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Lemma 3.1. Let G be a group of order p 2 q. Then G contains either a normal Sylow p- or a normal Sylow q-subgroup.

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Some Observations

In this chapter, we will investigate the structure of groups of order p 2 q in section 3.1. And we will give the criterion for checking whether ZC-3 holds for a group in section 3.2.

3.1 Groups of Order p 2 q

The following helps us to classify the structures of groups of order p 2 q.

Lemma 3.1. Let G be a group of order p 2 q. Then G contains either a normal Sylow p- or a normal Sylow q-subgroup.

Proof. Let m, n be the numbers of Sylow p- and Sylow q-subgroups of G respectively. We will show that either m = 1 or n = 1. By Sylow’s theorem, we know that

m | q, m ≡ 1 mod p, and

n | p 2 , n ≡ 1 mod q.

If p > q, then it can be easily deduced that m = 1, and hence G contains a normal Sylow p-subgroup.

12

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Suppose that q > p, and we suppose that n 6= 1. This implies that n = p or p 2 . First suppose that n = p, then p = kq + 1 for some k ≥ 1. Hence p > q, and this is a contradiction. So we have n = p 2 , and hence there are p 2 (q −1) = p 2 q −p 2 elements of order q. Therefore, the remaining p 2 elements form a normal Sylow p-subgroup.

By Lemma 3.1, we know that a group of order p 2 q must be of the following forms: C p

2

o C q , (C p × C p ) o C q , or C q × P with P a group of order p 2 .

We denote F p = Z/pZ to be the field of p elements.

Lemma 3.2. If G = (C p × C p ) o C q with q | p − 1, then we may write G as hx, y, z : x p = y p = z q = 1, zxz −1 = x u , zyz −1 = y v , xy = yxi,

where u, v ∈ F p such that u q = v q = 1.

Proof. Let C p × C p = hai × hbi. Define S i = hab i i\{1} , i = 0, . . . , p − 1, and S p = hbi\{1}. Suppose that z ∈ G is an element of order q, and we let [S i ] = {z −j S i z j | j ∈ Z}. Notice that z −q S i z q = S i for all i, so we have

|[S i ]| = 1 or q.

Suppose that q 6= 2. As q | p − 1, and there are p + 1 S i ’s, there must be two number m, n such that z −1 S m z = S m and z −1 S n z = S n . Let x ∈ S m , y ∈ S n , and we have proved the case.

Now suppose that q = 2. If (az) 2 = 1, then we have zaz = a −1 , and we may just choose x = a. If (az) 2 6= 1, then we choose x = (az) 2 , since zxz = (az)x(az) −1 = x. In both cases, there must be an element y ∈ C p ×C p , and y 6∈ hxi such that zyz ∈ hyi by using the facts that |[S i ]| is either 1 or 2 for all i and that p + 1 is even. Therefore, the proof is complete.

3.2 Criterion

To prove that ZC-3 holds for a group, the following proposition is a crucial

criterion which is from [Val94, Lemma 4].

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Proposition 3.3. Let G be a finite group, G 0 a subgroup of G, and H a finite subgroup of V (ZG). Suppose that there exists an isomorphism φ : H → G 0

such that, for all h ∈ H and for all complex irreducible characters χ of G, χ(h) = χ(φ(h)). Then G 0 ∼ H.

To show Proposition 3.3, we shall use the following lemmas.

Lemma 3.4. Let F ⊂ E be a field extension. Let M i ∈ M n (F ), i = 1, . . . , k, and let

M i X = 0, X ∈ M n×1 (E)

be a system of linear equations. Suppose that 0 6= Y =

 y 1

.. . y n

 ∈ M n×1 (E) is a solution of the system.

1. There exists a nonzero solution C ∈ M n×1 (F ).

2. Let {v 1 , . . . , v l } be an F -basis of

{X ∈ M n×1 (F ) : M i X = 0, i = 1, . . . , k} . Then {v 1 , . . . , v l } spans

{X ∈ M n×1 (E) : M i X = 0, i = 1, . . . , k}

over E.

Proof. Let F 0 be the F -subspace of E spanned by y 1 , . . . , y n , and let b 1 , . . . , b j

be an F -basis of F 0 . We may write y i = c i1 b 1 + . . . + c ij b j , where c it ∈ F , for all i, t. Now

0 = M i Y = b 1 M i

 c 11

.. . c n1

 + · · · + b j M i

 c 1j

.. . c nj

 ,

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and b 1 , . . . , b j is linearly independent over F imply M i

 c 1s

.. . c ns

 = 0 for all

i, s. Since Y 6= 0, we can choose a nonzero C =

 c 1s

.. . c ns

 , and the first claim is proved.

For the second part, let D =

 d 1

.. . d n

 ∈ M n×1 (E) be any solution of the system. Let F 00 be the F -subspace of E spanned by d 1 , . . . , d n , and let e 1 , . . . , e r be an F -basis of F 00 . Write d i = f i1 e 1 + . . . + f ir e r , where f it ∈ F , for all i, t. Then

0 = M i D = e 1 M i

 f 11

.. . f n1

 + · · · + e r M i

 f 1r

.. . f nr

 ,

which implies M i

 f 1s

.. . f ns

 = 0 for all i, s. Hence

 f 1s

.. . f ns

 ∈ span F (v 1 , . . . , v l ), and this implies that D ∈ span E (v 1 , . . . , v l ).

Lemma 3.5. Let R be an Artinian ring with unity and let 0 6= u ∈ R. Then u is either a zero divisor or a unit.

Proof. Suppose that u is not a zero divisor. Consider the decending chain of left ideals:

Ru ⊃ Ru 2 ⊃ . . . ⊃ Ru n ⊃ . . . .

Since R is Artinian, there exists k ∈ N such that Ru k = Ru k+1 . So u k =

vu k+1 for some v ∈ R, and hence (1 − vu)u k = 0. Now the hypothesis that

u is not a zero divisor implies vu = 1, so u has a left inverse. By considering

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the decending chain of right ideals:

uR ⊃ u 2 R ⊃ . . . ⊃ u n R ⊃ . . . ,

we can show that u has a right inverse similarly. Therefore, u is a unit.

Lemma 3.6. Let K be an infinite field and let f (x 1 , . . . , x n ) ∈ K[x 1 , . . . , x n ] be a polynomial of n indeterminants. If f (y 1 , . . . , y n ) = 0 for all y i ∈ K, then f (x 1 , . . . , x n ) ≡ 0.

Proof. We prove this lemma by induction on n. The case that n = 1 is easy, since there are finitely many roots in K for a polynomial of positive degree. Suppose the lemma is true for all n ≤ k. When there are k + 1 indeterminants, we may write f (x 1 , . . . , x k+1 ) as

f (x 1 , . . . , x k+1 ) = f m (x 1 , . . . , x k )x m k+1 + . . . + f 0 (x 1 , . . . , x k ).

If we fix y 1 , . . . , y k ∈ K, then f (y 1 , . . . , y k , y k+1 ) = 0 for all y k+1 ∈ K implies that f i (y 1 , . . . , y k ) = 0 for all i. Also notice that the choices of y 1 , . . . , y k are arbitrary, so we have f i (x 1 , . . . , x k ) ≡ 0 by induction hypothesis. Therefore f (x 1 , . . . , x k , x k+1 ) ≡ 0.

Lemma 3.7. Let F ⊂ E be infinite fields and G any finite group. Suppose that two finite subgroups H 1 , H 2 ≤ U (F G) are given. If H 1 , H 2 are conjugate in EG, then H 1 , H 2 are conjugate in F G.

Proof. This proof is slightly modified from [PMS84, Lemma 5]. Suppose that u ∈ U (EG) such that u −1 h i u = h 0 i where h i ∈ H 1 and h 0 i ∈ H 2 . Write x =

n

X

i=1

x i g i ∈ EG where g i ∈ G and |G| = n, and we define a bijection f :

EG → M n×1 (E) by f (x) =

 x 1

.. . x n

 . Consider the functions F i : F G → F G defined by F i (x) = h i x − xh 0 i . They are clearly linear functions since

F i (cx + y) = h i (cx + y) − (cx + y)h 0 i

= c(h i x − xh 0 i ) + (h i y − yh 0 i ) = cF i (x) + F i (y)

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for all c ∈ F and x, y ∈ F G. Let M i be the matrix representation of F i with respect to the basis {g i } n i=1 . Then the equations F i (x) = 0 give a system of linear equations

M i X = 0, X =

 x 1

.. . x n

 , M i ∈ M n (F ) ⊂ M n (E), i = 1, · · · , |H 1 |.

Let

V F = {v ∈ F G | h i v = vh 0 i for all i}.

Then

f (V F ) = {X ∈ M n×1 (F ) | M i X = 0 for all i}.

Since f (u) ∈ M n×1 (E) is a solution of all equations M i X = 0, by Lemma 3.4, it follows that f (V F ) 6= {0}, and we may choose v 1 , · · · , v l ∈ M n×1 (F ), l ≥ 1 such that all elements in f (V F ) are F -linear combinations of v 1 , · · · v l and all solutions in M n×1 (E) are E-linear combinations of v 1 , · · · , v l . We are also given that there exists r i ∈ E such that f (u) = P r i v i in EG.

Suppose, by way of contradiction, that for all elements in V F are zero divisors. Let T be the regular representation of EG, and let S = T ◦ f −1 , S(v i ) = L i . Define the polynomial φ(x 1 , . . . , x l ) ∈ F [x 1 , . . . , x l ] by

φ(x 1 , · · · , x l ) = det X x i L i 

.

We want to show that φ ≡ 0. For any element X =

l

X

i=1

y i L i ∈ f (V F ),

φ(y 1 , . . . , y l ) = det(

l

X

i=1

y i L i )

= det(

l

X

i=1

y i S(v i ))

= det(S(X))

= det(T (f −1 (X))) = 0

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since f −1 (X) is a zero divisor in F G. This implies that φ ≡ 0 by Lemma 3.6.

But

φ(r 1 . . . r l ) = det(

l

X

i=1

r i S(v i )) = det(T (u)) 6= 0

since u is a unit in EG, and this is a contradiction. Hence we have proved that there exists t ∈ V F which is not a zero divisor. Since F G is Artinian it follows from Lemma 3.5 that t must be invertible in F G, and this proves the lemma.

The following lemma is modified from [RS83, page 258].

Lemma 3.8. The following statements are equivalent.

1. Two finite subgroups H 1 and H 2 of G are conjugate in QG.

2. There exists an isomorphism φ : H 1 → H 2 , and for every irreducible complex matrix representation T of G, there exists an invertible matrix U such that U −1 T h U = T φ(h) for all h ∈ H 1 .

Proof. (2) ⇒ (1): We first write down the Wedderburn decomposition of the complex group algebra CG,

CG

∼ η

= ⊕ n i=1 M k

i

(C),

where M k

i

(C) is the matrix ring which is isomorphic to the sum of iso- morphic CG-submodules of CG corresponding to an irreducible representa- tion T i . Since by assumption, there exists an invertible matrix U i such that U i −1 T i (h)U i = T i (φ(h)). Now let α = η −1 (U 1 , . . . , U n ). Then α −1 H 1 α = H 2 . Finally, by Lemma 3.7, the fact that H 1 is conjugate to H 2 in CG implies that H 1 is conjugate to H 2 in QG.

(1) ⇒ (2) is trivial.

With these lemmas at hand, we are able to prove Proposition 3.3

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Proof of Proposition 3.3. Let T be an irreducible representation of G. Then

T and T ◦ φ can be regarded as representations of H. Let χ and χ φ be

the characters afforded by T and T ◦ φ respectively. Then χ| H = χ φ | H by

assumption, and hence T is equivalent to T ◦ φ as representations of H by

Proposition 2.15. That is, there exists an invertible matrix U such that

U −1 T h U = T φ(h) for all h ∈ H. Now apply Lemma 3.8, and the result

follows.

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