Page 85 - Ellingham, Mark, Mariusz Meszka, Primož Moravec, Enes Pasalic, 2014. 2014 PhD Summer School in Discrete Mathematics. Koper: University of Primorska Press. Famnit Lectures, 3.
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mož Moravec: Some Topics in the Theory of Finite Groups 73
Suppose sq = 1. Denote P = 〈a 〉 and Q = 〈b 〉. Then a b = a k and b a = b for some
integers k and . Therefore a k −1 = [a ,b ] = b − +1. Since the orders and b are coprime, it
follows that [a ,b ] = 1, hence G =∼ Cp × Cq ∼= Cpq .
Now let sq = p , that is, let q divide p −1. We still have a b = a k . By induction, a bs = a k s .
Since |b | = q , we conclude that k q ≡ 1 mod p . There are exactly q solutions to this equa-
tion; if k is one of them, the others are powers of k . By replacing b by a power of itself we
see that all these solutions give rise to the same group, namely, a group with presentation
〈a ,b | a p = b q = 1, a b = a k 〉
for some k satisfying k q ≡ 1 mod p , k ≡ 1 mod p .
More on finite p -groups will be discussed later on. We conclude with two useful
lemmas which are of similar nature:
Lemma 3.2.28 (The Frattini argument) Let G be a group and H a finite normal sub-
group. If P is a Sylow p -subgroup of H , then G = NG (P)H .
PROOF. For g ∈ G we have P g ≤ H and P g = Ph for some h ∈ H . Thus g h−1 ∈ NG (P).
Lemma 3.2.29 If P is a Sylow p -subgroup of a finite group G and NG (P) H G , then
H = NG (H ).
PROOF. Clearly P H NG (H ). By Frattini’s argument we have that NG (H ) = NNG (H)(P)H .
But NNG (H)(P) ≤ NG (P) ≤ H , hence the result.
3.2.5 An estimate of the number of finite groups
In this short section we derive a rough bound for the number of of groups of order n.
Lemma 3.2.30 A group G of order n can be generated by a set of at most log2 n elements.
PROOF. Choose a non-trivial element g 1 ∈ G , and let G1 = 〈g 1〉. If G1 = G , then stop.
Otherwise choose g 2 ∈ G − G1 and let G2 = 〈g 1, g 2〉. Repeat the procedure until we find
g 1, . . . , g k ∈ G such that G = 〈g 1, . . . , g k 〉.
We prove that |Gi | ≥ 2i for all i = 1, . . . , k ; this suffices to prove our lemma. The proof
is by induction on i , the case i = 1 being obvious, Suppose that |Gi | ≥ 2i . Since |Gi | di-
vides |Gi +1| and Gi = Gi +1, we have |Gi +1| ≥ 2|Gi | ≥ 2i +1, as required.
Proposition 3.2.31 The number of groups of order n is at most n n log2 n .
Suppose sq = 1. Denote P = 〈a 〉 and Q = 〈b 〉. Then a b = a k and b a = b for some
integers k and . Therefore a k −1 = [a ,b ] = b − +1. Since the orders and b are coprime, it
follows that [a ,b ] = 1, hence G =∼ Cp × Cq ∼= Cpq .
Now let sq = p , that is, let q divide p −1. We still have a b = a k . By induction, a bs = a k s .
Since |b | = q , we conclude that k q ≡ 1 mod p . There are exactly q solutions to this equa-
tion; if k is one of them, the others are powers of k . By replacing b by a power of itself we
see that all these solutions give rise to the same group, namely, a group with presentation
〈a ,b | a p = b q = 1, a b = a k 〉
for some k satisfying k q ≡ 1 mod p , k ≡ 1 mod p .
More on finite p -groups will be discussed later on. We conclude with two useful
lemmas which are of similar nature:
Lemma 3.2.28 (The Frattini argument) Let G be a group and H a finite normal sub-
group. If P is a Sylow p -subgroup of H , then G = NG (P)H .
PROOF. For g ∈ G we have P g ≤ H and P g = Ph for some h ∈ H . Thus g h−1 ∈ NG (P).
Lemma 3.2.29 If P is a Sylow p -subgroup of a finite group G and NG (P) H G , then
H = NG (H ).
PROOF. Clearly P H NG (H ). By Frattini’s argument we have that NG (H ) = NNG (H)(P)H .
But NNG (H)(P) ≤ NG (P) ≤ H , hence the result.
3.2.5 An estimate of the number of finite groups
In this short section we derive a rough bound for the number of of groups of order n.
Lemma 3.2.30 A group G of order n can be generated by a set of at most log2 n elements.
PROOF. Choose a non-trivial element g 1 ∈ G , and let G1 = 〈g 1〉. If G1 = G , then stop.
Otherwise choose g 2 ∈ G − G1 and let G2 = 〈g 1, g 2〉. Repeat the procedure until we find
g 1, . . . , g k ∈ G such that G = 〈g 1, . . . , g k 〉.
We prove that |Gi | ≥ 2i for all i = 1, . . . , k ; this suffices to prove our lemma. The proof
is by induction on i , the case i = 1 being obvious, Suppose that |Gi | ≥ 2i . Since |Gi | di-
vides |Gi +1| and Gi = Gi +1, we have |Gi +1| ≥ 2|Gi | ≥ 2i +1, as required.
Proposition 3.2.31 The number of groups of order n is at most n n log2 n .
