18 Application: Groups of Order Pq

18 Application: groups of order pq

∼ Recall. If G is a group, |G| = 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= {e}. 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 | 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 | 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 |G| = pq, and q | (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

ϕ: K → Aut(N) be a homomorphism. The semidirect product of N and K with respect to ϕ is the group N oϕ K such that

N oϕ K = N × K as sets. Multiplication in N oϕ K is given by

(a1, b1) · (a2, b2) := (a1(ϕ(b1)(a2)), b1b2)

67 18.4 Note.

1) If ϕ is the trivial homomorphism then N oϕ K = N × K.

2) If (a, b) ∈ N oϕ K then (a, b)−1 = (ϕ(b−1)(a−1), b−1)

3) N oϕ K contains subgroups

N ∼= {(a, e) | a ∈ N} and K ∼= {(e, b) | b ∈ K}

We have N C N oϕ K, and N ∩ K = {(e, e)}.

18.5 Examples. ∼ 1) Notice that Aut(Z/3Z) = Z/2Z. Let

ϕ: Z/2 → Aut(Z/3Z) be the isomorphism. Check: ∼ Z/3Z oϕ Z/2Z = GT

2) In general if p is a prime then we have ∼ Aut(Z/pZ) = Z/(p − 1)Z

For any n | (p − 1) there is a unique cyclic subgroup H ⊆ Aut(Z/pZ) of order n. Let ϕ: Z/nZ → Aut(Z/pZ)

be any homomorphism such that Im(ϕ) = H. Then Z/pZ oϕ 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

ϕ: Z/2Z → Aut(N)

such that ϕ(1) = inv. We obtain in this way a group N oϕ Z/2Z of order 2|N|. 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 D∞. The group D∞ 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 homomorphisms such that fg = idH then ∼ G = Ker(f) oϕ H where ϕ: H → Aut(Ker(f)) is given by

ϕ(b)(a) := g(b)ag(b)−1 for a ∈ Ker(f), b ∈ H.

Proof. Exercise.

18.7 Proposition. Let p, q be prime numbers such that p > q and q | (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 oϕ Z/qZ where ϕ: 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 f|Q : Q → G/P . Notice that Ker(f|Q) = Ker(f) ∩ Q = P ∩ Q = {e}, so f|Q is a monomorphism. In addition |Q| = q = |G/P |, so f|Q is an isomorphism. As a consequence we have a homomorphism

−1 (f|Q) g : G/P / Q,→ G

Since fg = idG/P by Proposition 18.6 we obtain ∼ G = P oϕ G/P ∼ for some homomorphism ϕ: G/P → Aut(P ). Also, since P = Z/pZ and ∼ G/P = Z/qZ we get ∼ G = Z/pZ oϕ Z/qZ for some ϕ: Z/qZ → Aut(Z/pZ). ∼ ∼ If ϕ is the trivial homomorphism then G = Z/pZ × Z/qZ = Z/pqZ. If ϕ is non-trivial then G is a non-abelian group. Notice that since q | (p − 1) such non-trivial homomorphism exists by (18.5).

It remains to show that for any two non-trivial homomorphisms ϕ, ψ : Z/qZ → Aut(Z/pZ) we have an isomorphism ∼ Z/pZ oϕ Z/qZ = Z/pZ oψ 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= {e} and the only normal subgroups of G are G and {e}.

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 {e} = 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 {0} ⊆ Z/2Z ⊆ Z/4Z ⊆ D4

Another composition series of D4:

{0} ⊆ Z/2Z ⊆ Z/2Z × Z/2Z ⊆ D4

77 19.10 Theorem (Jordan – H¨older). If G 6= {e} 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

{e} = G0 ⊆ ... ⊆ Gk = G and {e} = H0 ⊆ ... ⊆ Hl = G

then k = l and there is a bijection σ : {1, . . . , k} → {1, . . . , k} such that for i = 1, . . . , k we have an isomorphism ∼ Gi/Gi−1 = Hσ(i)/Hσ(i)−1

Proof. Exercise (or see Hungerford p. 111).

Upshot. If G 6= {e} is a ﬁnite group then G can be obtained by taking successive extensions of simple groups as follows.

1) Take a composition series

{e} = G0 ⊆ G1 ⊆ G2 ⊆ ... ⊆ Gk = G

2) For every i = 1, . . . , k we have a short exact sequence

1 −→ Gi−1 −→ Gi −→ Gi/Gi−1 −→ 1

where Gi/Gi−1 is a simple group. Therefore

G1 is an extension of G1/G0 by G0

G2 is an extension of G2/G1 by G1 ......

G = Gk is an extension of Gk/Gk−1 by Gk−1

78 The grand plan for classifying all ﬁnite groups (The H¨olderProgram)

1) Classify all ﬁnite simple groups.

2) For any two groups N, K describe all possible extensions of K by N.

Good news: part 1) is done.∗ See

R. Solomon, A brief history of the classification of the finite simple groups, Bulletin AMS 38 (3) (2001), 315-352. M. Aschbacher, The status of the classification of the finite simple groups, No- tices AMS 51(7) (2004) , 736-740.

79 20 Simple groups

Recall. A group G 6= {e} is a simple group if the only normal subgroups of G are G and {e}.

Note. If G is a simple group then any non-trivial homomorphism f : G → H is a monomorphism.

∼ 20.1 Proposition. If G is an abelian group then G is simple iﬀ G = Z/pZ for some prime p.

Proof. Exercise.

∼ 20.2 Proposition. If G is a simple p-group then G = Z/pZ.

Proof. If G is a p-group then Z(G) 6= {e} by (16.4). We have Z(G) C G, so if G is simple we must have G = Z(G). Therefore G is a simple abelian p-group, ∼ and so G = Z/pZ.

20.3 Lemma. There are no non-abelian simple groups of order prm where p is a prime, r ≥ 1, p - m and prm - m!.

Proof. Assume that G is a simple, non-abelian group of such order. We must have m > 1 (since if m = 1 then G is a p-group). Let P be a Sylow p-subgroup of G. Consider the action of G on the left cosets G/P :

G × G/P → G/P, a · bP = (ab)P

This action deﬁnes a homomorphism

%: G → Perm(G/P )

80 where Perm(G/P ) is the group of all permutations of the set G/P . Since G 6= P this homomorphism is non-trivial, and so, since G is a simple group, % is a monomorphism. Therefore G can be identiﬁed with a subgroup of Perm(G/P ). By Lagrange’s Theorem (7.4) we obtain that |G| divides |Perm(G/P )|. Since |Perm(G/P )| = m! this gives prm | m! which contradicts assumptions of the lemma.

20.4 Theorem. There are no non-abelian simple groups of order < 60.

Proof. Check: If 1 ≤ n < 60, and n 6= 30, 40, 56 then n is of the form prm for some prime p, and r, m ≥ 1 such that p - m and prm - m!. By Lemma 20.3 we obtain then that a non-abelian group G of order n < 60 may be simple only if n = 30, 40 and 56.

Assume that |G| = 30 = 2·3·5. We will show that G cannot be a simple group.

We argue by contradiction. Assume that G is simple and let s3 be the number of Sylow 3-subgroups of G. We have

s3 | 10 and s3 ≡ 1 (mod 3)

It follows that either s3 = 1 or s3 = 10. Since G is simple s3 6= 1, so s3 = 10.

Notice that if P , P 0 are two distinct Sylow 3-subgroups of G then P ∩P 0 = {e}. We obtain:

– G contains 10 Sylow 3-subgroups. – each Sylow 3-subgroup contains 2 elements of order 3.

It follows that G contains 20 elements of order 3.

By a similar argument we obtain that

– G must contain 6 Sylow 5-subgroups.

81 – each Sylow 5-subgroup contains 4 elements of order 3. so we have 24 elements of order 5 in G. This is however impossible, since 20 + 24 > 30 = |G|.

In a similar way one can show that if |G| = 40 or |G| = 56 then G is not a simple group (exercise)

Next goal: there are inﬁnitely many non-abelian simple ﬁnite groups. In particular there is a simple group of order 60.