CAPÍTULO II: REVISIÓN DE LA LITERATURA
4. Marco teórico
4.1. Dramatización
4.1.1. Objetivos de la dramatización
A classification problem of finite groups of ordern (up to isomorphism) for every n is certainly very difficult. Even for simple groups, it was a long standing and famous problem as mentioned already. In this section we shall discuss some partial classifications including the cases of special orders.
(I) A classification of finite abelian groups
Fortunately, one can classify all finite abelian groups, with a help of Theorem 1.2.9.
Theorem 3.4.1 LetGbe a finite abelian group, say|G|=p`1
1 p`22· · ·p`tt.
Then,
(1) every Sylowpi-subgroup ofG is normal, and hence Gis the direct
product of its Sylow subgroups.
(2) Each Sylowp-subgroupP ofGcan be expressed as a direct product of cyclic subgroups:
P =ha1i × · · · × hasi,
where{a1, . . . , as}is a basis forP and(pe1, . . . , pes)is the type in-
variant ofP. It means that the numbersand the orderspe1, . . . , pes
of the direct factors (where one may assume that e1 ≥ · · · ≥ es)
are uniquely determined by P.
(3) G can be expressed as a direct product of cyclic groups:
G=ha1i × ha2i × · · · × hasi,
where o(ai)|o(ai+1), i = 1,2, . . . , s−1.
From this classification, one can show that the converse of Lagrange’s Theorem 1.1.4 is true for any abelian group.
(II) A classification of groups whose all Sylow subgroups are cyclic
The simplest case is for the group of order pq where q < p are primes.
Theorem 3.4.2 Let q < p be primes and G a group of order pq. (1) If q -p−1, then G∼=Zpq, the cyclic group of order pq.
(2) If q |p−1, then either G∼=Zpq or
G∼=ha, b|ap =bq = 1, b−1ab=ari, (a nonabelian group) where r 6≡1 (modp) but rq ≡1 (modp).
Proof: LetP ∈Sylp(G) andQ∈Sylq(G). By Sylow theorems, PEG
always. If Gis abelian or ifq -p−1,then we also haveQEG. (In fact,
q-p−1 iffQEG.) Since P and Qare of prime order, they are cyclic. It follows that G∼= Zpq, the cyclic group of order pq; (1) and the first
assertion of (2) hold.
Now assume that q| p−1 and G is not abelian. ThenQ 5G and
nq(G) = p. Since G/P ∼= Q, by H¨older’s theorem, G has the defining
relations as shown above. To prove the uniqueness of the nonabelian group of order pq, we need to show that the isomorphic type of the group
K =ha, b|ap =bq = 1, b−1ab=ari
is independent of the choice of r, where r 6≡ 1 (modp) but rq ≡
1 (mod p). However, it follows from number theory thatxq ≡1 (mod p)
has exactly q distinct solutions in Z∗
p. Let r be one of them such that
r 6≡ 1 (modp). Then 1, r, r2, . . . , rq−1 are all the distinct solutions of
xq ≡ 1 (modp). Now, for any 1 ≤ s ≤ q−1, one can show that the
group
Ks =hu, v |up =vq = 1, v−1uv =ur
s
i
is isomorphic to K via the map u 7→ a, v 7→ bs. This completes the
proof. ¤
In particular, q = 2, then G is either cyclic or isomorphic to the dihedral groupD2p. Note that forn = 8 = 23 there are two nonabelian
groups up to isomorphism: the Hamilton quaternion groupQ8 and the dihedral group D8. For n = 12, there are exactly three nonabelian groups up to isomorphism, see Example 3.4.5.
However, the nonabelian group of order pq such that q | p−1 is unique up to isomorphism. We denote this group by Fpq.
Classifying groups of special orders 113
Definition 3.4.1 A finite group Gis called aFrobenius groupif Ghas a proper subgroup H satisfying (1) NG(H) =H and (2) H ∩Hg = 1
for any g /∈H. The subgroup H is called the Frobenius complement of
G, and the setK =G−Sg∈G(Hg− {1}) is called theFrobenius kernel
of G.
Another definition of a Frobenius group will be given later in terms of a permutation group, see Volume two, Definition 3.4.1.
By a theorem due to Frobenius (see Theorem 5.4.11 in Chapter 5), the Frobenius kernelK of a Frobenius groupGis a normal subgroup of
Gand satisfiesG=HK andH∩K ={1}. Also, Frobenius conjectured that the Frobenius kernel of a Frobenius group is nilpotent. (For the definition of a nilpotent group, see the next chapter.) Thompson (1959) proved this conjecture affirmatively.
Example 3.4.1 (1) Let U be a group of units in a finite field F. Define
ta,b :F →F; ta,b(x) =ax+b for a∈U, b∈F
and let G = {ta,b | a ∈ U, b ∈ F}. Then G is Frobenius with
the Frobenius complement H = {ta,0 | a ∈ U} ∼= U, all ro- tations, which is a point stabilizer of 0. It Frobenius kernel is
K ={t1,b | b∈F} ∼= (F,+), all translations.
(2) Dihedral group D2n with odd n : H ∼= Z2 with reflection and
K ∼=Zn all rotations.
(3) The groupsFpqare Frobenius with Frobenius kernelP ∈Sylp(Fpq),
where q|p−1.
For more knowledge about Frobenius groups see Volume two, Chap- ter 6, Section 3. (See also Dixon and Mortimer (1996), Section 3.4.)
Next we give a classification for finite groups all of whose Sylow subgroups are cyclic. First, we need the following lemma.
Lemma 3.4.3 LetG be a group. If bothG0/G00 andG00/G000 are cyclic,
Proof: It suffices to prove that ifG000 = 1 thenG00 = 1 by considering
G/G000 instead of G. If G000 = 1, then G00 is cyclic. Let G00 = hbi. By
N/C-theorem, G/CG(G00) is isomorphic to a subgroup of Aut(G00) =
Aut(hbi), and hence is abelian. So, CG(G00)≥G0, or equivalently G00 ≤
Z(G0). Now, by using the assumption that G0/G00 is cyclic, one can
show that G0 is abelian. It means G00= 1. ¤
Now, we generalize Theorem 3.4.2 as follows.
Theorem 3.4.4 Let Gbe a finite group whose all Sylow subgroups are cyclic. If G is abelian, then G is cyclic; if G is nonabelian, then G is metacyclic defined by the following defining relations:
G=ha, bi, am =bn = 1, b−1ab=ar, (3.9) where
((r−1)n, m) = 1 (mod m), r6≡1 (mod m) but rn≡1 (modm), |G|=nm.
Proof: Firstly, by Corollary 2.5.4 G is solvable. Note that a direct product of cyclic groups of coprime orders is also cyclic. So,Gis cyclic if it is abelian.
Now assume that G is not abelian. Note that the property – “all Sylow subgroups are cyclic” – of a group is inherited to its subgroups and quotient groups. We have that G/G0, G0/G00, . . . are cyclic since
they are abelian. Applying Lemma 3.4.3,G00 =G000. SinceGis solvable,
we have G00 = 1. Hence G0 and G/G0 are cyclic. Let G0 = hai ∼= Z
m
and G/G0 = hbG0i ∼= Z
n. Since G0 EG, we may assume b−1ab = ar
where r 6≡ 1 (mod m). Since bn ∈ hai and b−nabn = arn
, we have
rn ≡ 1 (mod m). Since G0 is cyclic, we have G0 = h[a, b]i = har−1i. Since G0 = hai ∼= Z
m, we have (r−1, m) = 1. Let bn = aj. We have
b−1ajb =aj, implying thatarj =aj,a(r−1)j = 1. By (r−1, m) = 1, we
have aj = 1, hence bn = 1. Finally, we claim that (n, m) = 1. Suppose
that a prime number p | (n, m). Then ham/p, bn/pi is an elementary
abelian group of order p2, contradicting the fact that the Sylow p- subgroup of G is cyclic.
Conversely, by H¨older’s theorem, Eq.(3.9) does give a metacyclic group all of whose Sylow subgroups are cyclic. ¤
Classifying groups of special orders 115
Differently from the case of order pq, a classification of the isomor- phic types of nonabelian groups mentioned in Theorem 3.4.4 is more difficult. Certainly, it might depend on the choice of the number r. Recall that every group of order ofpqr is solvable for any three distinct primes p, q, r. (See Theorem 2.3.3.) In general, we have the following corollary.
Corollary 3.4.5 Let G be a finite group of square-free order. Then G
is solvable and metacyclic with defining relations given in Eq.(3.9).
Example 3.4.2 Classify groups Gof order 30 = 2·3·5.
Solution: The number 30 = 2·3·5 is a multiple of three distinct primes and hence all Sylow subgroups of G are cyclic. If G is abelian, then G∼=Z30. If G is nonabelian, by Theorem 2.4.7, G has a normal subgroup N of order 15. By Exercise 2.2.5, N is cyclic. By H¨older’s Theorem, G is generated by two elements; say a and b. and Ghas the following defining relations:
G=ha, b:a15 =b2 = 1, b−1ab=ari,
where r2 ≡1(mod 15) and r 6≡1(mod 15). Its three distinct solutions
r≡ −1, or ±4(mod 15) define three groups:
G = ha, b:a15=b2 = 1, b−1ab=a−1i,
G = ha, b:a15=b2 = 1, b−1ab=a4i,
G = ha, b:a15=b2 = 1, b−1ab=a−4i.
Since G0 is cyclic, G0 = h[a, b]i. By computation, [a, b] = a−2, a3 and
a−5, respectively. In these three cases, |G0| = 15, 5 and 3, respec-
tively. Therefore, these three groups are not isomorphic one another. In summary, there are four distinct groups up to isomorphism. ¤
Example 3.4.3 Classify groups Gof order 42 = 2·3·7.
Solution: IfGis abelian, thenG∼=Z42. IfGis nonabelian, by Theo- rem 2.4.7,G has a normal subgroupN of order 21. By Theorem 3.4.2,
IfN ∼=Z21, lettingN =hai, sinceZ∗21 has three elements of order 2, namely,α1 :a7→a−1, α2 :a7→a8 andα3 :a7→a−8, H¨older’s Theorem gives that up to isomorphism
G=ha, bi, a21 = 1, b2 = 1, b−1ab=ar,
where r=−1, 8 or−8.
If N ∼= F21, letting N = ha, b | a7 = b3 = 1, b−1ab = a2i, G is an extension of N by Z2. Assume that G= hN, ci. Then c2 = 1 and the conjugate action of c on N is an automorphism γ of N of order ≤ 2. If γ = 1, then G = N ×Z2. If o(γ) = 2, we shall prove that G is isomorphic to the following group:
G=ha, b, ci, a7 =b3 =c2 = 1, b−1ab=a2, c−1ac=a−1, cb=bc. In fact, this group is a semidirect product of N by Z2 with respect to a map α:c7→γ0 ∈Aut(N), where γ0 maps a to a−1 and fixesb.
To prove this claim, it suffices to show that any automorphism γ
of N of order 2 is conjugate to γ0 in Aut(N). By computation, γ has the form: γ : a 7→ a−1, b 7→ bai for 0 ≤ i ≤ 6. Take σ : a 7→ a,
b 7→ba2i. Thenσ is an automorphism of N. Then σ−1γσ =γ
0, detail computations are omitted. In total, there are six distinct groups up to
isomorphism. ¤
(III) A classification of some metacyclic groups
For a classification of metacyclic groups, one may start with a clas- sification for the extension of Zn by Z2.
First, we consider the case where n is an odd prime power.
Theorem 3.4.6 Let p be an odd prime and m a positive integer. Let
G be a nonabelian group and letG be an extension ofZpm byZ2. Then
G is isomorphic to the dihedral group:
D2pm =ha, b|ap m
= 1, b2 = 1, b−1ab=a−1i. (3.10)
Proof: By H¨older’s Theorem 3.3.2,G has the following defining rela- tions:
G=ha, bi, apm
Classifying groups of special orders 117
where pm, t and r satisfy
r2 ≡1 (mod pm), t(r−1)≡0 (mod pm).
By G being nonabelian, we have r ≡ −1 (mod pm). It follows that
pm |2t, and hencepm | t, then we get the dihedral group D
2pm. ¤
Theorem 3.4.7 Let p be an odd prime and m a positive integer. Let a nonabelian group G be an extension of Z2pm by Z2. Then G is iso-
morphic to one of
(1) (the dihedral group) D4pm = ha, b | a2p m
= 1, b2 = 1, b−1ab =
a−1i,
(2) (the generalized quaternion group)
Q4pm =ha, b|a2p m
= 1, b2 =apm
, b−1ab=a−1i.
Proof: By the same argument as in Theorem 3.4.6, Ghas the follow- ing defining relations:
G=ha, bi, a2pm
= 1, b2 =at, b−1ab=ar,
where 2pm, t and r satisfy
r2 ≡1 (mod 2pm), t(r−1)≡0 (mod 2pm).
By G being nonabelian, we have r ≡ −1 (mod 2pm). It follows that
2pm | 2t and pm | t. So we get either t = pm or 2pm. In the latter
case, it is the dihedral group. In the former case, it is the generalized quaternion group.
Note that elements outside the cyclic group hai are of order 2 or 4 depending on the group G being dihedral or generalized quaternion. So, these two groups are not isomorphic. This completes the proof. ¤
Next, let n = 2m be an even prime power. Note that Aut(Z
2m) has
0, 1, or 3 elements of order 2 if m≤ 1, m= 2, or m ≥3, respectively. So, for general n, we have
Theorem 3.4.8 Let m ≥ 2 be an integer. Let a nonabelian group G
be an extension of Z2m by Z2. Then G is isomorphic to one of
(1) (the dihedral group) D2m+1 = ha, b | a2
m
= 1, b2 = 1, b−1ab =
a−1i,
(2) (the generalized quaternion group)
Q2m+1 =ha, b|a2 m = 1, b2 =a2m−1 , b−1ab=a−1i, (3) ha, b|a2m = 1, b2 = 1, b−1ab=a1+2m−1 i;m≥4,
(4) (the semi-dihedral group)ha, b|a2m
= 1, b2 = 1, b−1ab=a−1+2m−1
i
;m≥4.
Proof: By H¨older’s Theorem 3.3.2,G has the following defining rela- tions:
G=ha, bi, a2m
= 1, b2 =at, b−1ab=ar,
where 2m, t and r satisfy
r2 ≡1 (mod 2m), t(r−1)≡0 (mod 2m).
ByGbeing nonabelian, we haver ≡ −1 or ±1 + 2m−1 (mod pm), the
latter two cases can happen only when m ≥3. It follows that 2m | 2t,
and hence 2m−1 | t, that is t = 0 or t = 2m−1. Now we consider the three cases separately.
(i) r ≡ −1 (mod 2m): In this case we get the dihedral group (1)
and the generalized quaternion group (2) depending on b2 = 1 or b2 =
a2m−1
. By the same reason as Theorem 3.4.6, these two groups are not isomorphic.
Note that the following cases (ii) and (iii) happen only whenm≥3. So, when m= 2 we have only the above two groups.
(ii)r≡1+2m−1(mod 2m): In this case, we havet2m−1 ≡0 (mod 2m)
which implies that t is even. Let t = 2s. Since m ≥ 3, there is a j
satisfying j(1 + 2m−2) +s≡0 (mod 2m−1). Let b
1 =baj. Then b2 1 =b2(b−1ajb)aj =b2aj(2+2 m−1) =a2[j(1+2m−2)+s] = 1.
Now the generators {a, b1} satisfy the relations in the group (3), re- placingb1 byb.
Classifying groups of special orders 119
(iii)r≡ −1+2m−1(mod 2m): In this case (−2+2m−1)t ≡0 (mod 2m),
that is (−1+2m−2)t≡0 (mod 2m−1). It follows thatt ≡0 (mod 2m−1). Thusb2 = 1 or a2m−1 . If b2 =a2m−1 , letting b1 =ba, we have b2 1 = (ba)2 =b2(b−1ab)a =b2a−1+2 n−2 a=a2n−2 a2n−2 = 1,
Thus we get the group (4).
Finally, we shall show that the mentioned four groups are not iso- morphic, and we assume thatm ≥3. It is east to see that in these four cases G0 =h[a, b]i. We calculate [a, b] and get
[a, b] =a−1b−1ab=
a−2 for the groups (1), (2),
a2m−1
for the group (3),
a−2+2m−1
for the group (4).
So, we have |G0|= 2 for (3), and |G0|= 2m−1 for the others. It follows that the group (3) is not isomorphic to any one of the rest. To prove the rest three groups are not isomorphic, we calculate the square of the elements of the formbai outside hai. We have
(bai)2 =b2(b−1aib)ai =
1 for the group (1),
a2m−1
for the group (2),
ai2m−1
for the group (4).
This shows that the subgroup of order 2m in G is unique, and outside
this subgroup hai, all elements are of order 2 in the group (1), order 4 in the group (2), and some are of order 2 and the others are of order 4 in the group (4). Therefore, all the four groups are not isomorphic one
enother. ¤
For general n, we determine the semidirect product of Zn by Z2, which is always a split extension.
Theorem 3.4.9 Let G be a semidirect product of Zn byZ2 where n >
4. If G is nonabelian, then G has the following presentation:
where r 6≡ 1(mod n) and r2 ≡ 1(mod n). Different values of r give non-isomorphic groups.
Let n = 2α1pα2
2 · · ·pαss, where p2, . . . , ps are distinct odd primes, α1 is a non-negative integer, andα2, . . . , αs are positive integers. Then the
number f(n) of non-isomorphic types of the groups is given by
f(n) = 2s−1−1 if α 1 ≤1, 2s−1 if α 1 = 2, 2s+1−1 if α 1 ≥3.
Proof: By H¨older’s Theorem,Gis isomorphic to a semidirect product of Zn by Z2 since r is an involution in Z∗n. It suffices to show that
different values of r give non-isomorphic groups. Assume that r1 6≡
r(mod n) and
G1 =ha1, b1 |an1 =b21 = 1, b−11a1b1 =ar11i.
We prove thatGandG1 are not isomorphic. If they were, andσ:G→
G1 were an isomorphism, then aσ =ai1 for someiwith (i, n) = 1, since there is only one cyclic subgroup of order n in both G and G1. (Proof of this fact is left to the reader.) Assumebσ =aj
1b1. Then (bσ)−1aσbσ = (aσ)r. However, by computation, (bσ)−1aσbσ = b−1
1 a−1jai1aj1b1 = air11 = (aσ)r1, a contradiction.
Finally, the calculation for f(n) is left to the reader. ¤
Example 3.4.4 Determine extensions of Z30 byZ2.
Solution: Note that 30 = 2·3·5. By Theorem 3.4.9, there are three nonabelian split metacyclic groups of order 30. They are:
G=ha, b:a30 =b2 = 1, b−1ab=a−1i, (3.11)
G=ha, b:a30=b2 = 1, b−1ab=a11i, (3.12)
G=ha, b:a30=b2 = 1, b−1ab=a19i. (3.13) Next, we determine non-split metacyclic groupsG. By Theorem 3.3.2 it has the following presentation:
Classifying groups of special orders 121
where r=−1,11,19(mod 30) and i(r−1) = 0(mod 30).
For the same reason as above, different values ofrgive non-isomorphic groups. By calculations, if r=−1 theni= 15; ifr= 11 then 3|i; and if r= 19 then 5|i.
When r=−1,i= 15, we have the group
G=ha, b:a30= 1, b2 =a15, b−1ab=a−1i, (3.14) which is not isomorphic to the group (3.11).
When r= 11, i= 3,6,9, . . ., we have the groups
G=ha, b:a30= 1, b2 =ai, b−1ab=a11i.
Replacing b bybaj, we get (baj)2 =ai+12j So, ifi= 3,9,15, . . ., choose
suitablej, the above group is isomorphic to the following:
G=ha, b:a30= 1, b2 =a3, b−1ab=a11i; (3.15) if i= 6,12, . . ., choose suitable j, the above group is isomorphic to the group (3.12), which is split.
When r= 19, i= 5,10,15. . ., we have the groups
G=ha, b:a30= 1, b2 =ai, b−1ab=a19i.
Replacing b by baj, we get (baj)2 = ai+20j So, for the same reason as