## Chapter 6 Main result

### In this chapter, we will complete our proof of Theorem 1.1. We still assume that v 6= 1 in F p as in the previous chapter. Let H be a finite subgroup of V (ZG). As we have discussed in chapter 4, it is enough to verify the case that |H| = pq. It is clear that H is not abelian, otherwise H is cyclic, and would contain an element of order pq. But this contradicts to Lemma 4.2 because G contains no element of order pq.

### Lemma 6.1. Suppose that u 6= v in F p , and let s ∈ H be an element of order p. If s ∼ x ^{i} y ^{j} for some i, j, then either i = 0 or j = 0.

### Proof. Let t ∈ H be an element of order q. Since H is not abelian, it must be a Frobenius group. Let 1 6= w ∈ hui such that tst ^{−1} = s ^{w} . Then we have x ^{i} y ^{j} ∼ s ∼ s ^{w} ∼ x ^{iw} y ^{jw} , which implies that x ^{i} y ^{j} is conjugate to x ^{iw} y ^{jw} in G by Corollary 2.13. But this contradicts to u 6= v unless i = 0 or j = 0.

### Let s ∈ H be an element of order p. Then s ∼ x ^{i} y ^{j} for some i, j by Lemma 4.2. We may assume that i 6= 0, and we may further assume that s ∼ xy ^{j} by replacing s with an appropriate power of s.

### Remark 6.2. Notice that if u 6= v, then we have j = 0 by Lemma 6.1.

### Moreover, we have that zxy ^{j} z ^{−1} = (xy ^{j} ) ^{u} whether u equals to v or not.

### 27

### CHAPTER 6. MAIN RESULT 28 We know that all Frobenius groups of order pq are isomorphic, so we may choose a suitable t ∈ H of order q such that tst ^{−1} = s ^{u} . It is clear that the homomorphism from H to F , which sends s, t to x, z respectively, is an isomorphism.

### Lemma 6.3. Let s, t be as above. Then t ∼ z.

### Proof. Since t ∈ H is an element of order q, we know that t ∼ z ^{i} for some i by Lemma 4.2 and by the fact that any q-element of G is of the form z ^{i} x ^{m} y ^{n} which is conjugate to z ^{i} . It is enough to show that i ≡ 1 mod q, or equivalently, θ _{1} (t) = θ _{1} (z).

### Let ¯ : ZG → ZF be the unique ring homomorphism satisfying ¯ x = x,

### ¯

### y = 1, ¯ z = z. Notice that κ _{3} (¯ h) = ψ _{1,0} (¯ h) = ψ _{1,0} (h), and κ _{2} (¯ h) = θ _{1} (¯ h) = θ _{1} (h) for all h ∈ H.

### Now κ _{3} (¯ s) 6= I _{q} implies ord(¯ s) = p, and κ _{2} (¯ t) 6= 1 implies ord(¯ t) = q.

### So we have that |h¯ s, ¯ ti| = pq, which implies h¯ s, ¯ ti ∼ = F . By Lemma 5.4, t ∼ z ¯ ^{i} since κ _{2} (¯ t) = θ _{1} (t) = ω ^{i} , and ¯ s ∼ x ^{j} for some j since Trκ _{3} (¯ s) = Trψ _{1,0} (s) 6= 0. Moreover, we know that ¯ s ^{p} = ¯ t ^{q} = 1, and ¯ t¯ s¯ t ^{−1} = ¯ s ^{u} because

### ¯ is a homomorphism. Since ZC-3 holds for F by Lemma 4.1, there exists w ∈ QF such that w ^{−1} h¯ s, ¯ tiw = F .

### Notice that w ^{−1} ¯ tw = z ^{i} x ^{l} for some l since z ^{i} is conjugate to z ^{i} x ^{l} by Lemma 5.1. Also notice that w ^{−1} sw = x ¯ ^{k} for some k. Now ¯ s ^{u} = ¯ t¯ s¯ t ^{−1} = wz ^{i} x ^{l} x ^{k} x ^{−l} z ^{−i} w ^{−1} = wx ^{ku}

^{i}

### w ^{−1} = ¯ s ^{u}

^{i}