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LECTURE 6

Group homomorphisms and their kernels

PAVEL R

UZICKA

Abstract.We define and study the notions of a group homomorphism and the kernel of a group homomorphism. We prove that the kernels correspond to normal subgroups. We examine some examples of group homomorphisms that are based on geometric intuition.

6.1.Group homomorphisms.LetG= (G,·) andH= (H,·) be groups.

A (group) homomorphism?:G→His a map?from the setGtoHsuch that?(f·g) =?(f)·?(g), for allf,g?G. Lemma6.1.Let?:G→Hbe a group homomorphism,uGanduHre- spectively the units ofGandH. Then?(uG) =uHand?(g-1) =?(g)-1, for allg?G.

Proof.We have from the definition that

u

H·?(uG) =?(uG) =?(uG·uG) =?(uG)·?(uG)

Since the group operation is right cancellative, we infer thatuH=?(uG).

For an elementg?Gwe have that

?(g)·?(g-1) =?(g·g-1) =?(uG) =uH=?(g)·?(g)-1. SinceHis left-cancellative we infer that?(g-1) =?(g)-1.? A (group) embeddingis an one-to-one group homomorphism. We say that a groupGcan be embeddedinto a groupHif there is a group embedding

G→H.

A (group) isomorphismis a group homomorphism that is both one-to-one and onto. GroupsGandHare called isomorphicprovided that there is a group isomorphismG→H. For each groupGlet us denote by1Gthe identity mapG→G. The map is clearly a (group) homomorphism and we will call the identity isomorphism ofG. Lemma6.2.A group homomorphism?:G→His an isomorphism if and if there is a group homomorphismψ:H→Gsuch thatψ◦φ=1Gand The Lecture and the tutorial took place in Mal´a strana, roomS11, on November 13, 2018.
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φ◦ψ=1H. That is, a group homomorphism is an isomorphism if and only if it has an inverse. Proof.(?) It follows fromψ◦φ=1Gthatφis one-to-one. Fromφ◦ψ=1H we infer thatφmapsGontoH. (?) Since?is a one-to-one map fromG ontoH, eachh?Hhas a uniqueg?Gwith?(g) =h. We defineψ(h) =g. From?(ψ(h)) =?(g) =hwe get that?◦ψ=1H. From the choice of ψ(?(g)) as the unique?-preimage?(g), we see thatψ(?(g)) =g, for all g?G. Thereforeψ◦?=1G. Letfandhbe arbitrary elements ofH.

Since?is a homomorphism, we have that

ψ(f·h) =ψ((?◦ψ)(f)·(?◦ψ)(h)) =ψ(?(ψ(f))·?(ψ(h)))) =ψ(?(ψ(f)·ψ(h))) = (ψ◦?)(ψ(f)·ψ(h)) =ψ(f)·ψ(h). It follows thatψ:H→Gis a group homomorphism.?

We say that groupsGandHare

isomorphic, which we denote byG?H, if there is an isomorphismG→H. Observe that the inverse to an isomor- phism is again an isomorphism and a composition of isomorphisms gives an isomorphism. It follows that the binary relation?defined on the class of all groups is symmetric and transitive. Since each group is isomorphic to itself via the identity isomorphism,?is an equivalence relation. Obviously, a group isomorphismG→Htransfers properties of the group Gto properties ofH. Thus saying that some (group) property is unique up to isomorphism means that the property determines a groupup to its isomorphism class (i.e, the block of?). Given a setXwe denote bySXthe set of all one-to-one maps fromX ontoX. The set is equipped with the binary operation◦of composition of maps and thus it forms a group called the symmetric group of the setXand denote bySX. Clearly, for finite setsXandYthe groupsSXandSYare isomorphic if and only if the setsXandYhave the same size. In particular, ifXis ann-element set, thenSX?Sn. Theorem6.3(Cayley).Every group can be embedded into a symmetric group of its underlying set. Proof.LetG= (G,·) be a group. For eachf,g?Gwe setλ(f)(g) = f·g. Thus we have defined a mapλ(f):G→Gfor allf?G. From the left cancellativity of the group operation it follows thatλ(f)(g)?=λ(f)(h) wheneverg?=h, hence the matλ(f) is one-to-one. The left divisibility of the group operation implies thatλ(f) mapsGontoG. Thereforeλcan be regarded as a map fromGtoSG. Since λ(f·g)(h) = (f·g)·h=f·(g·h) =λ(f)(λ(g)(h)) = (λ(f)◦λ(g))(h), for allf,g,h?G, and soλ(f·g) =λ(f)◦λ(g), the map is a group ho- momorphismλ:G→SG. Letudenote the unit ofG. Iff?=ginG, then (6.1)λ(f)(u) =f·u=f?=g=g·u=λ(g)(u),

Group homomorphisms and their kernels 3

in particularλ(f)?=λ(g). We conclude thatλis a group embedding.? Corollary6.4.A finite group embeds intoSn, wherenis the size of the group. Remark6.5.The mapλ:G→SGis called aleft translationinG. Simi- larly we can define a right translation, sayρ, byρ(f)(g) =g·f-1(we need the inverse offto makeρan homomorphism) and prove that it induces another embeddingρ:G→SG. Observe that in the proof of Theorem 6.3 we only needed the left cancellativity, the left divisibility, and the existence of a right unit (respectively the right cancellativity, theright divisibility, and the existence of a left unit if we argue usingρinstead ofλ). This leads to an elegant solution of Exercies 2.4. Lemma6.6.Let?:G→Hbe a group homomorphism. The following hold true: (a)LetKbe a subgroup of the groupG. Then ?(K) :={?(k)|k?K} is the universe of a subgroup ofH. Moreover if?is an epimorphism and the subgroupKis normal inG, the image?(K)is a normal subgroup ofH. (b)LetLbe a subgroup of the groupH. Then -1(L) :={g?G|?(g)?L} is a universe of a subgroup ofG. Moreover ifLis a normal subgroup of the groupH, then?-1(L)is a normal subgroup ofG. Proof.We prove the two parts (a) and (b) separately. (a) Letg,h??(K). There arek,l?Ksuch thatg=?(k) and h=?(l). Computing thatg·h-1=?(k·l-1)??(K), we prove that?(K) is the universe of a subgroup ofH. Suppose thatKis a normal subgroup ofGand that?is an epimorphism. Then for everyh?Hthere isg?Gwith?(g) =h.

Applying the normality ofKinGwe get that

for allk?K. Therefore?(K)?H. (b) Letg,h??-1(L). From ?(g·h-1) =?(g)·?(h)-1?L, we infer thatg·h-1??-1(L), hence?-1(L)is a subgroup of the groupG. Suppose thatLis a normal subgroup ofH. Letk??-1(L) and g?G. We compute that ?(g·k·g-1) =?(g)·?(k)·?(g)-1??(g)·L·?(g)-1?L,

4P. R°UZICKA

sinceL?H. We conclude that?-1(L)is a normal subgroup of G. Lemma6.7.lemma:podgrupy homomorfismus Let?:G→Hbe a group homomorphism. The mappingK?→?(K)is a bijection between the set of all subgroupsKofGsuch thatker??Kand the set of all subgroups of ?(G). Proof.Observe thatK??-1(?(K)) for every subsetK?G. LetKbe a subgroup containing ker?. Letg??-1(?(K)). Then?(g)??(K), and so there isk?Ksuch that?(g) =?(k). It follows that?(k-1·g) =uH, hencek-1·g?ker?, henceg?k·ker?. Sincek?Kand ker??K, we conclude thatg?K. ThereforeK=?-1(?(K)). Clearly?(?-1(L)) =Lfor everyL??(G). According to Lemma 6.6 the maps

K?→?(K)andL?→?-1(L)

are mutually inverse bijections between the set of all subgroups ofGcon- taining ker?and the set of all subgroups of?(g).?

6.2.Kernels of group homomorphisms.

Definition6.8.Let?:G→Hbe a group homomorphism. Akernelof the homomorphism?is the set ker?:={g?G|?(g) =uH}, whereuHdenotes the unit ofH. Observe that the kernel of a homomorphism contains the unit ofG, and so it is non-empty. However, even more holds true: Lemma6.9.The kernel of a group homomorphism?:G→His a normal subgroup ofG.

Proof.Ifg,h?ker?, then

?(g·h-1) =?(g)·?(h)-1=uH·u-1 H=uH, and sog·h-1?ker?. Therefore ker?is a subgroup ofG.

Letk?ker?andg?G. Then

?(g·k·g-1) =?(g)·?(k)·?(g-1) =?(g)·uH·?(g)-1=uH. Thereforeg·k·g-1?ker?, and so the subgroup ker?is normal due to

Lemma 5.1.?

LetNbe a normal subgroup of a groupG. The mapπG/N:G→G/N defined byg?→N·g=g·Nis a group homomorphism1, indeed G/N(g·h-1) =N·g·h-1=N·g·N·h-1= (N·g)·(N·h)-1,

1Note that sinceN?G, we have thatN·g=g·N, for allg?G.

Group homomorphisms and their kernels 5

for allg,h?G. By the definition,

Therefore

Corollary6.10.Normal subgroups correspond to kernels of group homo- morphisms. Example6.11.As in Example 5.6 letRdenote the group of all rotations of a cube. We showed thatRis isomorphic to the group of permutations S

4. A numbering of vertices of the cube induces an embeddingα:R→S8.

Similarly a numbering of edges and a numbering of faces of thecube induce embeddingsβ:R→S12andγ:R→S6, respectively. H G C DE F B A HG C DE F B A

Figure 1.Cubes with tetrahedrons

Observe that we can inscribe two regular tetrahedrons into the cube as in Figure 1. Each rotation either maps each of the tetrahedronsonto itself or exchange them. This induces a homomorphism fromRto the two-element group of permutations of the two tetrahedrons. Up to an isomorphism, this map corresponds to a homomorphismS4→S2. The kernel of this homo- morphism consists of the rotations mapping each of the tetrahedrons onto itself. On one hand, if we restrict to one of the tetrahedrons, say the blue oneACFH, these are exactly the rotations of the tetrahedron. On the other hand these are the exactly the rotations of the cube corresponding to even permutations of its diagonals. We conclude that the kernel of this homo- morphism is isomorphic to the alternating group of permutationsA4. Color axes of faces of the cube as in Figure 2. Each rotation ofthe cube induces a permutation of these axes. Since a rotation over one of them induces a transposition of the remaining two, each permutation of the axes is induced by a rotation of the cube. Thus we get a homomorphism from the groupRonto the group of all permutations of the three-element set of the axes, up to an isomorphism, corresponding to an epimorphismS4→S3.

6P. R°UZICKA

Figure 2.Axis of faces of the cube

The kernel of the homomorphism consists of the rotations that map all three axes onto themselves. These are the identity rotation and the flips over the exes. Each of the flips exchange two pairs of diagonals of the cube. Therefore the flips correspond to permutations of the type?0,2,0,0?. It follows that the kernel of the corresponding homomorphismS4→S3is the four-element group

2V={υ4,(1,2)·(3,4),(1,3)·(2,4),(1,4)·(2,3)}. The groupVis

indeed a normal subgroup ofS4. Let us note thatVis the only non-trivial normal subgroup ofA4andVandA4are only non-trivial normal subgroups ofS4. (0,0,-1)(-1,0,0) (0,1,0) (0,-1,0) (1,0,0)(0,0,1) (0,0,0)

Figure 3.The cube in a coordinate system

Finally, let us insert the cube into the3-dimensional real vector space so that the center of the cube corresponds do the zero vector andthe centers of faces to the vectors of a canonical basis ofR3and their inverses, as depicted

2The group in question was namedVierergruppe(= four-group) by Felix Klein. That

is why it is often denoted byV.

Group homomorphisms and their kernels 7

in Figure 3. We can view the rotations of the cube as restrictions of one-to- one linear mapsR3→R3. The matrices of these linear maps with respect to the canonical basis have one non-zero entry in each line and each column and the non-zero entries are1or -1. Moreover the determinant of each of these matrices is equal to1. On the other hand every matrix with the properties corresponds to some rotation of the cube. In thisway we have defined an embeddingR→GL(3,R). There are48matrices inGL(3,R)that have one non-zero entry in each line and each column and the non-zero entries are1or -1and these ma- trices form a group (with an operation of the matrix multiplication). They matrices correspond to linear mapsR3→R3whose restriction to the cube map bijectively vertices to vertices and edges to edges. Letus call such maps symmetriesof the cube and let us denote bySthe48-element group they form. Associating to each symmetry its matrix with respect to a canonical basis ofR3, we extend the embeddingR→GL(3,R)above to an embedding

μ:S→GL(3,R).

Observe that all non-zero real numbers with the operation ofmultiplication form a group. We denote the group by(R?,·). The subset{1,-1}is a universe of a two-element subgroup, sayC2, of(R?,·). The multiplication table of the groupC2is

·1-1

1-1 -1 11 -1 Since the derminant of a product of square matrices is a product of their determinants, the mapdet: GL(3,R)→(R?,·)that assigns to a regular matrix its non-zero determinant is a group homomorphism. Let us denote δ:= det◦μ. Note that the image ofδis the subgroupC2of(R?,·), and so we have a group epimorphismδ:S→C2. We can sketch the situation as follows:

SGL(3,R)

C

2(R?,·).

δdet

The matrices in the image ofμcorresponding to rotations are exactly those with determinant equal to1. We conclude thatRis the kernel ofδ. This corresponds toRbeing a normal subgroup ofS, which follow also from [S:R] = 2(cf. Exercise 4.3). Finally let us note that symmetries thatare not rotations are called reflections. They form a coset ofRinSand the determinants of matrices corresponding to reflections are all equal to-1.

8P. R°UZICKA

Exercises

Exercise6.1.Decide whether all3-cycles are conjugated inA4. Exercise6.2.Letα:R→S8,β:R→S12,γ:R→S6, be as in Exam- ple 6.11. (i)Prove thatα(R)?A8. (ii)Decide whetherβ(R)?A12,γ(R)?A6. Exercise6.3.Letδ:R→S2andε:R→S3be as in Example 6.11. (i)Find kernels of the group homomorphismsδandε. (ii)Show thatε(R) =S3. Exercise6.4.Describe all conjugacy classes of the groupSof all symme- tries of a cube. Compute characteristic polynomials and Jordan canonical forms of corresponding matrices. Exercise6.5.Analyze the group of all rotations and the group of all sym- metries of (i)a square. (ii)a regular tetrahedron. Exercise6.6.LetGbe a group. Let us define a mapγ:G→SGby g?→[h?→gh=g·h·g-1]. •Prove thatγ(g)is a permutation ofGfor everyg?G. •Prove thatγ:G→SGis a group homomorphism. •Prove thatkerγ=Z(G); the centrum of the groupG.quotesdbs_dbs21.pdfusesText_27

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