Chapter 1. Background Material: Group Theory

The aim of this chapter is to introduce formally two constructions of the theory of groups: semi-direct products and presentations of groups. Later on in the lecture we will relate semi-direct products with a 1st and a 2nd cohomology group. Presentations describe groups by generators and relations in a concise way, they will be useful when considering concrete groups, for instance in examples.

References:

[Hum96] J. F. H��������,A course in group theory, Oxford Science Publications, The Clarendon Press, Oxford University Press, New York, 1996.

[Joh90] D.L. J������,Presentations of groups, London Mathematical Society Student Texts, vol. 15, Cambridge University Press, Cambridge, 1990.

1 Semi-direct Products

The semi-direct product is a construction of the theory of groups, which allows us to build new groups from old ones. It is a natural generalisation of the direct product.

Definition 1.1 (Semi-direct product)

A groupG is said to be the (internal or inner) semi-direct product of a normal subgroup N ú G by a subgroupH §G if the following conditions hold:

(a) G“NH; (b) NXH “ t1u.

Notation: G“N¸H. Example 1

(1) A direct product G1 ˆG2 of two groups is the semi-direct product of N :“ G1 ˆ t1u by H :“ t1u ˆG2.

(2) G“S3 is the semi-direct product of N “C3 “ xp1 2 3qyúS3 und H “C2 “ xp1 2qy§S3. HenceS3–C3¸C2.

Notice that, in particular, a semi-direct product of an abelian subgroup by an abelian subgroup need not be abelian.

7

(3) More generallyG“S�(�•3) is a semi-direct product ofN“A�úS�byH “C2 “ xp1 2qy.

Remark 1.2

(a) If G is a semi-direct product of N byH, then the 2nd Isomorphism Theorem yields G{N “HN{N–H{HXN “H{t1u –H

and this gives rise to a short exact sequence

1›ÑN ›ÑG›ÑH ›Ñ1�

Hence a semi-direct product ofN byH is a special case of an extension ofN by H.

(b) In a semi-direct productG “N¸H of N byH, the subgroup H acts by conjugation on N, namely@�PH,

θ�: N ›Ñ N

� fiÑ ���^´1,

is an automorphism of N. In addition θ��¹ “θ�˝θ�¹ for every �� �¹ PH, so that we have a group homomorphism

θ: H ›Ñ AutpNq

� fiÑ θ�.

Proposition 1.3

With the above notation, N�H andθ are sufficient to reconstruct the group law on G.

Proof : Step 1. Each �PGcan be written in a unique way as�“��where�PN,�PH: indeed by (a) and (b) of the Definition, if�“��“�¹�¹ with�� �¹PN,�� �¹PH, then

�^´1�¹“�p�¹q^´1PNXH“ t1u� hence�“�¹ and�“�¹.

Step 2. Group law: Let�1 “�1�1� �2 “�2�2 PG with�1� �2PN,�1� �2PH as above. Then

�1�2 “�1�1�2�2“�1�loooomoooon1�2p�^´1₁

θ�1p�2q

�1q�2“ r�1θ�1p�2qs ¨ r�1�2s�

With the construction of the group law in the latter proof in mind, we now consider the problem of constructing an "external" (or outer) semi-direct product of groups.

Proposition 1.4

LetNandH be two arbitrary groups, and letθ:H ›ÑAutpNq� �fiÑθ�be a group homomorphism.

SetG:“NˆH as a set. Then the binary operation

¨: ` GˆG ›Ñ G p�1� �1q�p�2� �2q˘

fiÑ p�1� �1q ¨ p�2� �2q:“ p�1θ�1p�2q� �1�2q

defines a group law onG. The neutral element is1G “ p1N�1Hqand the inverse ofp�� �q PNˆH isp�� �q^´1 “ pθ_�^´1p�^´1q� �^´1q.

FurthermoreG is an internal semi-direct product of N0 :“Nˆ t1u –N byH0 :“ t1u ˆH–H.

Proof : Exercise 1, Exercice Sheet 1.

Definition 1.5

In the context of Proposition 1.3 we say thatG is theexternal (or outer) semi-direct product ofN byH w.r.t. θ, and we writeG“N¸θH.

Example 2

Here are a few examples of very intuitive semi-direct products of groups, which you have very prob- ably already encountered in other lectures, without knowing that they were semi-direct products:

(1) IfH acts trivially on N (i.e. θ� “IdN @�PH), then N¸θH “NˆH. (2) Let K be a field. Then

GL�pKq “SL�pKq ¸ diagpλ�1� � � � �1q PGL�pKq |λPK^ˆ(

�

wherediagpλ�1� � � � �1q is the diagonal matrix with (ordered) diagonal entriesλ�1� � � � �1.

(3) Let K be a field and let B:“

#˜_{˚ ˚}

0 ... ˚

¸

PGL�pKq +

p “ upper triangular matricesq�

U:“

#˜₁

...˚

0 1

¸

PGL�pKq +

p “ upper unitriangular matricesq�

T :“

#˜_λ

1 0

0 ... λ�

¸

PGL�pKq +

p “ diagonal matricesq�

Then B is a semi-direct product ofT byU.

(4) Let C� “ x�yand C� “ x�y (�� �PZ•1) be finite cyclic groups.

Assume moreover that� PZ is such that�^�”1 pmod �q and set θ: C� ›Ñ AutpC�q

�^� fiÑ pθ�q^�, whereθ�:C� ›ÑC�� �fiÑ�^�. Then

pθ�q^�p�q “ pθ�q^�´1p�^�q “ pθ�q^�´2p�^�2q “� � �“�^�� “�

since�p�q “�and�^�”1 pmod�q. Thus pθ�q^�“IdC� and θ is a group homomorphism. It follows that under these hypotheses there exists a semi-direct product ofC� byC� w.r.t. toθ.

Particular case: � •1, �“ 2 and � “ ´1 yield the dihedral group D2� of order 2� with generators� (of order�) and�(of order 2) and the relation θ�p�q “���^´1 “�^´1.

The details of Examples (1)-(4) will be discussed during the Präsensübung on Wednesday, 11th of April.

2 Presentations of Groups

Idea: describe a group using a set of generators and a set of relations between these generators.

Examples: p1q C� “ x�y “ x�|�^� “1y 1generator: � 1relation: �^� “1 p2q D2� “C�¸θC2 (see Ex. 2(4)q 2generators: �� �

3relations: �^� “1� �² “1� ���^´1 “�^´1

p3q Z“ x1Zy 1generator: 1Z

no relation (ù"free group") To begin with we examine free groups and generators.

Definition 2.1 (Free group / Universal property of free groups)

LetX be a set. A free group of basisX (orfree group onX) is a groupF containingX as a subset and satisfying the following universal property: For any group G and for any (set-theoretic) map

� :X ›ÑG, there exists a unique group homomorphism˜� :F ›ÑG such that˜�|X “�, or in other words such that the following diagram commutes:

X G

F

�:“���

� ö

D!˜�s.t.˜�|X“˜�˝�“�

Moreover,|X|is called therank ofF. Proposition 2.2

IfF exists, thenF is the unique free group of basisX up to a unique isomorphism.

Proof : Assume F¹ is another free group of basisX.

Let�:X ãÑF be the canonical inclusion ofX inF and let�¹:X ãÑF¹ be the canonical inclusion of X inF¹.

X F¹ F

�

�¹ D!˜�

D! ˜�¹ By the universal property of Definition 2.1, there exists:

- a unique group homomorphism�˜¹:F ›ÑF¹ s.t. �¹“˜�¹˝�; and - a unique group homomorphism˜�:F¹›ÑF s.t. �“˜�˝�¹.

X F

F

�

�

˜�˝�˜¹

IdF Thenp˜�˝�˜¹q|X “�, but obviously we also haveIdF|X “�. Therefore, by uniqueness, we have˜�˝�˜¹“IdF.

A similar argument yields �˜¹˝˜� “ IdF¹, hence F and F¹ are isomorphic, up to a unique isomorphism, namely˜�with inverse�˜¹.

Proposition 2.3

IfF is a free group of basisX, then X generates F.

Proof : Let H :“ xXy be the subgroup of F generated by X, and let �H :“ X ãÑ H denote the canonical inclusion ofXinH. By the universal property of Definition 2.1, there exists a unique group homomorphism

�rH such that �rH˝�“�H :

X H

F

�

�H

ö D!�rH

Therefore, lettingκ:HãÑF denote the canonical inclusion ofHinF, we have the following commutative diagram:

X H F

F

�

�H κ

�rH IdF

κ˝�rH

Thus by uniquenessκ˝�rH “IdF, implying that�rH:H›ÑF is injective. Thus F “ImpIdFq “Impκ˝�rHq “Imp�rHqÑH and it follows thatF “H. The claim follows.

Theorem 2.4

For any setX, there exists a free groupF with basis X.

Proof : Set X :“ t�α |α PIuwhereI is a set in bijection with X, set Y :“ t�α |α PIuin bijection withX but disjoint fromX, i.e. XXY “ H, and letZ :“XYY.

Furthermore, setE:“î₈

�“0Z^�, whereZ⁰ :“ tp qu(i.e. a singleton), Z¹ :“Z,Z²:“ZˆZ,� � � ThenE becomes a monoid for the concatenation of sequences, that is

p�1� � � � � ��q looooomooooon

PZ^�

¨ p�looooomooooon₁¹� � � � � �_�¹q

PZ^�

:“ p�loooooooooooomoooooooooooon1� � � � � ��� �₁¹� � � � � �_�¹q

PZ^�`�

�

The law¨is clearly associative by definition, and the neutral element is the empty sequencep q PZ⁰. Define the followingElementary Operations on the elements ofE:

Type (1): add in a sequencep�1� � � � � ��qtwo consecutive elements�α� �α and obtain p�1� � � � � ��� �α� �α� ��`1� � � � � ��q

Type (1bis): add in a sequencep�1� � � � � ��qtwo consecutive elements�α� �α and obtain p�1� � � � � ��� �α� �α� ��`1� � � � � ��q

Type (2): remove from a sequencep�1� � � � � ��qtwo consecutive elements�α� �α and obtain p�1� � � � � ���ˇ�α��ˇα� ��`1� � � � � ��q

Type (2bis): remove from a sequencep�1� � � � � ��qtwo consecutive elements�α� �α and obtain p�1� � � � � ����ˇα�ˇ�α� ��`1� � � � � ��q

Now define an equivalence relation„onE as follows:

two sequences inE are equivalent :ñ the 2nd sequence can be obtain from the 1st sequence through a succession of Elementary Operations of type (1), (1bis), (2) and (2bis).

It is indeed easily checked that this relation is:

– reflexive: simply use an empty sequence of Elementary Operations;

– symmetric: since each Elementary Operation is invertible;

– transitive: since 2 consecutive sequences of Elementary Operations is again a sequence of Elementary Operations.

Now setF :“E{ „, and writer�1� � � � � ��sfor the equivalence class ofp�1� � � � � ��qinF “E{ „.

Claim 1: The above monoid law onE induces a monoid law on F.

The induced law onF is: r�1� � � � � ��s ¨ r�₁¹� � � � � ��¹s “ r�1� � � � � ��� �₁¹� � � � � �¹�s.

It is well-defined: ifp�1� � � � � ��q „ p�1� � � � � ��qandp�¹₁� � � � � ��¹q „ p�¹₁� � � � � �_�¹q, then p�1� � � � � ��q ¨ p�₁¹� � � � � �_�¹q “ p�1� � � � � ��� �₁¹� � � � � �_�¹q

„ p�1� � � � � ��� �1¹� � � � � ��¹q via Elementary Operations on the 1st part

„ p�1� � � � � ��� �1¹� � � � � ��¹q via Elementary Operations on the 2nd part

“ p�1� � � � � ��q ¨ p�₁¹� � � � � �_�¹q

The associativity is clear, and the neutral element isrp qs. The claim follows.

Claim 2: F endowed with the monoid law defined in Claim 1 is a group.

Inverses: the inverse of r�1� � � � � ��s P F is the equivalence of the sequence class obtained from p�1� � � � � ��qby reversing the order and replacing each�α with�α and each�α with�α. (Obvious by definition of„.)

Claim 3: F is a free group onX.

LetG be a group and� :X›ÑGbe a map. Define

p�: E ›Ñ G

p�1� � � � � ��q fiÑ �p�1q ¨ ¨ ¨ ¨ ¨�p��q, where� is defined onY by�p�αq:“�p�_α^´1qfor every �αPY.

Thus, ifp�1� � � � � ��q „ p�1� � � � � ��q, thenp�p�1� � � � � ��q “p�p�1� � � � � ��qby definition of� onY. Hence

� induces a map

rp�: F ›Ñ G

r�1� � � � � ��s fiÑ �p�1q ¨ ¨ ¨ ¨ ¨�p��q,

By constructionp� is a monoid homomorphism, therfore so isrp�, but sinceF and Gare groups,rp� is in fact a group homomorphism. Hence we have a commutative diagram

X G

F

�

� ö

rp�

where�:X ›ÑF� � fiÑ r�sis the canonical inclusion.

Finally, notice that the definition ofrp� is forced if we wantrp� to be a group homorphism, hence we have uniqueness ofrp�, and the universal property of Definition 2.1 is satisfied.

Notation and Terminology

¨ To lighten notation, we identify r�αs P F with �α, hence r�αs with �_α^´1, and r�1� � � � � ��s with

�1¨ ¨ ¨�� inF.

¨ A sequence p�1� � � � � ��q P E with each letter �� (1 § � § �) equal to an element �α� P X or

�α^´1� is called a word in the generators t�α | α P Iu. Each word defines an element of F via:

p�1� � � � � ��qfiÑ�1¨ ¨ ¨��PF. By abuse of language, we then often also call �1¨ ¨ ¨��PF a word.

¨ Two words are calledequivalent :ñ they define the same element of F.

¨ Ifp�1� � � � � ��q PZ�ÑE (�PZ•0), then �is called thelength of the word p�1� � � � � ��q.

¨ A word is said to bereducedif it has minimal length amongst all the words which are equivalent to this word.

Proposition 2.5

Every group G is isomorphic to a factor group of a free group.

Proof : Let S :“ t�α P G | α P Iu be a set of generators for G (in the worst case, take I “ G). Let X :“ t�α |α PIube a set in bijection withS, and let F be the free group onX. Let �:X ãÑF denote the canonical inclusion.

X G

F F{kerp˜�q

�

� D!˜�

���� �����

ö D!p˜�

By the universal property of free groups the map � : X ãÑ G� �α fiÑ �α

induces a unique group homomorphism ˜� : F ›Ñ G such that ˜� ˝�“ �. Clearly ˜� is surjective since the generators ofG are all Imp˜�q. Therefore the 1st Isomorphism Theorem yieldsG–F{kerp˜�q.

We can now consider relations between the generators of groups:

Notation and Terminology

Let S :“ t�α P G | α P Iu be a set of generators for the group G, let X :“ t�α | α P Iu be in bijection with S, and letF be the free group on X.

By the previous proof,G–F{N, whereN :“kerp˜�q (�α Ø�α “�αN via the homomorphismp˜�).

Any wordp�1� � � � � ��qin the�α’s which defines an element ofF inNis mapped inGto an expression of the form

�1¨ ¨ ¨��“1G � where ��:“image of�� inG under the canonical homomorphism.

In this case, the word p�1� � � � � ��q is called a relation in the groupG for the set of generators S.

Now let R:“ t�β|βPJu be a set of generators ofN as normal subgroup of F (this means thatN is generated by the set of all conjugates of R). Such a setR is called aset of defining relations ofG.

Then the ordered pair pX�Rq is called a presentationof G, and we write G “ xX |Ry “ xt�αuαPI | t�βuβPJy�

The group G is said to befinitely presented if it admits a presentation G “ xX |Ry, where both

|X|�|R|†8. In this case, by abuse of notation, we also often write presentations under the form G “ x�1� � � � � �|X||�1“1� � � � � �|R|“1y�

Example 3

The cyclic group C� “ t1� �� � � � � �^�´1u of order �PZ•1 generated by S :“ t�u. In this case, we have:

X “ t�u R“ t�^�u

F “ x�y – pC8�¨q C8 ˜�

›

ÑC�� �fiÑ�has a kernel generated by�^� as a normal subgroup. ThenC� “ xt�u | t�^�uy.

By abuse of notation, we write simplyC�“ x� |�^�yor alsoC� “ x� |�^� “1y.

Proposition 2.6 (Universal property of presentations)

Let G be a group generated by S “ t�α |α P Iu, isomorphic to a quotient of a free group F on X “ t�α |α PIuin bijection with S. Let R :“ t�β |βPJu be a set of relations inG.

Then G admits the presentation G “ xX | Ry if and only if G satisfies the following universal property:

X H

G

�

� ö

�

For every group H, and for every set-theoretic map � : X ›Ñ H such that

˜�p�βq “ 1H @ �β P R, there exists a unique group homomorphism � : G ›Ñ H such that �˝� “�, where � : X ›Ñ G� �α fiÑ �α, and˜� is the unique extension of� to the free group F onX.

Proof : "ñ": Suppose that G “ xX | Ry. Therefore G – F{N, where N is generated by R as normal subgroup. Thus the condition˜�p�βq “1H @ �β PR implies thatNÑkerp˜�q, since

˜�p��β�^´1q “˜�p�qloomoon˜�p�βq

“1H

˜�p�q^´1 “1H @�β PR�@�PF�

Therefore, by the universal property of the quotient, ˜� induces a unique group homomorphism

� :G –F{N ›Ñ H such that � ˝π “˜�, where π : F ›Ñ F{N is the canonical epimorphism.

Now, if�:X›ÑF denotes the canonical inclusion, then� “π˝�, and as a consequence we have

�˝�“�.

"": Conversely, assume that G satisfies the universal property of the statement (i.e. relatively to X�F�R). SetN:“R for the normal closure ofR. Then we have two group homomorphisms:

�: F{N ›Ñ G

�α fiÑ �α

induced by˜� :F ›ÑG, and

ψ: G ›Ñ F{N

�α fiÑ �α

given by the universal property. Then clearly �˝ψp�αq “ �p�αq “ �α for each α P I, so that

�˝ψ“IdG and similarlyψ˝�“IdF{R. The claim follows.

Example 4

Consider the finite dihedral groupD2� of order 2� with 2§�†8. We can assume that D2� is generated by

� :“ rotation of angle 2π

� and�:“ symmetry through the origin inR²�

Then x�y –C�ÑG,x�y –C2 and we have seen that D2�“ x�y ¸ x�y with three obvious relations

�^� “1,�² “1, and���^´1 “�^´1.

Claim: D2� admits the presentation x�� �|�^�“1� �² “1� ���^´1“�^´1y.

In order to prove the Claim, we let F be the free group on X :“ t�� �u, R :“ t�^�� �²� ���^´1�u, NúF be the normal subgroup generated byR, andG :“F{N so that

G“ x�� �|�^�“1� �² “1� � � �^´1� “1y� By the universal property of presentations the map

�: t�� �u ›Ñ D2�

� fiÑ �

� fiÑ �

induces a group homomorphism

�: G ›Ñ D2�

� fiÑ �

� fiÑ �,

which is clearly surjective sinceD2�“ x�� �y. In order to prove that� is injective, we prove that G is a group of order at most 2�. Recall that each element of G is an expression in �� �� �^´1� �^´1, hence actually an expression in �� �, since �^´1 “ �^�´1 and �^´1 “ �. Moreover, ���^´1 “ �^´1 implies�� “�^´1�, hence we are left with expressions of the form

�^��^� with 0§�§�´1and 0§�§1� Thus we have|G|§2�, and it follows that � is an isomorphism.

Notice that if we remove the relation�^� “1, we can also formally define aninfinite dihedral group D8 via the following presentation

D8:“ x�� �|�² “1� ���^´1“�^´1y� Theorem 2.7

Let G be a group generated by two distinct elements, � and �, both of order 2. Then G – D2�, where 2§�§8. Moreover,�is the order of �� inG, and

G“ x�� � |�²“1� �²“1�p��q^�“1y� (�“ 8simply means "no relation".)

Proof : Set�:“�� and let�be the order of�.

Firstly, note that �• 2, since �“ 1 ñ �� “1 ñ � “ �^´1 “ � as �² “ 1. Secondly, we have the relation���^´1 “�^´1, since

���^´1 “loomoon�p�

“1G

�q�^´1“��^´1“�^´1�^´1 “ p��q^´1“�^´1� ClearlyG can be generated by�and �as�“�� and so� “��.

Now,H:“ x�y –C� andH úG since

���^´1“�^´1PH and ���^´1“�PH (or because|G:H| “2)� SetC:“ x�y –C2.

Claim: �RH.

Indeed, assuming�PH yields�“�^� “ p��q^� for some0§�§�´1. Hence 1“�² “�p��q^� “ p��q^�´1�“ p��looomooon¨ ¨ ¨�q

length�´1

�p��looomooon¨ ¨ ¨�q

length�´1

�

so that conjugating by�, then�, then� � �, then�, we get1“�, contradicting the assumption that�p�q “2.

The claim follows.

Therefore, we have proved that G“HC and HXC “ t1u, so that G “H¸C “D2� as seen in the previous section.

Finally, to prove thatGadmits the presentationx�� �|�² “1� �²“1�p��q^�“1y, we apply the universal property of presentations twice to the maps

�: t��� ��u ›Ñ x�� �|�² “1� �² “1�p��q^�“1y

�� fiÑ �

�� fiÑ ��

and

�: t��� ��u ›Ñ G“ x�� �|�^�“1� �²“1� ���^´1 “1y

�� fiÑ �

�� fiÑ �� .

This yields the existence of two group homomorphisms

� :G“ x�� �|�^�“1� �²“1� ���^´1 “1y›Ñ x�� �|�²“1� �² “1�p��q^�“1y and �:x�� �|�²“1� �²“1�p��q^�“1y›ÑG“ x�� �|�^�“1� �²“1� ���^´1 “1y such that�� “Idand��“Id. (Here you should check the details for yourself!)