will see that Q[G]

has only one component which is not a division ring.

Recall that S_{3} = D_{2·3}, G(p, 2, 1) = D_{2p} and G(p, 2, 2) = Q_{4p}. Let ε_{n} be a primitive
n-th root of 1 in C, i =√

−1 and G_{27} be either of the two nonabelian groups of order 27.

Denote

a,b F

the quaternion algebra over a field F defined by a and b. It is well-known that

• Q[G27] ' Q ⊕ 4Q(ε3) ⊕ M_{3}(Q(ε3)) ([LP09, Lemma 3.1]).

• Q[Q8×C_{p}] ' 4Q ⊕ 4Q(εp) ⊕ HQ⊕ M_{2}(Q(εp)) (Section I.5).

• Q[G(p, 2, 1)] = Q[D2p] ' 2Q ⊕ M2(Q(αp)) ([HLP90, 3.4 Theorem]).

• Q[G(p, 2, 2)] = Q[Q4p] ' 2Q ⊕ Q(i) ⊕

α_{2p}^{2} −4,−1
Q(α2p)

⊕ M_{2}(Q(αp)) ([JdR16,
Exam-ple 3.5.7]).^{4}

• Q[G(7, 3, 1)] ' Q ⊕ Q(ε3) ⊕ M_{3}(Q(√

−7)) ([Aro94, Proposition 5.1]).

Note that α_{2n} is a real number and |α_{2n}| < 2 for n ≥ 3. Thus, X^{2}= (α_{2n}^{2} − 4)Y^{2}− Z^{2}
has no solutions (X ,Y, Z) in Q(α2n)^{3}\ {(0, 0, 0)}. It follows that

α_{2n}^{2}−4,−1
Q(α2n)

is a division ring for n ≥ 3 by [JdR16, Example 2.1.7(3)].

We list the groups below which we do not know whether their integral group rings have MJD or not.

Q_{8}×C_{p} p> 41 with ord_{p}(2) ≡ 0 (mod 4)

C_{p}o C_{n}^{k} (p, n, k) = (1 (mod 4), 2, ≥ 3), (7, 3, ≥ 2)
Tab. I.2: Remaining groups.

In Section I.5, we know that Q[Q8×C_{p}] has only one Wedderburn component which
is not a division ring. This is also true for Cpo C_{n}^{k} with (p, n, k) = (5 (mod 8), 2, ≥ 3)
and (7, 3, ≥ 2) by Theorem II.4.19 in the next chapter. Thus, most rational group algebras
have only one Wedderburn component which is not a division ring if their integral group
rings have MJD, except for groups C_{p}o Cn^{k} with (p, n, k) = (1 (mod 8), 2, ≥ 3). Based
on the above observation, we have the following conjecture.

Conjecture 2. If Z[G] has MJD, then Q[G] has at most one Wedderburn component which is not a division ring.

4Note that Q_{4n}/Q^{0}_{4n}' C_{2}×C_{2}if n is even and Q_{4n}/Q^{0}_{4n}' C_{4}if n is odd. It is a typo in [JdR16, Example
3.5.7] that the commutative components of Q[Q4n] is 4Q for “all” n.

In this chapter, we will classify integral group rings satisfying the nilpotent decomposition property for a class of finite groups. This property holds for groups which are already known to have MJD. In fact, this is the motivation of this chapter.

In the first section, we introduce the concepts of Shoda pairs, strong Shoda pairs, crossed products and cyclic algebras. We also present some results which we will need throughout the whole chapter. For the next section, we give the concept of the nilpotent decomposition property, called ND, for rational group algebras which comes from the study of the MJD problem. This property leads to a special kind of groups, called SN groups. A group is called SSN if all its subgroups are SN groups. Finite SSN groups have a nice description given by C.-H. Liu and D.S. Passman. We use this description to study the ND property for SSN groups. However, it is not easy to check whether the ND property holds or not for a given SSN group. For convenience, a Wedderburn component of a semisimple ring (Theorem I.1.1) is called a matrix component if it is not a division ring. Clearly, the ND property holds if the rational group algebra of a finite group has at most one matrix component. Finite groups such that their rational group algebras have no matrix components are well-known. In the last two sections, we study the ND property for SSN groups and classify finite SSN groups such that their rational group algebras have only one matrix component. In fact, we conjecture that these two concepts are equivalent:

a rational group algebra has the ND property if and only if it has at most one matrix component. This is a joint work with Eric Jespers.

54

### II.1 Prerequisite

Before we start, we need to have a nice presentation for primitive central idempotents in Q[G] for some kinds of groups. The following notations are referred to [JdR16, Section 3.3 and 3.4]. Let G be a finite group and N be a normal subgroup of G. Then put

ε (G, N) = (

Ge if G = N

∏M/N∈M (G/N)( eN−M) if G 6= N,e

whereM (G/N) is the set of all minimal normal nontrivial subgroups of G/N andG, ee N, eM are defined in Section I.1. Note that the above product is well-defined since M/N is nor-mal in G/N which implies that M is nornor-mal in G and eN−Meis central in Q[G]. Moreover, N eeM= eMand it follows that eN−Meis an idempotent. Thus, ε(G, N) is a product of central idempotents of Q[G] so it is also a central idempotent of Q[G].

For α ∈ Q[G], the G-centralizer of α is defined by CenG(α) = {g ∈ G | gα = αg}

which is a subgroup of G. Let K, H be two subgroups of G with KE H and T be a right
transversal of Cen_{G}(ε(H, K)) in G. Then one has a central element of Q[G]:

e(G, H, K) =

## ∑

t∈T

ε (H, K)^{t}.

Note that e(G, H, K) is independent of the choice of T .

Definition II.1.1. A pair (H, K) of subgroups of a group G is called a Shoda pair if it satisfies the following conditions:

(S1) KE H,

(S2) H/K is cyclic and

(S3) for every g ∈ G \ H, there exists h ∈ H so that (h, g) = h^{−1}g^{−1}hg∈ H \ K.

Definition II.1.2. A pair (H, K) of subgroups of a group G is called a strong Shoda pair if it satisfies the following conditions:

(SS1) KE H E NG(K),

(SS2) H/K is cyclic and a maximal abelian subgroup of N_{G}(K)/K,
(SS3) for every g ∈ G \ N_{G}(K), ε(H, K)g^{−1}ε (H, K)g = 0.

The following result shows that every strong Shoda pair is a Shoda pair and it leads to a primitive central idempotent.

Proposition II.1.3 ([OdRS04, Proposition 3.3]). The following conditions are equivalent for a pair(H, K) of subgroups of a group G:

(i) (H, K) is a strong Shoda pair;

(ii) (H, K) is a Shoda pair, HE NG(K) and the G-conjugates of ε(H, K) are orthogo-nal.

Moreover, if the previous conditions holds, thenCen_{G}(ε(H, K)) = N_{G}(K) and e(G, H, K)
is a primitive central idempotent of Q[G].

If (H, K) is a strong Shoda pair of G, we can know when Q[G]e(G, H, K) is commu-tative via K.

Lemma II.1.4. Let (H, K) be a strong Shoda pair of G. Then the Wedderburn component
Q[G]e(G, H, K) is commutative if and only if K ⊇ G^{0}.

Proof. If K ⊇ G^{0}, then H and K are normal in G. Thus, e(G, H, K) = ε(H, K) and
g^{−1}h^{−1}ghε(H, K) = ε(H, K) so Q[G]e(G, H, K) is commutative. Conversely, assume that
Q[G]e(G, H, K) is commutative. For g ∈ G, we have

g∈ K ⇔ gε(H, K) = ε(H, K)

since (H, K) is a strong Shoda pair and we apply [OdRS04, Lemma 3.2(2)]. Let g ∈ G^{0}.
By assumption, ge(G, H, K) = e(G, H, K). Note that e(G, H, K)ε(H, K) = ε(H, K). Thus,
gε(H, K) = ge(G, H, K)ε(H, K) = e(G, H, K)ε(H, K) = ε(H, K). Hence, g ∈ K.

The next theorem will be widely used in the following sections.

Theorem II.1.5 ([OdRS04, Theorem 4.7]). Let G be a metabelian finite group and let A
be a maximal abelian subgroup of G containing G^{0}. The primitive central idempotents of
Q[G] are the elements of the form e(G, H, K), where (H, K) is a pair of subgroups of G
satisfying the following conditions:

(i) H is a maximal element in the set {B ≤ G | A ≤ B and B^{0}≤ K ≤ B} and
(ii) H/K is cyclic.

Note that in Theorem II.1.5 the subgroup A can be arbitrarily chosen and every pair (H, K) is a strong Shoda pair of G.

RemarkII.1.6. The above theorem gives a character-free method to write down primitive central idempotents of the rational group algebra of a metabelian group. The concrete construction of primitive central idempotents is important to study the unit group of group algebras. To do that for the rational group algebra of arbitrary finite group, there is a classical method in [Yam74, Proposition 1.1] by using complex irreducible characters.

However, to compute the character table of a finite group, it would be a task of exponential growth with respect to the order of the group. So it could be better to have a character-free

construction. Indeed, such a construction of some finite groups has been developed. For abelian groups, one can see [PW50], [AA69] and [GJP96, Theorem VII.1.4]; for nilpotent groups, one can see [JLP03, Theorem 2.1]. The authors of the paper [OdRS04] also give a character-free construction for abelian-by-supersolvable groups and monomial groups, namely that all complex characters are induced from a linear character of some subgroup.

For non-monomial groups, there are some results in [Lin20].

We give a concept about the crossed product. The following resource refers to [JdR16, Section 2.6]. The reader can also refer to [Pas86, Section 5].

Definition II.1.7. Let R be a ring and G a group. A crossed product R ∗ G of G over R is an associative ring which has a set of invertible elements {ug: g ∈ G}, a copy of G, such that

R∗ G = ⊕_{g∈G}Ru_{g}

is a free left R-module. The multiplication is defined by the following rules:

u_{g}r= αg(r)u_{g} and u_{g}u_{h}= f (g, h)u_{gh},
for g, h ∈ G, r ∈ R such that the associativity holds where

α : G → Aut(R) and f : G × G →U (R)

are two maps and α_{g}= α(g). The map α is called the action and the map f is called the
twistingof the crossed product. We also denote

R∗ G = (R, G, α, f ).

Note that the associativity of R ∗ G is equivalent to the assertions that, for all g, h, k ∈ G,

α_{g}α_{h}= ι_{f}_{(g,h)}α_{gh} and f(g, h) f (gh, k) = αg( f (h, k)) f (g, hk),

where ι_{u}denotes the inner automorphism of R defined by ι_{u}(x) = uxu^{−1}for u ∈U (R).

Group rings are special cases of crossed products if the action and twisting are trivial,
that is if α_{g}= 1 and f (g, h) = 1 for all g, h ∈ G. Next, we introduce another special types
of crossed products.

Definition II.1.8. Let E/F be a finite Galois extension and let G = Gal(E/F). There is a natural action α : G → Aut(E). Let f : G × G →U (E) such that

f(σ , τ) f (σ τ, ρ) = α(σ )( f (τ, ρ)) f (σ , τρ),

where σ , τ, ρ ∈ G. Thus, (E, G, α, f ) forms a crossed product (where ι_{u}= id_{E} for each

u∈U (E)). This is called a classical crossed product and simply denoted by (E/F, f ) since G and α are determined by E/F. Here, we can identify α(σ ) = σ .

Assume that G is cyclic. Let G = hσ i and |G| = n. Put 1 = u_{1}, u = u_{σ} and define

f(σ^{i}, σ^{j}) =

( 1 if i + j < n a if i + j ≥ n

for some a ∈U (F) where 0 ≤ i, j < n. One can check that (E/F, f ) is a classical crossed
product since σ fixes the images of f . A crossed product of this type is called a cyclic
algebra and denoted by (E/F, σ , a) or (E, σ , a) since E/F is a Galois extension which
implies F = E^{σ}, the fixed field of σ . We also denote (E/F, a) if σ is clear from the
context.

Recall that an F-algebra is called a central algebra if its center is exactly the field F.

We list some properties which we will need later.

Proposition II.1.9 ([JdR16, Theorem 2.6.3 and Proposition 2.6.7]). Let E/F be a finite Galois extension.

(i) Every classical crossed product (E/F, f ) is a finite dimensional central simple F-algebra.

(ii) A cyclic algebra (E/F, σ , a) ' M_{n}(F) if and only if a ∈ N_{E/F}(E) where the integer
n = |Gal(E/F)| and N_{E/F}(E) consists elements of the norm N_{E/F}(x) for x ∈ E
where N_{E/F}(x) = ∏σ ∈Gal(E/F )σ (x). In particular, (E/F, σ , 1) ' M_{n}(F).

Note that if N is a normal subgroup of G, then G/N can act on N by conjugation in
G: g · x = x^{g} for g ∈ G and x ∈ N. Moreover, if we fix a right transversal T of N in G,
then there is a well-defined map φ : G/N → G such that p ◦ φ = id_{G/N} where p is the
natural homomorphism G → G/N. The following proposition describes the structure of a
Wedderburn component via a strong Shoda pair.

Proposition II.1.10 ([OdRS04, Proposition 3.4]). Let (H, K) be a strong Shoda pair of a
finite group G. Let h= [H : K], N = N_{G}(K), n = [G : N], H/K = hxi. Denote the action
of N/H ' (N/K)/(H/K) on H/K by (x^{i})^{a}for a∈ N/H and x^{i}∈ H/K. Then

Q[G]e(G, H, K) ' Mn(Q(εh) ∗ N/H) where the action and twisting are given by

action: ε_{h}^{a}= ε_{h}^{i} if x^{a}= x^{i}
twisting: f(a, b) = ε_{h}^{j} if a^{0}b^{0}= x^{j}(ab)^{0}

for a, b ∈ N/H and integers i, j where a^{0}, b^{0}, (ab)^{0}are fixed preimages of a, b, ab under the
natural homomorphism N/K → N/H.

### II.2 Nilpotent Decomposition Property, SN and SSN Groups

We present the concept of the nilpotent decomposition property in the first subsection.

This property leads to a type of groups, called SN groups and SSN groups. We give the definitions of these groups and present some properties of them in Subsection II.2.2. The classification of finite SSN groups is also mentioned. In Subsection II.2.3, we give some examples of SN and SSN groups. We also discuss the nilpotent decomposition property for those examples.

II.2.1 Nilpotent Decomposition

Let G be a finite group. Recall that if G has MJD then reduced degrees of Q[G] are at most three (Theorem I.2.11). Using this result, one can show that if G has MJD, then each nilpotent element of Z[G] can be decomposed in Z[G] under the Wedderburn decomposition of Q[G]. More precisely, one has the following theorem.

Theorem II.2.1 ([HPW07, Theorem 8]). Let G be a finite group such that Z[G] has MJD.

Let

Q[G] = A1⊕ A_{2}⊕ · · · ⊕ A_{m}

be the Wedderburn decomposition with A_{i}' M_{s}_{i}(D_{i}), i = 1, . . . , m, where the D_{i}’s are
division Q-algebras. If z = z1+ z_{2}+ · · · + z_{m}∈ Z[G] is nilpotent, with each zi∈ A_{i}, then
z_{i}∈ Z[G] for all i.

One has an immediate consequence.

Corollary II.2.2 ([HPW07, Corollary 9]). Assume that Z[G] has MJD. If z ∈ Z[G] is a nilpotent element, then ze∈ Z[G] for all central idempotents e ∈ Q[G].

In fact, the condition “MJD” is not necessary. For instance, nilpotent elements of
Z[A4] can be decomposed in Z[A4] since Q[A4] ' Q ⊕ Q(ε3) ⊕ M_{3}(Q) has only one
Wed-derburn component which is not a division ring. But this integral group ring do not have
MJD by Theorem I.4.9. We now formalize this concept.

Definition II.2.3. A finite group G is said to have the nilpotent decomposition property (ND) if for every nilpotent element α ∈ Z[G] and every central idempotent e ∈ Q[G], we have αe ∈ Z[G].

Example II.2.4. We claim that D_{12} is the smallest finite group without ND. If G is an
abelian finite group, then G has ND since 0 is the only nilpotent element of Z[G]. If G is
nonabelian and |G| ≤ 11, then G is either S3, Q8, D8 or D10. Those groups have MJD so
they have ND. Denote

D_{12} = ha, b | a^{6}= b^{2}= 1, bab^{−1}= a^{−1}i.

Let e = ea^{3}−ea∈ Q[D12] and α = (1 − b)a(1 + b) ∈ Z[D12]. Then e is a central idempotent
since ha^{3}i and hai are normal in D_{12}. Moreover, α is nilpotent since 1 − b^{2}= 0. However,

α e = 1

2(1 + a^{3})(a − a^{−1})(1 + b) 6∈ Z[D12].

Thus, D_{12} does not have ND. Hence for nonabelian groups of order 12, Q_{12} has both ND
and MJD, A_{4}has ND but it does not have MJD, and D_{12}has neither ND nor MJD.

We desire to know which finite groups have the ND property. Clearly, if Q[G] has
at most one Wedderburn component which is not a division ring, then G has ND
au-tomatically. For convenience, we say that Q[G] has a matrix component if Q[G] has a
Wedderburn component isomorphic to M_{n}(D) for some division ring D with n > 1.
Re-call that ord_{n}(m) is the multiplicative order of m modulo n for n, m ∈ N and gcd(n, m) = 1.

So we have the following consequence by Theorem I.1.3 and Section I.7.

Lemma II.2.5. If Q[G] has at most one matrix component, then G has ND. Thus, the following groups haveND:

(i) abelian groups;

(ii) Q_{8}× E × A where E is an elementary abelian 2-group and A is an abelian group of
odd order m such thatord2(m) is odd;

(iii) Q_{8}×C_{p}where p is an odd prime such thatord_{2}(p) is even.

(iv) groups in Table I.1 where those groups have MJD.

We present some results below for G having ND. Those results come from several papers on the MJD problem. We replace the MJD assumption by ND and present the proofs here.

Lemma II.2.6 ([HPW07, Corollary 10-12]). Let G be a nonabelian finite group. If G has ND, then the following properties hold.

(i) There is no nontrivial normal subgroup N of G and y, z ∈ G such that both y^{2}= 1
and yzyN 6= zN.

(ii) Either every element of order 2 is central orZ (G) is cyclic.

(iii) Let N be a nontrivial normal subgroup of G. If H is a subgroup of G such that H∩ N = 1, then Q[H] has no nonzero nilpotent elements.

Proof. (i) Prove by a contradiction. Suppose such N, y, z exist. Consider the nilpotent element α = (1 − y)z(1 + y) = z + zy − yz − yzy of Z[G]. The condition yzyN 6= zN implies that α 6= 0. Note that y 6∈ N otherwise yzyN = yzN = Nyz = Nz = zN. Moreover, z 6∈ N for

otherwise z ∈ N implies that yzy = (yzy^{−1})y^{2}∈ N since N is normal and y^{2}= 1. Now, we
have α bN/|N| ∈ Z[G] since bN/|N| is a central idempotent. As |N| 6= 1 and z 6∈ N, it follows
that either z ∈ zyN, z ∈ yzN or z ∈ yzyN. As y 6∈ N, by the assumption yzyN 6= zN, only
z∈ yzN is possible. Then z^{−1}yz∈ N. Since N is normal, we obtain y ∈ N, a contradiction.

(ii) Suppose that there is a non-central element y of order 2 and the center of G is not cyclic. Then [y, G] 6= 1 and there are two nontrivial central subgroups with trivial intersection. So at least one of two central subgroups does not contain [y, G], say N.

Choose z ∈ G such that [y, z] 6∈ N. Then we have yzyN 6= zN, a contradiction with (i).

(iii) Let e = bN/|N|. If Q[H] has some nonzero nilpotent elements, then we can take a
nilpotent element α in Z[H] \ |N|Z[H]. However, H ∩ N = 1 implies that Nh ∩ Nh^{0}= ∅
for h 6= h^{0}∈ H. It follows that eα 6∈ Z[G].

There are groups G having ND with non-cyclic center, for example, Q8× C_{2}. The
following result is an immediate consequence by Lemma II.2.6(iii).

Corollary II.2.7 ([HPW07, Corollary 13]). If H is a finite group such that Q[H] has some nonzero nilpotent elements, then the group H× N does not have ND for any nontrivial finite group N.

Proof. If H × N has ND, then it contradicts to Lemma II.2.6(iii).

The next result extends Lemma II.2.6(i) and it is a key ingredient to classify 3-groups and {2, 3}-groups having MJD. This result is just Theorem I.4.7(i) by replacing MJD by ND and we present the proof here for the reader.

Proposition II.2.8 ([LP09, Proposition 2.5]). Let G have ND and let NE G. If Y ≤ G, then either Y ⊇ N or Y N E G.

Proof. Suppose H = Y N is not normal in G. We claim that Y ⊇ N. Let g ∈ G be such that
H^{g}6= H. We have Y^{g}6⊆ H since N^{g}= N ⊆ H. Choose y ∈ Y with y^{g}6∈ H. Consider the
nilpotent element α = (1 − y)gbY ∈ Z[G] and let e = bN/|N|. Because, G has ND, we have
α e ∈ Z[G]. Note that bY bN = bH|Y ∩ N| and αe = (1 − y)g bH|Y ∩ N|/|N|. As y^{g}6∈ H, we
have that the two cosets gH and ygH are disjoint. Hence, we must have |Y ∩ N|/|N| ∈ Z.

Thus, N = Y ∩ N ⊆ Y .

II.2.2 SN Groups and SSN Groups

We recall the following definition which has appeared in Theorem I.4.7.

Definition II.2.9. A finite group G is said to have SN if for any subgroup Y of G and any normal subgroup N of G, we have either Y ⊇ N or Y NE G.

Thus, by Proposition II.2.8, we have the following conclusion.

Corollary II.2.10. If a finite group has ND, then it has SN.

The converse of above corollary is false, see Example II.2.21. Note that the dihedral
group D_{12}does not have SN by considering Y = hbi and N = ha^{3}i where a, b are given in
Example II.2.4. Thus, D_{12}does not have ND by Corollary II.2.10. This argument reduces
the computation in Example II.2.4.

Note that the ND property may be not inherited by quotient groups because in general nilpotent elements of an integral group rings can not be lifted via an epimorphism of groups. The reader can refer to [HPW07, p. 117] for an example. The following lemma shows another difference between ND and SN.

Lemma II.2.11. Let NE G. If G has SN, then so does G/N.

Proof. Let Y /N and M/N be two subgroups of G/N and M/N E G/N for some sub-groups Y, M of G. Then M is normal in G. If M/N 6⊆ Y /N, then M 6⊆ Y . Thus, Y ME G since G has SN. Hence, (Y /N)(M/N) = Y M/NE G/N, as desired.

For SN groups, we have the following result (cf. Lemma II.2.6 and Corollary II.2.7).

Lemma II.2.12 ([LP16, Lemma 2.1]). Let G be a finite group with SN, and let 1 6= NE G.

(i) If N is not cyclic, then G/N is either abelian or Hamiltonian.

(ii) If H ≤ G and H ∩ N = 1, then NHE G and H is either abelian or Hamiltonian. In particular, this holds if|H| are |N| are relatively prime.

(iii) If H is a subgroup of G generated by elements of prime order, then NHE G.

A group Q acting on another group P is said to be reducible if P = P_{1}× P_{2}for some
nontrivial Q-stable subgroups P_{1}, P_{2} of P. Otherwise, Q acts irreducibly on P. A special
type of non-nilpotent group with SN has been classified.

Lemma II.2.13 ([LP16, Lemma 2.4]). Let G = PQ be a finite group with P a normal
p-subgroup of G for some prime p, and Q a p^{0}-subgroup that acts nontrivially on P. If G
hasSN, then

(i) P is an elementary abelian p-group and Q acts irreducibly on P.

(ii) If, in addition, Q does not act faithfully on P, then P is cyclic of order p, and Q is a
cyclic q-group for some prime q6= p. Furthermore, |Q| ≥ q^{2}.

Definition II.2.14. A finite group G is said to have SSN if every subgroup of G has SN.

There is an SN group which is not SSN, see Example II.2.20. Finite SSN groups has been classified in [LP16]. We present the classification below. This classification will be used in Section II.3 and Section II.4. Recall that a group is said to have NCN if all its non-cyclic subgroups are normal. For p-groups, one has the following result.

Proposition II.2.15 ([LP16, Proposition 2.2]). Let G be a finite p-group for some prime p. Then G hasSSN if and only if G has NCN.

A complete classification of finite p-groups with NCN can be found in [BJ09, Theo-rem 1.1]. For nilpotent groups which are not p-groups, one has the following result.

Proposition II.2.16 ([LP16, Proposition 2.3]). Let G be a finite nilpotent group that is not a p-group for any prime p. Then G has SSN if and only if G is either abelian or Hamiltonian.

For solvable SSN groups that are not nilpotent, one has the following result.

Theorem II.2.17 ([LP16, Theorem 2.7]). Let G be a finite solvable group that is not nilpotent. Then G hasSSN if and only if G is one of the following two types of groups:

(i) G = P oQ where P is a normal elementary abelian p-subgroup of G for some prime
p, Q is a cyclic p^{0}-subgroup of G which acts faithfully on P, and every nontrivial
subgroup of Q acts irreducibly on P.

(ii) G = P o Q where P is a normal subgroup of G of order p for some prime p, Q is a
cyclic q-group for some prime q6= p with |Q| ≥ q^{2}, and Q does not act faithfully on
P.

There is exactly one non-solvable finite SSN group by the following result.

Theorem II.2.18 ([LP16, Theorem 3.3 and Corollary 3.4]). The alternating group A_{5}of
degree 5 is the unique nonabelian simple group with SSN. Moreover, A_{5} is the unique
non-solvable finite group withSSN.

II.2.3 Examples

We give some examples of SN groups and SSN groups and illustrate that each of which has ND or not.

Example II.2.19. (i) Let

G= (C_{2}×C_{2}) oC3= ha, b, c | a^{2}= b^{2}= c^{3}= 1, ab = ba, a^{c}= b, b^{c}= abi.

This action is faithful and every nontrivial subgroups of C_{3} acts irreducibly on C_{2}× C_{2}.
So G has SSN by Theorem II.2.17(i). Note that G ' A_{4}and G has ND.

(ii) If P is cyclic, then |Q| may not be a prime. For instance, let
G= C_{5}o C4= ha, b | a^{5}= b^{4}= 1, bab^{−1}= a^{2}i.

It easy to check that G has SSN since it is the type of groups in Theorem II.2.17(i).

Furthermore, we can use Lemma II.4.2 to check

Q[G] ' 2Q ⊕ Q(i) ⊕ M4(Q) and thus G has ND.

(iii) We also have an example of Theorem II.2.17(i) that both |P| and |Q| are primes.

Let

G= C_{11}o C5= ha, b | a^{11}= b^{5}= 1, bab^{−1}= a^{3}i.

Then G has SSN since it is the type of groups in Theorem II.2.17(i). Moreover,

Q[G] ' Q ⊕ Q(ε5) ⊕ M_{5}(Q(α))

by Lemma II.4.2 where α = ε_{11}+ ε_{11}^{3} + ε_{11}^{4} + ε_{11}^{5} + ε_{11}^{9} and Q(α) = Q(√

−11). Thus, G has ND.

Example II.2.20. We give an example to illustrate that there is a finite SN group which does not have SSN. This example will also show that the ND property can not be inherited by subgroups. Let

G= (C_{3}×C_{3}) oC8= ha, b, c | a^{3}= b^{3}= c^{8}= 1, ab = ba, a^{c}= b, b^{c}= abi.

We illustrate that G has SN but it does not have SSN. It is easy to check that each nontrivial
normal subgroup N of G must contain ha, bi. Thus, G/N is cyclic and then Y NE G for
every subgroup Y ≤ G. So G has SN. Consider the subgroup G_{1}= ha, b, c^{4}i = (C_{3}×C_{3}) o
C_{2}. Since a^{c}^{4} = a^{2}, N_{1}= hai is a normal subgroup of G_{1}. Let Y = hbc^{4}i. Then Y 6⊇ N_{1}.
Moreover, (bc^{4})^{c}^{4} = b^{2}c^{4}6∈ Y N_{1}. Thus, G_{1}does not have SN and then G does not have
SSN.

We can also conclude that G1 does not have ND by Proposition II.2.8. In particular, Q[G1] ' 2Q ⊕ 4M2(Q) has four matrix components (see [Gir06, Group Type 18/5 on p.

193]). However, G has ND since Q[G] has only one matrix component. To see this, we
first have Q[G] eG^{0}' Q[C8] ' 2Q⊕Q(i)⊕Q(ε8) where eG^{0}= bG^{0}/|G^{0}|. Note that G^{0}= ha, bi
and it is also a maximal abelian subgroup of G containing G^{0}. It is not hard to check that

193]). However, G has ND since Q[G] has only one matrix component. To see this, we
first have Q[G] eG^{0}' Q[C8] ' 2Q⊕Q(i)⊕Q(ε8) where eG^{0}= bG^{0}/|G^{0}|. Note that G^{0}= ha, bi
and it is also a maximal abelian subgroup of G containing G^{0}. It is not hard to check that