Chapter 4
Some Known Results and Simple Cases
We begin this chapter with some known results, which are useful to us for verifying ZC-3 for groups of order p2q. The following is from [Val94].
Lemma 4.1. Let G ∼= hai o X, where hai is a cyclic group of order n, and X is an arbitrary abelian group of order m with (m, n) = 1. Then ZC-3 holds for G.
We have shown that there is either a normal Sylow p- or a normal Sylow q-subgroup in a group of order p2q in Lemma 3.1. If it contains a normal Sylow q- or a normal cyclic Sylow p-subgroup, then Lemma 4.1 implies that ZC-3 holds for such a group. Hence, the remaining case we have to verify is nonabelian groups of the form (Cp× Cp) o Cq.
We collect two lemmas about the groups of the form (Cp × Cp) o Cq. Lemma 4.2. [Her06, Theorem 1.2] Suppose that the finite group G has a normal Sylow p-subgroup with an abelian complement. Then ZC-1 holds for G.
Lemma 4.3. [SSZ84] Let G = A o B with A abelian, B a nilpotent class two group, and (|A|, |B|) = 1. Then ZG = ZH implies G ∼= H.
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CHAPTER 4. SOME KNOWN RESULTS AND SIMPLE CASES 21 Let H be a finite subgroup of V (Z[(Cp× Cp) o Cq]). Then we have that
|H| divides p2q by Lemma 2.19. Hence |H| = 1, p, q, p2, pq or p2q. In order to reduce the cases, we also record the following lemma.
Lemma 4.4. [DJ96, Corollary 2.4] Let G be an extension of a nilpotent group A by a group X such that any finite p-subgroup of V (ZX) is conjugate to a p-subgroup of X in QG. If the order of A and X are relatively prime, then any finite p-subgroup of V (ZG) is conjugate to a p-subgroup of G in QG.
Let R be a ring with unity. We say that f is a normalized automor- phism of RG if f is an augmentation preserving automorphism, that is, if
ζ∗(f (u)) = ζ∗(u)
for all u ∈ ZG, where ζ∗ is the augmentation mapping defined in chapter 2.
In the following theorems, we denote Zp to be the set of p-adic integers and Qp to be the p-adic field.
Theorem 4.5. [HK02, Theorem 4.6] Let G be a finite group with a normal Sylow p-subgroup N containing its own centralizer, CG(N ) ⊂ N . Then for any normalized automorphism α of ZpG, the group α(G) and G are conjugate in the units of ZpG.
Theorem 4.6. ZC-2 holds for (Cp× Cp) o Cq.
Proof. Let H ≤ V (ZG) be a finite subgroup of order p2q. By Lemma 2.21, we have ZG = ZH, and it follows from Lemma 4.3 that G ∼= H. We can extend the isomorphism to the normalized automorphism α : ZpG → ZpH by Corollary 2.2. Theorem 4.5 implies that H = α(G) and G are conjugate in the units of ZpG ⊂ QpG.
Notice that Q ⊂ Qp, Lemma 3.7 implies that H = α(G) is conjugate to G in QG.
CHAPTER 4. SOME KNOWN RESULTS AND SIMPLE CASES 22 Now, the cases that |H| = p or q follow from Lemma 4.2, the case that
|H| = p2 follows from Lemma 4.4, and the case that |H| = p2q is done by Theorem 4.6. Therefore, we assume that |H| = pq, which is the remaining case to be verified.
First, we suppose that q - p − 1. We will see that (Cp× Cp) o Cq which is not abelian exists only if q | p + 1. Let z ∈ (Cp × Cp) o Cq be an element of order q, and let hzi act on (Cp× Cp) by conjugation. We denote G0 to be the set fixed by hzi. Then it is not difficult to verify that G0 is a subgroup of (Cp× Cp). If |G0| = p2, then (Cp × Cp) o Cq is abelian. If |G0| = p, then q | p2 − p, which is impossible since we assume that p, q are distinct prime numbers. Hence |G0| = 1, which implies q | p2− 1, and we have q | p + 1.
If p > q, then H must be abelian and hence cyclic. So there is an element of order pq, and the result follows from Lemma 4.2. The only case that p < q is the alternating group A4, which is a well known result. For convenience to the readers, we include a proof here.
Lemma 4.7. [PMS84, Corollary 1] Let G be a split extension A o X where A is an abelian normal p-group and X any group. If s ∈ U (1 + ∆Z(G, A)) is a torsion element, then the order of s divides |A|.
Theorem 4.8. ZC-3 holds for A4.
Proof. As discussed above, we may just assume that H is a finite group of V (ZA4) of order 6. Consider the canonical mapping ρ : ZA4 → Z[A4/A04] ∼= ZC3, and let θ = ρ|H. By Higman’s theorem (Theorem 2.18) we know that any torsion unit in ZC3 is trivial, and hence | ker θ| = 2 or 6. Notice that H can not be abelian, otherwise H contains an element of order 6 but A4 does not, contradicting to Lemma 4.2. Hence H ∼= S3 and contains four elements of order 2, which must be sent to 1 via θ, so | ker θ| = 6, or ker θ = H.
Now any element s ∈ H belongs to U (1 + ∆Z(A4, A04)), and hence has order dividing 4 by Lemma 4.7. This is a contradiction since H contains elements of order 3. Therefore, V (ZA4) has no finite subgroup of order 6.
