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18 Application: Groups of Order Pq 18 Application: groups of order pq ∼ Recall. If G is a group, jGj = p where p is a prime then G = Z=pZ. Goal. Classify all groups of order pq where p, q are prime numbers. 18.1 Proposition. If G is a group of order p2 for some prime p then either ∼ 2 ∼ G = Z=p Z or G = Z=pZ ⊕ Z=pZ. Proof. It is enough to show that G is abelian since then the statement follows from the classification of finitely generated abelian groups (14.7). Since G is a p-group by Theorem 16.4 we have Z(G) 6= feg. If Z(G) = G then G is abelian. If Z(G) 6= G then G=Z(G) is a group of order p and thus it is a non-trivial cyclic group. This is however impossible by Problem 8 of HW 1. 18.2 Proposition. If G is a group of order pq for some primes p, q such that p > q and q - (p − 1) then ∼ G = Z=pqZ Proof. If is enough to show that G contains an element of order pq. Let sp denote the number of Sylow p-subgroups of G. By the Third Sylow Theorem (17.5) we have sp j q and sp = 1 + kp Since q is a prime the first condition gives sp = 1 or sp = q. Since p > q the second condition implies then that sp = 1. Similarly, let sq be the number of Sylow q-subgroups of G. We have sq j p and sq = 1 + kq 66 The first condition gives sq = 1 or sq = p. If sq = p then the second condition gives p = 1 + kq, or p − 1 = kq. This is however impossible since q - (p − 1). Therefore we have sq = 1. We obtain that G has exactly one Sylow p-subgroup P (of order p) and exactly one Sylow q-subgroup Q (of order q). By the Second Sylow Theorem (17.3) every element of G of order p belongs to the subgroup P and every element of order q belongs to the subgroup Q. It follows that G contains exactly p − 1 elements of order p, exactly q − 1 elements of order q, and one trivial element (of order 1). Since for all p, q we have pq > (p − 1) + (q − 1) + 1 there are elements of G of order not equal to 1, p, or q. Any such element must have order pq. Note. If jGj = pq, and q j (p − 1) then G need not be isomorphic to Z=pqZ (take e.g. G = GT ). 18.3. Defintition. Let N, K be groups and let ': K ! Aut(N) be a homomorphism. The semidirect product of N and K with respect to ' is the group N o' K such that N o' K = N × K as sets. Multiplication in N o' K is given by (a1; b1) · (a2; b2) := (a1('(b1)(a2)); b1b2) 67 18.4 Note. 1) If ' is the trivial homomorphism then N o' K = N × K. 2) If (a; b) 2 N o' K then (a; b)−1 = ('(b−1)(a−1); b−1) 3) N o' K contains subgroups N ∼= f(a; e) j a 2 Ng and K ∼= f(e; b) j b 2 Kg We have N C N o' K, and N \ K = f(e; e)g. 18.5 Examples. ∼ 1) Notice that Aut(Z=3Z) = Z=2Z. Let ': Z=2 ! Aut(Z=3Z) be the isomorphism. Check: ∼ Z=3Z o' Z=2Z = GT 2) In general if p is a prime then we have ∼ Aut(Z=pZ) = Z=(p − 1)Z For any n j (p − 1) there is a unique cyclic subgroup H ⊆ Aut(Z=pZ) of order n. Let ': Z=nZ ! Aut(Z=pZ) be any homomorphism such that Im(') = H. Then Z=pZ o' Z=nZ. is a non-abelian group of order pn. 3) If N is any abelian group then the map inv: N ! N; inv(a) = a−1 68 is an automorphism of N. This gives a homomorphism ': Z=2Z ! Aut(N) such that '(1) = inv. We obtain in this way a group N o' Z=2Z of order 2jNj. Special cases: • If N = Z=nZ then this group is called the dihedral group of order 2n and it is denoted Dn. The group Dn is isomorphic to the group of all isometries ∼ of a regular polygon with n sides (exercise). In particular D3 = GT . • If N = Z then this group is the infinite dihedral group and it is denoted by D1. The group D1 is isomorphic to the free product Z=2Z ∗ Z=2Z (exercise). Recall. From HW 1: If G; H are abelian groups and f : G ! H, g : H ! G ∼ are homomorphisms such that fg = idH then G = Ker(f) ⊕ H. 18.6 Proposition. If G, H are groups and f : G ! H, g : H ! G are homo- morphisms such that fg = idH then ∼ G = Ker(f) o' H where ': H ! Aut(Ker(f)) is given by '(b)(a) := g(b)ag(b)−1 for a 2 Ker(f), b 2 H. Proof. Exercise. 18.7 Proposition. Let p, q be prime numbers such that p > q and q j (p − 1). Then, up to isomorphism, there are only two groups of order pq: 69 ∼ { the abelian group Z=pqZ = Z=pZ × Z=qZ { the non-abelian group Z=pZ o' Z=qZ where ': Z=qZ ! Aut(Z=pZ) is any non-trivial homomorphism. Proof. By the same argument as in the proof of Proposition 18.2 we get that G has only one Sylow p-subgroup. Call this subgroup P . We have P C G. Let Q be any Sylow q-subgroup. Consider the quotient map f : G ! G=P Take the restriction fjQ : Q ! G=P . Notice that Ker(fjQ) = Ker(f) \ Q = P \ Q = feg, so fjQ is a monomorphism. In addition jQj = q = jG=P j, so fjQ is an isomorphism. As a consequence we have a homomorphism −1 (fjQ) g : G=P / Q,! G Since fg = idG=P by Proposition 18.6 we obtain ∼ G = P o' G=P ∼ for some homomorphism ': G=P ! Aut(P ). Also, since P = Z=pZ and ∼ G=P = Z=qZ we get ∼ G = Z=pZ o' Z=qZ for some ': Z=qZ ! Aut(Z=pZ). ∼ ∼ If ' is the trivial homomorphism then G = Z=pZ × Z=qZ = Z=pqZ. If ' is non-trivial then G is a non-abelian group. Notice that since q j (p − 1) such non-trivial homomorphism exists by (18.5). It remains to show that for any two non-trivial homomorphisms '; : Z=qZ ! Aut(Z=pZ) we have an isomorphism ∼ Z=pZ o' Z=qZ = Z=pZ o Z=qZ (exercise). 70 18.8 Example. For any odd prime p there are two non-isomorphic groups of order 2p: { the cyclic group Z=2pZ { the dihedral group Dp. 71 H. U. Besche, B. Eick, E. A. O’Brien A millennium project: constructing small groups International Journal of Algebra and Computation 12(5) (2002) 623-644. 19 Group extensions and composition series 19.1 Definition. Let fi fi+1 ::: −! Gi −! Gi+1 −! Gi+2 −! ::: be a sequence of groups and group homomorphisms. This sequence is exact if Im(fi) = Ker(fi+1) for all i. 19.2 Definition. A short exact sequence is an exact sequence of the form f g 1 −! N −! G −! K −! 1 (where 1 is the trivial group). 19.3 Note. f g 1) A sequence 1 ! N −! G −! K ! 1 is a short exact sequence iff • f is a monomorphism • g is an epimorphism • Im(f) = Ker(g). 2) If H C G then we have a short exact sequence 1 −! H −! G −! G=H −! 1 Morever, up to an isomorphism, every short exact sequence is of this form: f g 1 / N / G / K / 1 =∼ = =∼ 1 / Ker(g) / G / G= Ker(g) / 1 76 19.4 Definition. If a group G fits into a short exact sequence f g 1 −! N −! G −! K −! 1 then we say that G is an extension of K by N. 19.5 Example. For any n > 1 the dihedral group Dn and the cyclic group Z=2nZ are non-isomorphic extensions of Z=nZ by Z=2Z. 19.6 Definition. A group G is a simple group if G 6= feg and the only normal subgroups of G are G and feg. 19.7 Example. Z=pZ is a simple group for every prime p. Note. A group G is simple iff it is not a non-trivial extension of any group. 19.8 Definition. If G is a group then a normal series of G is a sequence of subgroups feg = G0 ⊆ G1 ⊆ G2 ⊆ ::: ⊆ Gk = G such that Gi−1 C Gi for all i. A composition series of G is a normal series such that all quotient groups Gi=Gi−1 are simple. 19.9 Example. Take the dihedral group D4 = Z=4Z o Z=2Z. We have a composition series f0g ⊆ Z=2Z ⊆ Z=4Z ⊆ D4 Another composition series of D4: f0g ⊆ Z=2Z ⊆ Z=2Z × Z=2Z ⊆ D4 77 19.10 Theorem (Jordan { H¨older). If G 6= feg is a finite group then 1) G has a composition series. 2) All composition series of G are equivalent in the following sense. If we have composition series feg = G0 ⊆ ::: ⊆ Gk = G and feg = H0 ⊆ ::: ⊆ Hl = G then k = l and there is a bijection σ : f1; : : : ; kg ! f1; : : : ; kg such that for i = 1; : : : ; k we have an isomorphism ∼ Gi=Gi−1 = Hσ(i)=Hσ(i)−1 Proof.
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