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.

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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.

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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

, and hence u ≡ u i mod p. As the order of u in F p is q, we have i ≡ 1 mod q, or ¯ t ∼ z. Therefore, we have proved the lemma since θ 1 (t) = θ 1 (¯ t) = θ 1 (z).

Finally, we can prove our main result.

Proof of Theorem 1.1. We only have to verify that ZC-3 holds for G, which is the remaining case.

First, suppose that v 6= 1 in F p . Let s, t ∈ H be as above. By Remark

6.2, we know that the homomorphism φ : H → hxy j , zi with φ(s) = xy j ,

φ(t) = z is an isomorphism. By Lemma 5.1, Remark 5.2, and Lemma 6.3,

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CHAPTER 6. MAIN RESULT 29 we have

s m t n ∼ t n ∼ z n ∼ (xy j ) m z n = φ(s m t n )

as n 6= 0, and s ∼ xy j implies s m ∼ (xy j ) m . Hence h ∼ φ(h) for all h ∈ H. Therefore, Lemma 2.11 implies that χ(h) = χ(φ(h)) for all irreducible characters χ for all h ∈ H, and Lemma 3.3 yields the theorem.

If v = 1 in F p , then hyi is the center of G. It is clear that |hy, Hi| = pq or

p 2 q since hy, Hi is still a finite subgroup of V (ZG). If it is the first case, then

y ∈ H, and we may write H = hyi × hti for some t of order q. Lemma 4.2

and Lemma 5.1 implies that there exists w ∈ QG such that w −1 htiw = hzi,

and hence w −1 Hw = hy, zi, as desired. For the latter case, it follows from

Lemma 4.6 that there exists w ∈ QG such that w −1 hy, Hiw = G. Hence

w −1 Hw is a subgroup of G.

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