# Non-solvable SSN Groups

## II. Nilpotent Decomposition in Integral Group Rings

### II.4.3 Non-solvable SSN Groups

According to Theorem II.2.18, A5is the unique non-solvable finite SSN group.

Let A4= h(1, 2, 3), (1, 2)(3, 4)i and K = h(1, 2)(3, 4), (1, 3)(2, 4)i be the Klein group. primitive central idempotent of Q[A5] (see [JdR16, Example 3.4.5]). Let

α =a(1, 2)(3, 4)(1 − a)b Hence, αe has support b with the coefficient

1

It follows from Theorem II.2.18 and the argument above that

Corollary II.4.18. If a group has both ND and SSN, then it is solvable.

Now, combining Theorem II.2.17, Lemma II.4.2, Proposition II.4.7, Proposition II.4.15 and Corollary II.4.18, we can get the following result.

Theorem II.4.19. Let G be a non-nilpotent SSN group. Then Q[G] has only one matrix component if and only if G is one of the following:

(i) G = hxipo hyinwherehyinacts faithfully onhxip;

(ii) G = Cnpo Cqwhere n≥ 2, pn− 1 = (p − 1)q and Cqacts nontrivially on Cnp;

(iii) G = hxipo hyiqk where k≥ 2, y acts nontrivially on x, yqacts trivially on x and one of the following holds:

(a) q 6= 2 and vq(p − 1) = 1;

(b) q = 2 and either k = 2, or k > 2 and p ≡ 5 (mod 8).

Here p, q are distinct primes and n, k ∈ N.

In particular, for groups of type(i) and (iii), the matrix components are isomorphic to Mn(Q(εp)σ) and Mq(Q(εp)σ) where σ ∈ Aut(Q(εp)/Q) such that σ (εp) = εipif xy= xi; for the group of type (ii), it is Mq(Q(εp)). Moreover, rational group algebras of type (i) and (ii) have only one commutative component; for type (iii), it has k − 1 non-commutative components which are division algebras.

We present a conjecture below based on Corollary II.3.14 and Corollary II.4.5.

Conjecture 3. Let G be a finite group such that Q[G] has some nonzero nilpotent ele-ments. If G has ND, then Q[G] has only one matrix component.

This conjecture implies Conjecture 2 by Theorem II.2.1.

Multiplicative Jordan Decomposition

Based on the work discussed in Chapter I, the remaining groups to solve the MJD problem are Q8×Cpfor prime p > 41 with 4 | ordp(2) and G(q, n, k) = Cqo Cnkwith the action describing in Theorem I.4.3(iv) for prime q such that either q ≡ 1 (mod 4), n = 2, k≥ 3, or q = 7, n = 3, k ≥ 2.

Problem 1. Determine groups Q8×Cpand G(q, n, k) as above such that MJD holds.

Note that the rational group algebra Q[Q8×Cp] has only one matrix component. Ac-cording to Theorem II.4.19(iii), Q[G(q, n, k)] has only one matrix component for those cases above except q ≡ 1 (mod 8), n = 2 and k ≥ 3. Thus, Conjecture 2 holds if and only if Z[G(q, 2, k)] does not have MJD for q ≡ 1 (mod 8) and k ≥ 3.

We remark here that groups Q8×Cpand G(q, n, k) are isomorphic to the multiplicative groups of some finite left braces ([Ced18, Theorem 7.4]). A left brace is a set B with two operations + and · such that (B, +) is an abelian group, (B, ·) is a group and

a· (b + c) + a = a · b + a · c

for all a, b, c ∈ B. The groups (B, +) and (B, ·) are called the additive group and the multiplicative group of the left brace B. According to [Ced18, Proposition 9.16], a group G is the multiplicative group of a left brace if and only if there exists a left ideal L of Z[G] such that the augmentation ideal ω (Z[G]) = G − 1 + L and G ∩ (1 + L) = 1. In particular, U (Z[G]) = ±GH where H = (1 + L) ∩ U (Z[G]) is a subgroup of U (Z[G]) and ±G ∩ H = 1.

Problem 2. Can we use this decompositionU (Z[G]) = ±GH to study the MJD problem?

Braces can be viewed as a generalization of Jacobson radical rings, namely rings without unity such that their Jacobson radicals are themselves. The notion of (left) braces was introduced by W. Rump [Rum07] in order to study non-degenerate involutive set-theoretic solutions of the Yang-Baxter equation. It was suggested by V.G. Drinfeld in [Dri92] to find all the set-theoretic solutions of the Yang-Baxter equation. The reader can consult the survey article [Ced18].

Up to now, we have focused on the MJD problem for Z[G]. It is natural to ask the same problem forOK[G] whereOK is the ring of integers of an algebraic number field K.

91

Similarly,OK[G] is said to have MJD if the semisimple part and the unipotent part of each unit ofOK[G] in K[G] are still inOK[G]. Clearly, ifOK[G] has MJD, then Z[G] has MJD sinceOK∩ Q = Z (see [IR90, Proposition 6.1.1]). However, the converse is false by the following result when G is nonabelian of order 27 (cf. Theorem I.4.8).

Theorem ([LP13, Theorem 3.12]). Let K 6= Q be an algebraic number field. If G is a finite3-group andOK[G] has MJD, then G is abelian.

Problem 3. Can we extend the above result to other finite groups?

It seems thatOK[G] is less likely to have MJD the largerOK is. For example, we have the following

Theorem ([HPW07, Theorem 3]). For G a finite group andO the ring of all algebraic integers, the group ringO[G] has MJD if and only if G is abelian.

Problem 4. Classify finite groups G such thatOK[G] has MJD where K 6= Q is an alge-braic number field.

We can also ask the MJD problem forR[G] where R is a subring of a perfect field F, see Definition I.2.5. For instance, we have the following

Problem 5. Classify finite groups G and primes p such that Zp[G] has MJD where Zpis the ring of p-adic integers.

Problem 6. Classify finite groups G and primes p such that Z(p)[G] has MJD where Z(p)

is the localization of Z at p, namely, Z(p)= {ab | a, b ∈ Z, p - b, gcd(a, b) = 1 if a 6= 0}.

Problem 7. Classify finite groups G and primes p such that (Z[1p])[G] has MJD where Z[1p] = {pan | a ∈ Z, n ∈ Z≥0}.

Problem 8. Classify finite groups G and primes p such that (Fq[x, x1/p, x1/p2, . . .])[G] has MJD where q is a power of p, Fqis a finite field of order q and Fq(x, x1/p, x1/p2, . . .) is a perfect field for an indeterminate x.

The study of MJD problem leads us to focus on semisimple units and unipotent units.

Let G be a finite group and let S (G), U (G) be subgroups of U (Z[G]) generated by semisimple units and unipotent units ofU (Z[G]), respectively. Both subgroups S (G) and U (G) are normal and, therefore, J (G) := S (G)U (G) is a normal subgroup of U (Z[G]). J (G) is called the Jordan subgroup of U (Z[G]). Clearly, J (G) = U (Z[G]) if and only if Z[G] has MJD. From this point of view, the quotient group U (Z[G])/J (G) tells how close Z[G] have MJD. Note that U (Z[G]) is finitely generated, thanks to C.L.

Siegel [Sie43] and A. Borel and Harish-Chandra [BH62]. Moreover, if α ∈U (Z[G]), then αn ∈J (G) for some n ∈ N. To see this, write its unipotent part αu= 1 + β for some nilpotent element β ∈ Q[G]. Choose n = k!mk where βk+1= 0 and mβ ∈ Z[G].

Then αun∈ Z[G]. Note that αun− 1 is a nilpotent element of Z[G]. It follows that αunlies inU (Z[G]) so does αsn. Thus, we have the following.

Theorem ([HP17, Theorem 4.3]). The quotient groupU (Z[G])/J (G) is a finitely gen-erated torsion group.

In view of [Kle87, Corollary 1.3], if G is of odd order, then U (Z[G])/S (G), and henceU (Z[G])/J (G), is a finite group.

Problem 9 ([HP17, Problem 1]). Classify the finite groups G of even order for which U (Z[G])/J (G) is a finite group.

Nilpotent Decomposition

Although the concept of ND property comes from the study of MJD problem, the ND property has its own interests. If F is a field (not necessarily perfect) such that char(F) does not divide |G|, then F[G] is semisimple. For a subring R of F, the group ring R[G]

is said to have ND if for every nilpotent element α ∈ R[G] and every central idempotent e∈ F[G], we have αe ∈ R[G]. Thus, we can ask the following problems.

Problem 10. Classify finite groups G and algebraic number fields K such that the group ringOK[G] has ND.

Problem 11. Classify finite groups G and primes p such that R[G] has ND where the ring R= Zp, Z(p), Z[1p] and Fq[x] where q is a power of p and x is an indeterminate.

Back to the integral group ring Z[G]. Denote {ei}i∈Ithe collection of primitive central idempotents of Q[G]. If α ∈ Z[G] is nilpotent, then α = ∑iαiwhere αi= αei. It follows that αiαj= αjαi= 0 for i 6= j. Thus, 1 + α = ∏i∈I(1 + αi). In other words, a unipotent unit of Z[G] is a product of unipotent units of Q[G]. As above, we can choose a positive integer n such that (1 + αi)n∈U (Z[G]) for all i. Thus, (1+α)nis a product of unipotent units of Z[G]. Denote pU (G) the subgroup of U (Z[G]) generated by primitive unipotent units, namely, units of the form 1 + β where β ∈ Z[G] is nilpotent such that β = β ei

for some i ∈ I. Then pU (G) is normal in U (Z[G]) and it is a subgroup of U (G). The above discussion shows thatU (G)/pU (G) is a torsion group. Clearly, if Z[G] has ND, then U (G)/pU (G) is a trivial group. Conversely, if U (G) = pU (G) and α ∈ Z[G]

is nilpotent, then 1 + α = ∏i∈I(1 + γi) where each term 1 + γi is a product of primitive unipotent units of the form 1 + β with β ∈ Z[G] and β ei = β . Observe that 1 + α = 1 + ∑i∈Iγi. By the uniqueness of Wedderburn decomposition, αei= γi∈ Z[G] for each i.

Hence, we have the following

Theorem. For a finite group G, Z[G] has ND if and only if U (G)/pU (G) is a trivial group.

Problem 12. Classify finite groups G such thatU (G)/pU (G) is a finite group.

Based on the study in Chapter II, the ND property relates to the number of matrix components. In general, we can ask

Problem 13. What is the connection between the groupU (G)/pU (G) and matrix com-ponents of Q[G]?

The ND property leads to finite SN groups. The class of SN groups is much bigger than the class of SSN groups. For example, every finite simple group has SN (cf. The-orem II.2.18) by definition. Moreover, the symmetric group Sn for n ≥ 5 has SN since the alternating group An is the only normal subgroup and it is of index 2. It would be interesting to characterize finite groups which has SN.

Problem 14. Classify finite SN groups.

We may classify rational group algebras with only one matrix component once we have the classification of finite SN groups.

Problem 15. Classify finite groups G such that Q[G] has only one matrix component.

It seems that we may also classify integral group rings with ND via SN groups. How-ever, it would be still a challenge to check the ND property because it is not easy to find

“non-obvious” nilpotent elements. Here we mean that an “obvious” nilpotent element of Z[G] is of the form (1 − y)gbY or bY g(1 − y) for some subgroup Y of G, y ∈ Y and g ∈ G. For example, G(p, 2, k) has SSN and Q[G(p, 2, k)] has at least two matrix components when p6≡ 5 (mod 8) and k ≥ 3. In [HP17, Theorem 4.1], Hales and Passi found a particular nilpotent element of Z[G(p, 2, k)] with p ≡ 3 (mod 4) to illustrate that this integral group ring does not have ND and thus it does not have MJD. This nilpotent element can not be used for Z[G(p, 2, k)] when p ≡ 1 (mod 8). We still do not know whether Z[G(p, 2, k)]

with p ≡ 1 (mod 8) has ND or not. So we have the following challenging problem which overlaps Problem 10.

Problem 16. Classify finite groups G such that Z[G] has ND.

Finally, we denote OMC to be the rational group algebra having at most one matrix component. We end this dissertation by giving the following picture which shows the relations between MJD, ND, SN, SSN and OMC, see also Subsection II.2.3 for details.

4

4

8

8

4

3

8

3

8

8

8

3

8

### ]

Fig. 1: Relations between MJD, ND, SN, SSN, OMC

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Notations:

2, MJD implies at most one matrix component, 53

3, ND implies at most one matrix com-ponent, 90

cyclic algebra, 58

element, 13 linear operator, 12 primitive, 93 unit, 13 unipotent part, 13 unit group, 7

Wedderburn decomposition, 8 reduced degree, 8

Wedderburn-Artin decomposition for Z[G] having MJD, 52 Yang-Baxter equation, 91