Notes on Group Theory

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Notes on Group Theory

Mark Reeder

March 7, 2014

1 Notation for sets and functions

For any setS, we write |S| for the number of elements in S if S is finite, and put |S| = ∞if S is infinite. The empty set is denotedS =_∅.

If f : S → T is a function, we write f(s) or fs for the value of f at an element s ∈ S. The set

imf =f(S) = {f(s) : s ∈ S}of these values is the image off andf−1(t) = {s ∈ S : f(s) = t}

for the fiber off over an elementt∈T. Thus,imf ={t∈T : f−1(t)6=_∅}.For any subsetT0 ⊂T, the setf−1(T0) ={s∈S : f(s)∈T}is the union of the fibersf−1(t)fort∈T.

A function f : S → T is injective if |f−1₍_t₎_{| ≤} ₁ _{for all} _t _∈ _T_{, in this case we sometimes write}

f :S ,→T to emphasize injectivity, and to indicate that we may identifySwith its image inT. A functionf : S → T is surjective if |f−1₍_t₎_{| ≥} ₁

for all t ∈ T, in this case we sometimes write

f :S T to emphasize surjectivity.

Finally,f is bijective if it is both injective and surjective, that is, if|f−1₍_t₎_| _{= 1}_{for all}_t _∈ _T_{. When}

this holds, we have|S| =|T|. We sometimes writef : S −→∼ T to indicate a bijection, orS ↔ T to mean that there exists a bijection betweenSandT.

2 Basic group theory

2.1 The definition of a group

Agroupis a setGtogether with a function∗ :G×G →G, assigning to each pair(a, b)of elements ofGanother elementa∗b∈G, satisfying the following three axioms:

G1 (associativity) We havea∗(b∗c) = (a∗b)∗c, for alla, b, c∈G.

G2 (existence of identity) There exists an elemente∈Gsuch thate∗a=a∗e=afor alla∈G. G3 (existence of inverses) For alla∈Gthere exists an elementa0 ∈Gsuch thata∗a0 =a0∗a =e,

whereeis an identity element as in axiom G2.

There is no requirement thata∗b=b∗afor alla, b∈G. If this property does hold, we say thatGis

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The element e of axiom G2 is unique: for if e0 is another identity element, then e0 = e0 ∗e = e by applying axiom G2 first toe, then toe0. For eacha∈G, the inverse elementa0 in axiom G3 is unique: for ifa00is another inverse element, we have

a00 =a00∗e=a00∗(a∗a0) = (a00∗a)∗a0 =e∗a0 =a0,

by applying successively axioms G2, G3 (fora0), G1, G3 (fora00) and finally G2 again.

We usually use multiplicative notation and abbreviatea∗basabora·b, and write1or1Ginstead of

e for the identity element of G, and a−1 instead ofa0 for the inverse element of a. For any positive integern, we writean_{for the product of}_a_{with itself}_n_{times, and}_a−n_{for the product of}_a−1_{with itself}

ntimes. Finally, we puta0 = 1.

IfGis abelian, and only in this case, we sometimes use additive notation, writinga∗b=a+b, denoting the identity element by0, and the inverse element ofaby−a.

Theorder ofGis the cardinality|G|, either a positive integer or∞. Any groupGof order one consists of the identity element only and is called thetrivial group.

We can extend the product of elements inGto subsets: IfSandT are subsets of a groupG, we define

ST ={st: s ∈S, t∈T}.

2.2 Group homomorphisms

The structure of a groupGis revealed by its subgroups and homomorphisms into other groups. Ahomomorphismof groupsG, G0 is a functionf :G→G0 satisfying

f(ab) = f(a)f(b) ∀a, b∈G.

This implies thatf(1G) = 1G0 and thatf(g−1) = f(g)−1for allg ∈G.

Thekernelof a homomorphismf :G→G0is the subset ofGdefined by

kerf :={g ∈G: f(g) = 1G0}.

For alla, b∈Gwe havef(a) =f(b)if and only ifab−1 ∈kerf. Hencef is injective iffkerf ={1G}.

Theimageoff is the set theoretic image, defined as above by

imf :=f(G) = {g0 ∈G: f−1(g0)6=_∅}.

There may be many homomorphisms between two given groups. We set

Hom(G, G0) ={homomorphismsf :G→G0}.

Anisomorphismf : G → G0 is a bijective group homomorphism. Thusf is an isomorphism if and only if kerf = {1G} and imf = G0. We sometimes write f : G

∼

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isomorphism. Two groups G, G0 are isomorphicif there exists an isomorphismf : G −→∼ G0. We writeG'G0 to indicate thatGandG0are isomorphic, without specifying any paticular isomorphism between them.

We sometimes abuse terminology and say thatGisoris a copy ofG0, when we really mean only that

G ' G0. For example, any two trivial groups are isomorphic, so we allow ourselves to say that the

trivial group is the unique group with one element.

Continuing in this vein, we say that a homomorphismf : G→ G0 istrivialifimf ={1G0}. This is

equivalent to havingkerf = G, so being a trivial homomorphism is the opposite of being an isomor-phism. Hence trivial homomorphisms are just as important as isomorphisms.

An isomorphism from a group to itself is called anautomorphism. We set

Aut(G) ={automorphisms of G}.

The set Aut(G) forms a group under composition, whose identity element is the identity automor-phism, which sendsg 7→g for allg ∈G.

Many automorphisms arise from within the group itself as follows. For each elementg in a groupG, the map

cg :G→G given by cg(x) =gxg−1 ∀x∈G

is an automorphism of G, called conjugation by g. Automorphisms of this form are called inner automorphisms. The functionc:G→Aut(G)sendingg 7→cg is a homomorphism.

2.3 Subgroups

AsubgroupofGis a subsetH ⊆Gwith the following three properties:

SG1 (closure)ab∈Hfor alla, b∈H.

SG2 (identity) The identity element ofGis contained inH. SG3 (inverses) For alla ∈H we havea−1 ∈H.

The subsets {1} and G are subgroups of G. All other subgroups of G, if any, are called proper subgroups. We writeH ≤Gto indicate thatH is a subgroup ofGwhich is possibly equal toGitself. We writeH < Gfor a subgroup which is not equal toG.

Lemma 2.1 LetGbe a group and letH be a nonemptyfinitesubset ofG. ThenH ≤ Gif and only if SG1 holds.

Proof: Lethbe an element of the nonempty setH. SinceH is finite, the powersh, h2_{, h}3_{, . . .} _must

eventually repeat, so we havehi = hj for some positive integersi < j. It follows that hj−i = 1, so SG2 holds, andh·hj−i−1

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This proof moved fromHto a particular kind of subgroup ofG. A groupC iscyclicif there exists an elementc∈C such thatC ={cn _: _n _∈

Z}. In this case we writeC =hci.

In an arbitrary groupG, any elementg ∈Gis contained in the cyclic subgroup

hgi={gn _: _n_∈

Z}.

Theorderofg is the order of the grouphgi. The order ofg is the smallest positive powermsuch that

gm = 1, if such anmexists. In this case, the order can be characterized by the useful property that for any integerd, we havegd _{= 1}_iff_m _|_d_{. If} _gd ₆_{= 1}_{for any nonzero integer}_d_{, we say the order of}_g _is

infinite.

More generally, ifSis any subset ofG, thesubgroup generated byS is the smallest subgrouphSiof

GcontainingS. More precisely,

hSi= \

S⊆H≤G

H

is the intersection of all subgroupsHofGwhich contain the subsetS.

2.4 Cosets and quotient spaces

Aleft cosetof a subgroupH < Gis a subset ofGof the formgH = {gh :h ∈ H}. Two left cosets are either equal or disjoint; we have

gH =g0H ⇔g−1g0 ∈H.

In particular, we havegH =Hif and only ifg ∈H. The set of left cosets ofH inGis denotedG/H, and is called thequotientofGbyH.

Aright cosetofHinGis a subset of the formHg ={hg : h∈H}. Two right cosets are either equal or disjoint; we have

Hg =Hg0 ⇔g−1g0 ∈H.

In particular, we haveHg = H if and only if g ∈ H. The set of right cosets of H in Gis denoted

G/H.

Acosetis a left or right coset. Any element of a coset is called arepresentativeof that coset. We have canonical bijectionsH → gHandH → Hg, sendingh7→ ghandh 7→hg, respectively. Hence ifH

is finite, all cosets have cardinality|H|.

There are an equal number (including infinity) of left and right cosets inG. We denote this number by

[G:H] =|G/H|=|H\G|,

and call it the index of H in G. If G is finite, then G is partitioned into [G : H] cosets, each of cardinality|H|. It follows that[G:H] =|G|/|H|. In particular we have

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Example 1: If|G| =pa prime, thenGhas no proper subgroups. Hence for any nonidentity element

g ∈Gwe haveG=hgi, soGis cyclic.

Example 2: The order of any element in a finite groupGdivides|G|. in particular, we haveg|G| = 1

for allg ∈ G. The smallest positive integere such thatge = 1is called the exponent ofG. If g has orderm, thenmdividese, which in turn divides|G|.

The converse of Lagrange’s theorem is false. The smallest counterexample is the groupA4of order12,

which has no subgroup of order6. However, the converse of Lagrange’s theorem is true for subgroups

Hof prime power order. This is part of the Sylow theorems, which we will prove later. However, one special case is easy:

Proposition 2.2 Any group of even order contains an element of order two.

Proof: SupposeGhas even order|G|= 2m. Pair the nonidentity elements with their inverses. Since there are2m−1such elements, at least one of them is paired with itself. This is a nonidentity element

g ∈Gsuch thatg =g−1. Thus,g is an element ofGof order two.

2.5 Normal subgroups and quotient groups

LetGbe a group and letH ≤ Gbe a subgroup. One attempts to define a group structure on the set

G/H by the rule:

gH∗g0H =gg0H, ∀g, g0 ∈G. (1) However, this rule is only well-defined when every left coset ofHinGis also a right coset ofHinG. The subgroupH is said to benormalinGifgH =Hg for allg ∈ H. On the level of elements, this means thatghg−1 ∈H for allg ∈ Gandh ∈ H. IfGis abelian, thenghg−1 =h, so every subgroup is normal inG. Thus, being normal inGis a weakening, with respect toH, of the abelian condition. We writeH / GorH _EGto indicate thatHis a normal subgroup ofG.

When, and only whenH _E G, the set G/H = H\Gbecomes a group under the operation given by (1). We callG/H thequotientofGbyH. It is a group of order equal to the index ofHinG:

|G/H|= [G:H].

Example: Thecenterof a groupGis the subgroup

Z(G) ={z ∈G: zg =gz ∀g ∈G}.

This is clearly a normal subgroup ofG. We will see the quotient group G/Z(G)appearing in several contexts. One useful fact is

Proposition 2.3 IfG/Z(G)is cyclic thenGis abelian.

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2.6 The first isomorphism theorem

Any group homomorphismf : G → G0 induces an isomorphism from a quotient ofGto a subgroup ofG0. More precisely, we have the following.

Theorem 2.4 (First isomorphism theorem) Letf : G → G0 be a group homomorphism with kernel

K = kerf. Then the kernelK = kerf is a normal subgroup ofG, and there is an isomorphism

¯

f :G/K −→∼ imf, given by f¯(gK) = f(g). (2)

Proof: It is a good exercise to check thatf¯is well-defined and bijective. It is useful to have this result in slightly more general form:

Theorem 2.5 Letf : G→G0 be a group homomorphism with kernelK = kerf. LetH be a normal subgroup ofGcontained inK. Then there is a surjective homomorphism

¯

f :G/H −→∼ imf, given by f¯(gH) = f(g), (3)

withker ¯f =K/H.

Proof: Again, this is a good exercise.

We often say thatf¯isinducedbyf or thatf factors throughG/H.

Conversely, every normal subgroupH _E G is the kernel of a surjective homomorphism fromGinto another group. Namely, thecanonical homomorphism

πH :G−→G/H, given by πH(g) = gH

is surjective withkerπH =H.

Example: Ifx, yare two elements ofG, thecommutator

[x, y] =xyx−1y−1

is an element ofGthat measures the failure ofx, y to commute. TheCommutator Subgroup

[G, G] =h[x, y] : x, y ∈Gi

is the subgroup generated by all commutators. Forg ∈ Gwe have g[x, y]g−1 = [gxg−1, gxg−1]. It follows that[G, G]/ G. The quotientG/[G, G]is called theabelianization ofG, and is often denoted

Gab, because of the following result.

Proposition 2.6 The quotientG/[G, G] of a groupGby its commutator subgroup is abelian. More-over,G/[G, G]is the largest abelian quotient of G, in the following sense: Iff : G → A is a homo-morphism fromGto an abelian groupA, then[G, G]<kerf, sof factors through a homomorphism

¯

f :G/[G, G]−→A.

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2.6.1 Exact sequences

A composition of group homomorphisms

G1

f1

−→G2

f2

−→G3

isexact atG2ifimf1 = kerf2. A sequence of group homomorphisms

. . .−→Gi−1

fi−1

−→Gi fi

−→Gi+1 −→. . .

is anexact sequenceif it is exact atGifor alli. Ashort exact sequenceis a sequence

1−→G1

f1

−→G2

f2

−→G3 −→1

with

kerf1 ={1}, imf1 = kerf2 'G1, imf2 =G3 'G2/G1.

2.7 The second isomorphism theorem (correspondence theorem)

The subgroups of a quotient groupG/H are related to subgroups ofGas follows.

Theorem 2.7 Let G be a group with normal subgroup H _E G, and let πH : G → G/H be the canonical homomorphism.

1. If K is any subgroup of Gcontaining H, then H _E K andK/H = πH(K) is a subgroup of

G/H.

2. Conversely, if J is any subgroup of G/H, then π−1_H (J)is a subgroup of G containingH as a normal subgroup andJ =π−1_H (J)/H.

3. K/H_EG/H if and only ifK _EG, in which case(G/H)/(K/H)'G/K.

Thus, we have a one-to-one correspondence

{subgroups ofGcontainingH} ↔ {subgroups ofG/H}

K → K/H

π_H−1(J) ← J,

and this correspondence preserves normal subgroups.

There is also a correspondence theorem for homomorphisms, the first part of which is just Thm. 2.5

above.

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1. Iff : G → G0 is a group homomorphism withH ≤ kerf thenf induces a well-defined homo-morphismf¯:G/H →G0,given byf¯(gH) = f(g).

2. If ϕ : G/H → G0 is any homomorphism, then ϕ◦πH : G → G0 is a homomorphism whose kernel containsH.

Thus, we have a one-to-one correspondence

Hom(G/H, G0)↔ {f ∈Hom(G, G0) : H ≤kerf} ϕ → ϕ◦πH

¯

f ← f

2.8 The third isomorphism theorem

The final isomorphism theorem concerns products of subgroups. IfHandK are subgroups of a group

G, the product HK = {hk : h ∈ H, k ∈ K}contains the identity, but it need not be closed under the operation inG, henceHKneed not be a subgroup ofG. However, it will be so under an additional condition.

ThenormalizerofHinGis the subgroup

NG(H) = {g ∈G: gHg−1 =H}.

We have NG(H) = G if and only if H E G. In general, NG(H) is the smallest subgroup of G

containingHas a normal subgroup.

Returning to our two subgroupsH, K inG, let us assume that

K ≤NG(H).

(This assumption holds automatically ifH _EG.) Then, forh, h0 ∈H andk, k0 ∈K, we have

(hk)(h0k0) = h(kh0k−1)·kk0

(where we insert() and·to help parse the product), and kh0k−1 belongs to H sincek ∈ NG(H), so

(hk)(h0k0)∈HK. Similarly,(hk)−1 =k−1h−1 = (k−1h−1k)·k ∈HK. HenceHK is a subgroup of

G. Since both H andK are contained inNG(H), it follows thatHK is also contained inNG(H). In

other words,His normal inHK.

This proves the first part of the following

Theorem 2.9 (Third Isomorphism Theorem) LetH andK be subgroups of a groupGand assume thatK ≤NG(H). Then

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2. H∩K is a normal subgroup ofK.

3. We have a group isomorphism K/(K∩H) ' HK/H, induced by the mapf : K → HK/H

given byk 7→kH.

Proof: We have already proved the first part, and the second part is easy. As for the third part, it is clear thatf is surjective, so it remains to determinekerf. Letk ∈ K. Thenf(k) = 1inHK/H iff

ι(k)∈H, which means thatk ∈H. Butk ∈K, so we havef(k) = 1iffk ∈H∩K, as claimed.

2.9 Direct products

LetHandK be groups with identity elements1H and1K. Then the direct product

H×K ={(h, k) : h∈H, k∈K}

is a group under the operation

(h, k)(h0, k0) = (hh0, kk0)

and identity element(1H,1K). The direct product of finitely many groupsG1, . . . , Gnis defined

simi-larly; we confine our discussion to the casen = 2.

Let G = H ×K, and write 1 = (1H,1K). The maps φ : H → G and ψ : K → G given by

φ(h) = (h,1K) andψ(k) = (1H, k) are injective homomorphisms. Their imagesH0 = φ(H) ' H

andK0 =ψ(K)'K are normal subgroups ofGsuch that

H0∩K0 ={1} and H0K0 =G.

Conversely, this can be used torecognizedirect products as follows.

Proposition 2.10 LetGbe a group with subgroupsHandK. Assume that 1. HandK are both normal inG;

2. H∩K ={1}; 3. G=HK.

ThenG'H×K, via the mapf :H×K →Ggiven byf(h, k) = hk.

Proof: Let h ∈ H, k ∈ K and parse the commutator [h, k] = hkh−1k−1 in two ways. On the one hand,[h, k] = (hkh−1)k−1 ∈ K since K _E G. On the other hand, [h, k] = h(kh−1k−1) ∈ H, since

H _E G. ButH∩K is trivial, so[h, k] = 1. Hencehandk commute for allh ∈ H andk ∈ K. It is now immediate thatfis a homomorphism, which is surjective by assumption 3. Finally, iff(h, k) = 1, we haveh=k−1 ∈H∩K ={1}, soh=k= 1. Thereforef is an isomorphism, as claimed. IfHandK are abelian groups then we often writeH⊕K instead ofH×K, in accordance our use of additive notation for abelian groups.

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2.10 Semidirect products (internal view)

Recall that a groupG with two normal subgroups H, K _E G is the direct productG = H×K iff

HK = G and H ∩K = {1}. This situation often occurs with the variation that only one of the subgroups is normal.

Definition 2.11 A group G is a semidirect product of two subgroups H, K ≤ G if the following conditions hold.

1. One of the subgroupsH and/orK is normal inG. 2. H∩K ={1}.

3. HK =G.

SupposeGis the semidirect product ofHandKand (say)His normal inG. On the setH×K, define a group law as follows:

(h, k)(h0, k0) = (hkh0k−1, kk0).

LetH_oK denote the setH×Kwith this group law.

Proposition 2.12 IfGis a semidirect product of two subgroupsH andK withH _EG, then the map

(h, k)7→hkis a group isomorphism

H_oK −→∼ G.

Proof: exercise.

2.11 Conjugacy

Forg, x∈G, let us set

g_x₌_gxg−1

.

Two elementsx, y ∈ GareconjugateinGif y = g_x_{for some}_g _∈ _G_{. Conjugacy is an equivalence}

relation onG, whose equivalence classes are theconjugacy classesofG. Thus any group is partitioned into conjugacy classes. We write

G_x₌_{g_x_: _g _∈_G}

for the conjugacy class ofxinG.

Some of our earlier notions can be expressed in terms of conjugacy. For example, we havex ∈Z(G)

if and only ifGx={x}. AndH _EGif and only ifHis a union of conjugacy classes inG.

Thecentralizerof a given elementx∈ Gis the subgroupCG(x) = {h ∈G : hx= x}consisting of

all elements inGwhich commute withx. This is generally not a normal subgroup ofG, so the quotient spaceG/CG(x)is not a group. However, we have

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Proposition 2.13 For everyx∈G, the mapg 7→g_x_{induces a well-defined bijection}

G/CG(x)

∼

−→g_x.

In particular, ifGis finite, we have

|G_x|_{= [}_G_:_C

G(x)] =

|G| |CG(x)|

.

Proof: Exercise

The last formula in Prop.2.13is very useful for computing the sizes of conjugacy classes and central-izers in finite groups. It implies for example that|G_x|_divides_|G|_{. Since}_G_{is the disjoint union of its}

conjugacy classes, we have

Corollary 2.14 (The class equation) LetGbe a finite group, letX1, . . . , Xkbe its conjugacy classes, choosexi ∈Xi for1≤i≤k, and letGi =CG(xi). Then we have

k

X

i=1

[G:Gi] =|G|.

Alternatively,

1

|G1|

+ 1

|G2|

+· · ·+ 1

|Gk|

= 1.

Corollary 2.15 If|G|is a power of a primepthenGhas nontrivial center.

Proof: We have Gpartitioned as G = Z(G)∪ {noncentral elements}. Every conjugacy class has size a power of p. This power is zero precisely for those classes consisting of a single element in Z(G) and every conjugacy class of noncentral elements has size a positive power of p. Hence

|{noncentral elements}|is divisible byp. As|G|is also divisible byp, it follows thatpdivides|Z(G)|. Since|Z(G)| ≥1, it follows that a positive power ofpdivides|Z(G)|.

3 Basic Examples of Groups.

We give brief introductions to the most fundamentally important groups here.

3.1 Cyclic groups

Let_Z be the group of integers under addition, with identity element0. Since addition of integers is commutative, the groupZis abelian. Using the division algorithm, one proves that for any subgroup

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H ≤ _Z, there is an integer n ≥ 0 such thatH = n_Z, the set of multiples of n. If n = 0 we have

H = {0}. Assume now that n ≥ 1. Then the quotient Z/nZ is finite of order n, and consists of

the cosetsn_Z,1 +n_Z, . . . ,(n−1) +n_Z. The subgroups of_Z/n_Zcorrespond to the subgroups of_Z containingn_Z. These are the subgroupsd_Z/n_Z, for positive integers d | n. Note that multiplication byd−1 induces an isomorphism

d_Z/n_Z'_Z/d−1n_Z.

Every subgroup is normal inZ, and we have

(_Z/n_Z)/(d_Z/n_Z)'_Z/d_Z.

Cyclic groups play an important role in any groupG. For each elementg ∈Gdetermines a homomor-phism

eg :Z→G, given by eg(n) = gn.

The image ofeg is the subgroup generated byg

imeg =hgi={gn: n∈Z}.

The kernel ofeg is a subgroup ofZ, hence is of the formmZ, for some integerm ≥0. Ifm= 0then

ghas andhgi '_Z. Ifm >0thenmis the order ofg.

A groupGiscyclicifG = hgifor someg ∈ G. That is,Gis cyclic iff there existsg ∈ Gsuch that every element ofGis a power ofg. We can also say thatGis cyclic iff there existsg ∈Gsuch that the homomorphismeg :Z→Gis surjective.

We have seen that every infinite cyclic group Gis isomorphic to _Z, and every finite cyclic group of ordern is isomorphic toZ/nZ. Usually we will encounter cyclic groups while we are working with

multiplicative notation, where we will letCndenote a generic cyclic group of ordern. Thus,Cn =hgi,

for any elementg ∈Gof ordern.

SinceCn ' Z/nZ, it is immediate from our discussion ofZ and its subgroups thatCn has a unique

subgroup of every orderddividingn, namelyhgd_{i '}_C

d. Every subgroup ofCnis of this form and we

have

Cn/Cd'Cn/d.

This is a complete description of all of the subgroups and quotients ofCn.

Example: Letm, nbe positive integers. The subgroup Cn[m] = {x ∈ Cn : xm = 1}ofCn is the

kernel of the homomorphismCn m

−→Cnsendingx7→xm, so it fits into the exact sequence

1−→Cn[m]−→Cn−→Cnm −→1.

Ifgis a generator ofCn, one can check thatCn[m] =hgn/di, whered= gcd(m, n). Thus,Cn[m]'Cd

andC_nm 'Cn/d.

A final remark on cyclic groups: The word “isomorphic” does not mean “equal”. We have_Z/n_Z'Cn,

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namely1 +n_Z. ButCn has no canonical generator. For ifhgi generatesCn then so does gk for any

integerkwithgcd(k, n) = 1(exercise...). For example, if

G=

1 0 0 1

,

0 1

−1 −1

,

−1 −1 1 0

thenG ' C3 and either nonidentity matrix generates G, but here is no natural preference for either

generator. The root of this issue is that the isomorphism_Z/n_Z−→∼ Cn induced byeg depends on the

choice of generatorg of Cn; a different choice would give a different isomorphism. An isomorphism

of this sort, which depends on one or more arbitrary choices, is callednoncanonical.

3.2 Finite abelian groups

Every finite abelian group is a direct product of cyclic groups. The first basic result in this direction is as follows.

Proposition 3.1 LetA, B, C be finite abelian groups fitting into the exact sequence

1−→A−→C −→π B −→1.

Assume that the orders ofAandBare relatively prime. ThenC'A×B.

Proof: Set m = |A| and n = |B|, so that |C| = mn. Let D be the set of elements in C whose order is relatively prime tom. ThenD∩A = {1}. I claim that π(D) = B. Letb ∈ B and choose

c∈Csuch thatπ(c) = b. Since|C|=mnwe have(cm)n = 1, so the order ofcm dividesn, which is relatively prime tom. Hencecm _∈ _D_{, and}_π₍_cm_{) =}_bm_{. But the map}_x _7→_xm _{is an automorphism of}

B, again sincegcd(m, n) = 1. Henceπ(D) =Bm ₌_B_{. It follows that}_C ₌_AD_{, and that}_π_maps_D

isomorphically ontoB. By Prop. ??we have

C =A×D'A×B.

However, a product decomposition of an abelian group need not be unique.

Proposition 3.2 Supposen1, n2, . . . , nk are relatively prime integers with product n = n1n2· · ·nk. Then

Cn1 ×Cn2 × · · · ×Cnk 'Cn.

Proof: Letgibe a generator ofCnifor eachi. I claim that the element(g1, g2, . . . , gk)∈Cn1×· · ·×Cnk

has ordern. We have(g1, g2, . . . , gk)n = (gn1, g2n, . . . , gkn) = (1,1, . . . ,1), sinceni |n for alli. And if

gm

i = 1for allithenni | mfor alli, son | m, since theni are relatively prime. Hencenis the order

of(g1, g2, . . . , gk).

We can get a unique decomposition of a finite abelian groupGas follows. For each primepletG(p)

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Theorem 3.3 LetGbe a finite abelian group of ordern. Then 1. G'Q

p|nG(p)is the direct product of its nontrivial subgroupsG(p).

2. For each prime p there exist unique positive integers e1 ≥ e2 ≥ · · · ≥ ek > 0 such that

e1+e2+· · ·+ekis the power ofpdividingnand

G(p)'Cpe₁ ×C_pe₂ × · · · ×C_p_ek.

Proof: Part 1 follows from Prop. 3.2, using induction on the number of primes dividing|G|. We will prove part 2 later, using modules over principal ideal domains. See Milne for an elementary proof.

Example 1: Ifn= p1p2· · ·pkis a product of distinct primes, then there is only one abelian group of

ordern, up to isomorphism, namelyCn. For the unique decomposition of Thm.3.3would be

G'Cp1 ×Cp2 × · · · ×Cpk,

which is isomorphic toCn, by Prop.3.2.

Corollary 3.4 A finite subgroup of the multiplicative group of a field is cyclic.

Proof: LetF be a field and letGbe a finite subgroup of the multiplicative groupF×. Ifnis a positive integer andg ∈Ghas order dividingn, theng is a root of the polynomialxn₋₁_{, which has at most}_n

roots inF. HenceGhas at mostnelements of order dividingn, for anyn ≥1. WriteG=Q

pG(p),

according to part 1 of Thm.3.3. EachG(p)has at mostpelements of order dividingp. HenceG(p)is

cyclic, by part 2 of Thm.3.3. NowGis cyclic, by Prop. 3.2.

Example 2: IfF is a finite field with|F|=q, thenF× 'Cq−1. Consequently, for any positive integer

mthe subgroupµm(F) = {x∈F : xm = 1}is also cyclic, of ordergcd(m, q−1):

µn(F)'Cgcd(m,q−1) (4)

3.2.1 Unit Groups

Letmbe a positive integer and considerZ/mZunder multiplication (that is, as a ring):

(k+m_Z)(k0+m_Z) =kk0+m_Z,

which is a well-defined operation. It is not a group operation, however, since the element0+m_Zhas no multiplicative inverse. In fact, from the Euclidean algorithm it follows thatk+m_Zhas a multiplicative inverse in_Z/m_Ziffgcd(k, m) = 1. Hence the set

(_Z/m_Z)× ={k+m_Z: gcd(k, n) = 1}

forms a group under multiplication inZ/mZ, with identity element1 +mZ. Since multiplication inZ

is commutative, the group(_Z/m_Z)×is abelian, of order

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Proposition 3.5 Let m = pr1 1 · · ·p

rk

k be the factorization of m into a product of powers of distinct primespi. Then

1.

(_Z/m_Z)×'

k

Y

i=1

(_Z/pri

i Z)

×

.

2. For any primepand integerr≥1we have

(_Z/pr_Z)×'

  

 

C(p−1)pr−1 'C_p₋₁×C_pr−1 ifp≥3

C2×C2r−2 ifp= 2andr≥2

1 ifp= 2andr= 1.

Proof: Part 1 follows from the Chinese Remainder Theorem. We prove part 2 for p ≥ 3and leave

p= 2as an exercise. Reduction modulopgives a surjective map

π : (_Z/pr_Z)× −→(_Z/p_Z)×.

By Prop. 3.2and Example 2 above, it suffices to show thatkerπis cyclic. Clearly1 +p∈kerπ. We will show that1 +phas orderpr−1 _modulo_pr_.

Recall thatp| p k

for all positive integersk < p. Now ifaandbare congruent integers modulo some powerp`_{, it follows that}_ap _≡_bp _mod_p`+1_{. Now using induction on}_` _≥₂_{we have}

(1 +p)p`−2 ≡1 +p`−1 mod p`,

which implies that1 +phas orderp`−1_modulo_`_{for any}_`_≥₂_.

3.3 The symmetric group

For any setX, the setSX of bijectionsσ :X →X fromX to itself forms a group under composition

whose identity element is the identity function e(x) ≡ x, called the symmetric group of X. The elements ofSX are usually calledpermutations, and the identity is thetrivial permutation.

Iff :X →Y is a bijection between two setsX andY, then we get an isomorphismSf :SX

∼

−→SY,

defined by

Sf(σ) =f ◦σ◦f−1.

IfX = {1,2, . . . , n}, we writeSn instead of SX. In the literature,Sn is sometimes called the

sym-metric group on n letters. This is a slight abuse of terminology, for the following reason. Sup-pose X is any finite set, with |X| = n. By labelling the elements of X we obtain a bijection

f : X → {1,2, . . . , n}, whence an isomorphism Sf : SX

∼

−→ Sn. However, this isomorphism is

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By counting permutations of{1,2, . . . , n}, we find that

|Sn|=n! = 1·2·3· · ·n.

For 1 ≤ k ≤ n, a k-cycle is an element of Sn obtained as follows. Choose distinct numbers

i1, i2, . . . , ik ∈ {1, . . . , n}. Then(i1 i2 . . . ik)is the element ofSnwhich sends

i1 7→i2 7→i3 7→ · · · 7→ik 7→i1

and fixes all numbers in{1, . . . , n} − {i1, . . . , ik}. Note that the cycles

(i1 i2 . . . ik), (iri1 i2 . . . ik−1), (ik−1 iki1 i2 . . . ik−2), · · · ,(i2i3 . . . iki1)

all give the same permutation of{1, . . . , n}. Hence we regard these cycles as equal.

We will see how to express each element of Sn as a product of cycles. Given σ ∈ Sn, define an

equivalence relation∼

σ on{1, . . . , n}as follows. Fora, b∈ {1, . . . , n}, we declare that

a∼

σ b ⇔ a=σ

i_b

for somei ∈ _Z. Let A1, . . . Aq be the equivalence classes (here q depends onσ), numbered so that

|A1| ≥ |A2| ≥ · · · ≥ |Aq|, and setλi =|Ai|for eachi. Thus, we have a set partition

{1, . . . , n}=

q

a

i=1

Ai

and a corresponding numerical partition

n =

q

X

i=1

λi,

where theλiare positive integers such thatλ1 ≥λ2 ≥ · · ·λq ≥1. Ifλ1 = 1then allλi = 1andσ =e

is the identity element ofSn. Assume thatλ1 >1and letp≤ qbe the largest index such thatλp >1.

Now choose one elementai ∈Aiarbitrarily, for eachi= 1,2, . . . , p. The cycle

σi = (ai σaiσ2ai . . . σλi−1ai)

preservesAiand acts there just asσdoes, while ignoring the numbers outsideAi. It follows that

σ=σ1σ2· · ·σp (5)

is a product of disjoint cycles. Disjoint cycles commute, as we will soon see. Hence the product (5) is independent of the order of the termsσ1, . . . , σp. Thecycle typeofσis the partition

λ(σ) = [λ1, λ2, . . . , λq]

recording the lengths of the cyclesσi, as well as the number of fixed-points ofσ, which isq−p.

For example, the elementσ∈S6 sending

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may be written as

σ= (143)(26),

and has cycle typeλ(σ) = [3,2,1]. Note that5, which is fixed, is omitted in the cycle decomposition ofσ, but is counted in the cycle type ofσ.

We multiply using the cycle decomposition by following the path of each number, starting with the right-most cycle. For example, we have

(143)(26)(465) = (142653).

Note that the cycle type ofτ = (143)(26)(465)isnot[3,3,2], because the cycles are not disjoint. We first have to make them disjoint, as we have done above. We obtained a single6-cycle(142653), so the cycle type ofτ inS6isλ(τ) = [6].

We illustrate the cycle types for all elements ofS3andS4, as follows.

λ σ

[3] (123),(321) [21] (12),(23),(13) [111] e

λ σ

[4] (1234),(1324),(1243),(1423),(1342),(1432) [31] (123),(321),(124),(142),(134),(143),(234),(432) [22] (12)(34),(13)(24),(14)(23)

[211] (12),(13),(14),(23),(24),(34)

[1111] e

The cycle types determines the conjugacy classes inSn.

Lemma 3.6 Letσ ∈Snbe any element and letτ = (i1i2 · · · ik)be a cycle inSn. Then

στ σ−1 = (σi1σi2 · · ·σik).

Proof: Letj ∈ {1,2, . . . , n}={σ1, σ2, . . . , σn}. Ifj =σipfor some1≤p≤k then

στ σ−1j =στ ip =σip+1,

where subscripts are read modulok. And ifj /∈ {σi1, . . . , σik}, thenτ σ−1j =σ−1j, soστ σ−1j =j.

Proposition 3.7 Elementsα, β ∈Snare conjugate if and only ifλ(α) =λ(β).

Proof: Supposeβ =σασ−1 for someσ ∈Sn. Letλ(α) = [λ1 λ2 . . . λq]be the cycle type ofα. Then

the cycle decomposition ofα is α = α1α2· · ·αp , where theαi are disjoint cycles of length λi, and

λp >1 =λp+1ifq > p. Now

β =σασ−1 =σα1σ−1·σα2σ−1· · ·σαpσ−1.

By Lemma 3.6, each βi := σαiσ−1 is aλi-cycle and theβi’s are disjoint cycles since σ is injective.

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Conversely, supposeλ(α) =λ(β) = [λ1λ2 . . . λq]. Then the disjoint cycle decompositions ofαandβ

have the formα =α1α2· · ·αp andβ =β1β2· · ·βp, where bothαi andβiareλi-cycles for1≤i≤q.

There areai, bi ∈ {1, . . . , n}such that

αi = (aiαai . . . αλi−1ai), and βi = (biβbi . . . βλi−1bi),

and

{αjai : 1≤i≤q, 1≤j ≤λi}={1,2, . . . , n}={βjbi : 1≤i≤q, 1≤j ≤λi}.

Hence there is a permutationσ ∈ Snsuch thatσαjai =βjbi for alli, j. In particular,σai =bi for all

i, and we have

βjbi =σαjai = (σασ−1)jbi

so thatβ =σασ−1, as claimed.

A permutation σ ∈ Sn is even if its cycle type λ(σ) contains an even number of even numbers;

otherwiseσisodd. Thus,(12)(34)and(123)are even, while(12)and(12)(345)are odd. We put

sgn(σ) =

(

+1 ifσ is even

−1 ifσ is odd

Note that{±1}is a group under multiplication.

Proposition 3.8 The functionsgn : Sn → {±1}is a surjective group homomorphism.

We will prove this in the next section, using determinants. TheAlternating GroupAnis defined as

An = ker sgn ={σ∈Sn : σis even}.

Thus,An/ Snand

|An|= 1₂n! = 3·4· · · · ·n.

For example, A4 has order 12; it consists of the eight 3-cycles, the three 22-cycles and the identity

element ofS4.

The conjugacy classes inAn are determined as follows. First,An consists of the even classes inSn.

However, two elements ofAnwhich are conjugate inSnneed not be conjugate inAn. Hence an even

Sn-class could break up into severalAn classes. To see when this happens, letσ ∈ Anand restriction

of the charactersgnto the centralizerCSn(σ)ofσ inSn. We have

CAn(σ) = ker[CSn(σ)

sgn

−→ {±1}],

so that the index

hσ := [CSn(σ) :CAn(σ)] =

(

2 if CSn(σ)6≤An

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Now the size of the conjugacy classσAn _in_A

nis given by

|σAn_|₌ |An| |CAn(σ)|

=

1 2|Sn| 1

hσ|CSn(σ)|

= hσ 2 · |σ

Sn_|.

It follows that ifCSn(σ)6≤Anthenσ

Sn _{is a single conjugacy class in}_A

nand ifCSn(σ)≤Anthenσ

Sn

breaks up into twoAn-conjugacy classes:

σSn ₌_σAn _∪_σ_¯An_,

whereσ¯is conjugate toσinSnbut not inAn.

For example, in S4 elements in the [22]-class contain 2-cycles in their centralizer (which is D4), so

the[22]is a single class inA4. But elements in the[31]-class generate their own centralizer, which is

therefore contained inA4. So the[31]class breaks up into two classes with four elements each. These

classes are mutually inverse. IfA4is viewed as rotations of the tetrahedron, then one class consists of

clockwise face rotations and the other class consists of counterclockwise face rotations.

3.4 Linear groups

LetV be a vector space over a fieldF. The setGL(V)of invertible linear transformationsT :V →V

forms a group under composition, called thegeneral linear groupofV, whose identity element is the transformationIV(v)≡v.

If V has finite dimension n over F and we choose an ordered basis {v1, . . . , vn} of V then each

T ∈GL(V)corresponds to an invertiblen×nmatrixAT whosejthcolumn is



   

a1j

a2j

.. . anj      ,

whereT(vj) =a1jv1+a2jvj +· · ·+anjvn. Matrix multiplication is defined so that

AS·AT =AST, for all S, T ∈GL(V).

It follows that sendingT 7→AT is a group isomorphism

GL(V)−→∼ GLn(F), (6)

whereGLn(F)is the group ofn ×n invertible matrices under matrix multiplication, whose identity

element is then×nidentity matrix

In =



   

1 0 . . . 0 0 1 . . . 0

..

. . .. 0

0 0 . . . 1

     .

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As was the case forSn, the isomorphism (6) is noncanonical, because it depends on a choice of ordered

basis{v1, . . . , vn}ofV. A1×1matrix is just a number, so

GL1(F) =F×

is the group of nonzero elements of the fieldF under multiplication. This groupF×appears inGLn(F)

as both a normal subgroup and a quotient. First, we have an injective homomorphism

F×,→GLn(F), given bya7→a·In=



   

a 0 . . . 0 0 a . . . 0

..

. . .. 0

0 0 . . . a



   

,

whose imageZ ' F×is the center ofGLn(F). TheProjective Linear GroupP GLn(F)is defined

as

PGLn(F) = GLn(F)/Z.

Elements ofP GLn(F)are no longer linear transformations, but they permute the lines in the vector

space Fn. Thus, the elements of PGLn(F) are transformations of the Projective Space Pn−1(F)

which is the set of lines inFn_.

Next, recall that an n × n matrix A is invertible if and only if its determinant det(A) is nonzero. Moreover, ifA andB are twon×n matrices, we have det(AB) = det(A)·det(B). Thus,det is a homomorphism

det : GLn(F)→F×.

TheSpecial Linear GroupSLn(F)is defined as

SLn(F) = ker det ={A∈GLn(F) : det(A) = 1}.

WhenF =_R, elements ofSLn(F)are the linear transformations ofRnwhich preserve volume.

Returning to generalF, consider the restriction ofdettoZ. Since

det(aIn) =an, (7)

it follows that

det(Z) =F×n

is the subgroup ofnth_{powers in}_F×

. Hence the composition

GLn(F)

det

−→F× −→F×/F×n

induces a surjective homomorphism

PGLn(F)

det

−→F×/F×n.

The latter group depends on the fieldF. For example, we have|_C×_/

C×n| = 1, while |R×/R×n| = 1

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It also follows from (7) that the intersection

Z1 =SLn(F)∩Z ={aIn : an = 1} 'µn(F),

whereµn(F) ={a ∈ F× : an = 1}is the subgroup ofnth roots of unity inF×. The groupZ1 is the

center ofSLn(F)and the quotient

PSLn(F) = SLn(F)/Z1 = ker det

is the image ofSLn(F)inPGLn(F). The centerZ1 'µn(F)ofSLn(F)also depends on the fieldF.

For example, we have|µn(C)|=n, while|µn(R)|= 1or2according asnis odd or even. For general

fieldsF,µn(F)is always finite of order dividingn.

In summary, we have four closely related groups GLn(F),SLn(F),PGLn(F),PSLn(F), related to

each other via the following commutative diagram with exact rows.

1 −−−→ µn(F) −−−→ F×

n

−−−→ F×n −−−→ 1

  y

1 −−−→ SLn(F) −−−→ GLn(F)

det

−−−→ F× −−−→ 1

  y

1 −−−→ PSLn(F) −−−→ PGLn(F)

det

−−−→ F×/F×n −−−→ 1

IfF is a finite field with|F|=q, we writeGLn(q),SLn(q),PGLn(q),PSLn(q)instead ofGLn(F),SLn(F),PGLn(F),PSLn(F).

Often in the literature one finds the abbreviation Ln(q) forPSLn(q). The orders of these groups are

given as follows.

|GLn(q)|=qn(n−1)/2(q−1)(q2−1)· · ·(qn−1)

|PGLn(q)|=|SLn(q)|=qn(n−1)/2(q2−1)(q3 −1)· · ·(qn−1)

|PSLn(q)|=|SLn(q)|/gcd(n, q−1)

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In the last line, recall from (4) thatgcd(n, q−1) = |µn(F)|. All of these orders then follow from the

calculation of|GLn(q)|, which can be done as follows. To have a matrixA ∈GLn(q), we can take any

of theqn₋₁_{nonzero vectors in}_Fn_{for the first column, then any of the}_qn₋_q_{vectors not in the line}

spanned by the first column, then any of theqn−q2 vectors not in the plane spanned by the first two columns, etc. This gives

|GLn(q)|= (qn−1)(qn−q)(qn−q2)· · ·(qn−qn−1)

=q1+2+···+(n−1)(qn−1)(qn−1−1)(qn−2−1)· · ·(q−1) =qn(n−1)/2(qn−1)(qn−1−1)(qn−2−1)· · ·(q−1),

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3.4.1 Conjugacy classes inGL2(F)

We assume a bit more familiarity with fields in this section. LetF be a field. Assume either thatF is finite or that2 6= 0in F. LetF¯ be a fixed algebraic closure ofF. We identifyF× with the center of

GL2(F), as in (—).

Aquadratic extensionofF is a fieldK such thatF ⊂K ⊂F¯andK is a two-dimensionalF-vector space. This meansK =F ⊕F λfor some (any)λ ∈ K −F. The elements1, λ, λ2 _{are then linearly}

dependent overF so we have a dependence relationλ2−Tλ+N= 0, whereT=T(λ)andN=N(λ)

(the norm and trace) are elements ofF. The roots of the polynomialx2₋_T_x₊_N_{are distinct; let}_λ_¯_be

the root other thanλ.

Using the basis{1, λ}the action ofK×onK by multiplication gives an embedding

ιλ :K× ,→GL2(F).

For an arbitrary elementa+bλ∈K×we have(a+bλ)·1 =a+bλand

(a+bλ)·λ =aλ+bλ2 =aλ+b(Tλ−N) = −bN+ (a+bT)λ.

Henceιλ is given explicitly by

ιλ(a+bλ) =

a −bN

b a+bT

.

The eigenvalues of this matrix area+bλanda+bλ¯; in particular these eigenvalues lie inK×.

A different choiceλ0 of λ ∈ K −F gives a new basis ofK overF, and the subgroups ιλ0(K×)and

ιλ(K×)are conjugate inGL2(F). We denote any of these subgroups byK×, with the understanding

that they are only determined up to conjugacy. IfLandK are distinct quadratic extensions ofF the groupsL× and K× are not conjugate in GL2(F) and any two members in the class L× or K× will

intersect inF×. This is because the eigenvalues of elements ofL× lie in Land similarly for K, and

K∩L=F.

Forg, h∈GL2(F)we writeg ∼hto mean thatgandhare conjugate inGL2(F).

Proposition 3.9 Letg ∈GL2(F)have eigenvaluesλ, µinF¯.

1. Ifλ=µ∈F×, then eitherg =

λ 0 0 λ

org ∼

λ 1 0 λ

.

2. Ifλ6=µ∈F×, theng ∼

λ 0 0 µ

∼

µ 0 0 λ

.

3. If λ 6= µ /∈ F× thenλandµ = ¯λbelong to a unique quadratic extension K andg ∼ ιλ(λ) ∼

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Proof: In case 1, the matrixg−λ·I2 has nonzero kernel. Ifg −λ·I2, choose any vectorv ∈ F2

such that the vectoru:= (g−λ·I2)v is nonzero. Since(g−λ·I2)2 = 0, the vectorsu, v are linearly

independent. Using the basisu, v, we haveg ∼

λ 1 0 λ

.

In case 2, we use the basis of eigenvectors ofgto see thatg ∼

λ 0 0 µ

.

In case 3, λ and λ¯ are the roots of the characteristic polynomial of g so they generate a quadratic extensionKofF. The elementιλ(λ)∈ ιλ(K×)has the same eigenvalues. Hencegandιλ(λ)are two

elements ofGL2(F)which are conjugate to

λ 0 0 ¯λ

inGL2(K). From the theory of rational canonical

form, or Hilbert’s theorem 90, it follows thatgandιλ(λ)are conjugate inGL2(F).

IfF has no quadratic extensions, for example ifF = _Cor ifF is the field of constructible numbers, then case 3 does not arise. IfF is finite, say|F| =q, thenF has only one quadratic extensionK, and

|K|=q2_{. In this case there are}₍_q2₋_q₎_/₂_{conjugacy classes of type 3.}

3.4.2 Snas a subgroup ofGLn(F)

The symmetric groupSn is (isomorphic to) a subgroup of GLn(F), as follows. Let ei ∈ Fn be the

vector with1in theithcomponent and zero in the other components. For each permutationσ ∈Sn, let

Aσ be then×nmatrix such thatAσ(ei) =eσ(i). Then forτ, σ ∈Snwe have

AτAσ(ei) = Aτ(eσ(i)) = eτ σ(i)=Aτ σei.

Hence the map

f :Sn →GLn(F) given by f(σ) = Aσ

is a homomorphism. It is easy to check that the mapf is injective and the image off consists of the matrices with a single entry= 1in each row and column and all other entries = 0. These are called

permutation matrices.

Supposeσ = (i1i2· · ·ir)is anr-cycle inSn. The matrixAσ sendsei1 7→ei2 7→ · · · 7→ eir 7→ ei1 and fixes the remainingej’s. Since the determinant of a matrix is unchanged by simultaneous interchanges

of rows and columns, we have

det(Aσ) = det

A0_σ 0 0 In−r

= det(A0_σ)

whereA0_σ is ther×rmatrix

A0_σ =

      

0 0 0 . . . 1 1 0 0 . . . 0 0 1 0 . . . 0

..

. ... . .. ... 0 0 0 . . . 1 0

       .

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Expanding along the top row, we compute

det(A0_σ) = (−1)r−1.

Hencedet(Aσ) = +1or−1according asris odd or even.

Now take a general σ ∈ Sn and write it as a product of disjoint cycles. We have Aσ = +1 or −1

according asσhas an even or odd number of even cycles. This agrees with our definition ofsgnabove, so we have shown that

det(Aσ) = sgn(σ).

Hencesgn = det◦f is a group homomorphism, as claimed in Prop. 3.8.

3.4.3 The Bruhat decomposition forGLn(F).

The invertible matrices are characterized as those whose determinant is nonzero. However, it is not ob-vious how to write down a general invertible matrix. That is, we have the locus, but not the parametriza-tion of invertible matrices. The Bruhat decomposiparametriza-tion remedies this by expressing a general invertible matrix in terms of simpler ones that can be parametrized.

We have seen that the group G = GLn(F) has the subgroup W ' Sn consisting of permutation

matrices. We also have the subgroupB < Gconsisting of upper triangular matrices with all diagonal entries nonzero.

Theorem 3.10 (Bruhat Decomposition) The groupG= GLn(F)is a disjoint union

G= a

w∈W

BwB,

whereBwB ={b1wb2 : bi ∈B}.

Proof: Let{e1, . . . , en}be the standard basis ofFn. ThenB is the subgroup of elementsb ∈ Gfor

whichbej is contained in the span of{ei : i < j}, for all1≤j ≤n.

Takeg ∈Gand write

ge1 =

n

X

i=1

giei,

and leti1 = max{i: gi 6= 0}. Defineb ∈Bby

bej =

(

g−1_i

1

ei1 − Pi1−1

i=1 giei

ifj =i1

ej ifj 6=i1.

Then

bge1 =b

i1 X

i=1

giei = i1−1 X

i=1

giei+gi1 ·g −1

i1 ei1 −

i1−1 X

i=1

giei

!

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Forj >1letg0_j be the coefficient ofei1 inbgej, and defineb

0 _∈_B_by

b0ej =

(

ej −gj0e1 ifj >1

e1 ifj = 1.

We then have

bgb0e1 =bge1 =ei1 and forj >1the coefficient ofei1 inbgb

0_e

j isgj0 −g

0

j = 0. Thus, the matrixbgb

0

has column1equal toei1 and rowi1 equal toe1.

We repeat the above procedure with column2of the matrixh:=bgb0. Write

he2 =

n

X

i=1

hiei,

and leti2 = max{i: hi 6= 0}. Note thathi1 = 0, soi1 6=i2. Defineb

00 _∈_B_by

b00ej =

(

h−1_i

2

ei2 − Pi2−1

i=1 hiei

ifj =i2

ej ifj 6=i2.

Then

b00he1 =b00ei1 =ei1, since i1 6=i2, and

b00he2 =b00

i2 X

i=1

hiei = i2−1 X

i=1

hiei+h−1i2 ·hi2 ei2 −

i2−1 X

i=1

hiei

!

=ei2.

Forj >2, leth0_j andh_j00 be the coefficients ofei1 andei2 inb 00_he

j, and defineb000 ∈B by

b000ej =

(

ej −h0je1 −h00je2 ifj >2

ej ifj = 1,2.

We have then

b00hb000e1 =b00he1 =ei1, b 00

hb000e2 =b00he2 =ei2 and forj >2the vectorsei1 andei2 each have coefficient= 0in the vector

b00hb000ej =b00hej −h0jei1 −h 00

jei2.

Thus, forj = 1,2, the matrixb00hb000 =b00bgb0b000 has columnj equal toeij and andi

th

j row equal toej.

Repeating this process up to j = n shows that there are elements b1, b2 ∈ B such that b1gb2 is the

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3.5 The dihedral groups

D

n

Areflectionis an isometry of a Euclidean spaceE whose set of fixed-points inE is a hyperplane. A

reflection groupis a group of isometries ofE which is generated by reflections. The dihedral groups are the reflection groups in two dimensions.

SupposeE is a plane. Each line`inEdetermines a reflectionrwith fixed-point set`. IfP is a point not on`, thenr(P)is the mirror image ofP with respect to`. We say thatris the “reflection about`”. Note thatris a nontrivial involution of the isometry group of the plane.

Suppose`and`0 are two lines in the plane, with reflectionsrandr0. If`and `0 are not parallel, then they meet in a point P, and the product rr0 is rotation about twice the angle at P from `0 to `. In particular, r and r0 commute precisely when` and `0 are perpendicular. If ` and `0 are parallel, the productrr0is translation by twice the perpendicular vector from`0 to`.

For any integern ≥ 1considernlines`1, . . . , `nin the plane meeting in a common point, with equal

angles (=π/n) between adjacent lines. Letri be the reflection about`i. Thedihedral groupDnis the

group generated by the reflectionsr1, . . . , rn. We defineD∞similarly, by taking a countable number

of parallel lines equally spaced apart (all meeting at infinity, with equal angle zero). Forn= 1we have just one line, with reflectionrand

D1 ={1, r} 'C2.

Forn= 2we have two perpendicular lines`1, `2, whose reflectionsr1, r2commute. Hence

D2 ={1, r1, r2, r1r2} 'C2×C2.

For n = 3 we have three lines `1, `2, `3 intersecting at the angle π/3. Let r, s be reflections about

adjacent lines. Thenrsis a rotation of order2π/3, hence has order three. The equation(rs)3 = 1can be written as

rsr =srs.

This elementrsr =srsis the third reflection. It follows thatD3 is generated by randsonly, and its

elements are

D3 ={1, r, s, rs, sr, rsr}.

The product of any two elements inD3 is completely determined by the three rules:

r2 = 1, s2 = 1, rsr=srs.

For example, we havers·rsr=rs·srs=r·rs=s.

It is a similar story for an arbitrary finiten ≥ 2. We again letr, sbe reflections about adjacent lines

`r, `s, so thatrsis a rotation by2π/nand hence has ordern. The equation(rs)n = 1can be written as

rsrs . . .

| {z }

nterms

=srsr . . .

| {z }

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The elementsrsis reflection about the lines(`r), which is the other line adjacent tos. It follows that

all reflections can be written in terms ofrands, ass(rs)i_{for some}₁_≤_i_≤_n_{and that the elements of}

Dnare

Dn ={(rs)i, s(rs)i : 1≤i≤n}

and|Dn| = 2n. The elementt = rsgenerates a cyclic subgrouphti ' Cn, consisting of all rotations

inDn, which has index two inDn. The reflections inDnare precisely the elements outside ofhti, and

for any reflectionr0, we haver0tr0 =t−1.

The subgroup lattice ofDncan be described as follows. For each divisormofn, we have first of all the

unique subgroupCm ≤ Cn = hti, as well asn/mcopies ofDm obtained as follows. Index the lines

byZ/nZ, and partition them according to the fibers of the natural mapπm : Z/nZ → Z/(n/m)Z.

LetDm(i)be the subgroup of Dn generated by the reflections about lines in the fiberπ−1m (i). Since this

fiber hasm equiangular lines, we indeed have D(mi) ' Dm. As iranges over Z/(n/m)Z, we obtain n/msubgroups D(mi), all containing the same cyclic subgroup Cm. In particular, we have n/1 = n

subgroupsD₁(i) 'D1, each generated by the reflections about one of the lines. Finally, if`|m |n, we

haveD(_`j) < Dm(i)iffπ_`−1(j)⊂πm−1(i), iffj ≡i mod mn.

The situation forD∞is similar but simpler. The elementt =rsnow has infinite order, hence generates

a copy of_ZinD∞, and for any reflectionr0 we again haver0tr0 =t−1. We leave the subgroupsD∞to

the exercises.

Returning to finiten, we can viewDnas subgroup ofGL2(R). Assume the lines intersect at(0,0)∈R2

and that the reflecting line`1is thex-axis. The reflectionr1has matrix

r1 =

1 0 0 −1

.

For1≤k < nlet`k+1be the line rotated from`counterclockwise bykπ/n. Then`2 is adjacent to`1

and the rotation(r1r2)k has matrix

(r1r2)k =

cos(2kπ/n) −sin(2kπ/n) sin(2kπ/n) cos(2kπ/n)

.

3.6 The quaternion and generalized quaternion groups

Q

4n

The generalized quaternion groups are best understood as subgroups of the group SL2(C) of 2×2

complex matrices with determinant =1. Let T be the subgroup of diagonal matrices in SL2(C). Its

normalizerN(T)inSL2(C)consists of two cosets ofT:

N(T) =T ∪wT, where w=

0 −1 1 0

.

The finite subgroups of N(T) are of two types: Those contained in T are cyclic. The generalized quaternion groupsare the finite subgroups ofN(T)which arenotcontained inT. LetQ < N(T)be such a subgroup. SinceQ6< T, it contains an element of the formwtfor somet ∈T. ReplacingQby

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s−1Qs, wheres2 =t, we may and shall assume thatt = 1, so thatw∈Q. Sincewhas order four, the order ofQis divisible by four.

The generalized quaternion groupQ4nis the unique subgroup ofN(T)containingwand having order

4n.

To seeQ4nexplicitly, letζk =eπi/n, which has order2nas an element ofC×. ThenQ4nis generated

by the two matrices

tn=

ζn 0

0 ζ_n−1

, and w=

0 −1 1 0

.

Note thattnhas order2n, thatwhas order four, and we have

wtnw−1 =t−1, w2 =tn=−I.

Thus, we have

Q4n={tinw

j _{: 0}_≤_i_≤₂_n₋₁_, _j _{= 0}_or₁_},

so that|Q4n|= 4n, as claimed. Note that

htni ∩ hwi=htnni=h−Ii,

where−I =

−1 0 0 −1

. This is the unique element of order two inQ4n, and the subgrouph−Ii is

normal inQ4n, with quotient

Q4n/h−Ii 'Dn.

Hence the subgroups of Q4n containing h−Ii are in bijection with the subgroups of Dn. Recall the

subgroups ofDnare cyclic rotations or dihedral. The subgroups ofQ4ncorresponding to cyclic

rota-tions are cyclic. These are precisely the subgroups of even order inQ4n. The odd-order subgroups are

cyclic and contained inhtni.

Fork = 1we haveQ4 = hwi ' C4. This is the only generalized quaternion group which is abelian.

Indeed, we haveZ(Q4n) = h−Iiwhenn≥2.

Fork = 2the groupQ8 is commonly known as thequaternion group(of order eight), and has different

notation. It is common to write

Q8 ={±1,±i,±j,±k},

where 1 = 1 0 0 1

, i=

√

−1 0 0 √−1

, j =

0 −1 1 0

, k=

0 √−1

√

−1 0

,

which have relationsi2 ₌_j2 ₌_k2 ₌₋₁_and

ij =k, jk =i, ki=j, ji =−k, kj =−i, ik =−j.

The proper subgroups ofQ8arehii,hji,hki,h−1i. We have

h−1i=hii ∩ hji ∩ hki

and this subgroup is both the centerZ(Q8)and the commutator subgroup [Q8, Q8]. The groupQ8 is

the simplest non-abelian group in which every subgroup is normal. It can be shown that any finite non-abelian group with all subgroups normal is isomorphic toQ8×A, whereAis abelian.

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3.7 p-groups, a first look

A finite groupGis ap-groupif the order ofGa power of a primep.

Each abelianp-group is a direct productG=Cpn₁ ×C_pn₂ × · · · ×C_pnk of cyclicp-groups, there being

one isomorphism class of such groups for every set of positive integers{n1, . . . , nk}. When allni = 1,

the groupCk

p =Cp×Cp× · · · ×Cpis calledelementary abelian of rankk. The dihedral groupsD2n

and generalized quaternion groupsQ2n are examples of nonabelian2-groups.

One cannot hope to classify allp-groups, except those whose orders are small powers ofp.

Proposition 3.11 Letpbe a prime and letGbe ap-group. 1. If|G|=pthenG'Cp.

2. If |G| = p2 _then _G_{is abelian. We have} _G _' _C

p2 if Gis cyclic and G ' C_p ×C_p if Gis not

cyclic.

3. If |G| = p3 then either G is one of two nonabelian groups or G is one of Cp3, C_p × C_p2 or

Cp×Cp×Cp.

Proof: We already noted that part 1 is a consequence of Lagrange’s theorem. We will prove part 2 here, and postpone the proof and a more detailed statement of part 3.

Assume|G| = p2_{. We have seen in Prop.} _2.15 _{that every} _p_{-group has a nontrivial center}_Z₍_G₎_{. By}

Lagrange’s theorem, we have|Z(G)| = por p2. If|Z(G)| = p then G/Z(G) has order p, hence is cyclic, soGis abelian, contradictingZ(G)6=G. HenceGis abelian.

The order of every element ofGalso dividesp2_{. If}_G_{has an element of order}_p2_then_G_'_C

p2. Assume

Ghas no element of orderp2. Then every nonidentity element ofGhas orderp. Chooseh, k ∈Gwith

h 6= 1andk /∈ hhi. The subgroupsH = hhiand K = hkihaver orderpand are both normal in the abelian groupG. NowHK is a subgroup ofGproperly containing H. Since [G: H] = p, it follows thatHK =G. Likewise,H∩K is a proper subgroup ofK, which has orderp, soH∩K ={1}. Now

by Prop.2.10we haveG'H×K 'Cp×Cp.

Prop. 2.15can be extended to prove the converse of Lagrange’s theorem forp-groups. First we need a lemma.

Lemma 3.12 IfA is a finite abelian group whose order is divisible by a primepthenA contains an element of orderp.

Proof: By induction, we may assume the result is true for groups of smaller order. Letb ∈ A have orderm >1, and letB =hbi. Ifp|mthenbm/p_{has order}_p_{. Assume}_p

-m. Thenpdivides|A/B|and |A/B|< |A|, soA/B has an element of orderp, by the induction hypothesis. This element is aBfor somea∈Asuch thata /∈B, butap _∈_B_{. Therefore}_ap ₌_br_{for some integer}_r_{. Since}_gcd(_{p, m}_{) = 1}_,

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we can writer =kp+`mfor integersr, `. The elementc=ab−kdoes not belong toBsincea /∈B, but sinceAis abelian we have

cp =apb−kp =br−kp =b`m = 1.

Hencec∈Ahas orderp.

As we will see in the next result, the lemma is true without the assumption thatA is abelian, but the proof is not as constructive.

Proposition 3.13 Let G be a finite group of order pr, where p is a prime. Then G has a chain of subgroups

1 =G0 < G1 < G2 <· · ·< Gr−1 < Gr =G such that for all0≤i < rwe have

1. |Gi|=pi;

2. Gi is a normal subgroup ofGandGi+1/Gi 'Cp; 3. Gi+1/Gi is contained in the center ofG/Gi.

Proof: We argue by induction on r. By Prop. 2.15, the centerZ(G)is a nontrivial abelianp-group. By Lemma3.12, there exists a subgroup G1 < Z(G) of orderp. Since G1 is central in G we have

G1/ G. The groupG=G/G1has orderpr−1. Applying the induction hypothesis toG, there is a chain

of subgroups

1 =G0 < G1 < G2 <· · ·< Gr−2 < Gr−1 =G

such that for all0≤i < r−1we have|Gi|=pi andGi/ GandGi+1/Giis contained in the center of

G/Gi.

By the Correspondence Theorem applied toG/G1there are normal subgroupsGi EGsuch that

Gi =Gi/G1.

Moreover, the canonical projectionG→Ginduces isomorphisms

G/Gi

∼

−→G/Gi

which restrict to isomorphisms

Gi+1/Gi

∼

−→Gi+1/Gi

for each0≤i < r. It follows thatGi+1/Giis contained in the center ofG/Gi, as claimed.

Thus, everyp-group has a tower of normal subgroups whose quotients are cyclic of orderp. Despite this apparent simplicity, the number of isomorphism classes of groups of orderpr _{grows rapidly with}

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|G| number of groups

2 1

22 ₂

23 5

24 15

25 ₅₁

26 267

27 2 328 28 _{56 092}

29 10 494 213 210 49 487 365 422

It has been determined1_{that the total number of all groups of order}_≤₂₀₀₀_is_{49 910 529 484}_{, so over}

99%of these groups have order210.

3.8 Simple groups

A groupG is simple if G has no normal subgroups other than {1} and Gitself. Such groups have remarkable properties. For example,

Every homomorphism from a simple group to another group is either injective or trivial.

For iff :G→G0 is a nontrivial homomorphism from a simple groupGinto some other groupG0then

f is automatically injective, sincekerf is a normal subgroup ofG.

Likewise,

IfGis nonabelian simple, then the centerZ(G) ={1}and the commutator[G, G] =G.

For bothZ(G)and[G, G]are normal subgroups of G. AsGis nonabelian, we have Z(G) 6= Gand

[G, G]6={1}, so we must haveZ(G) = {1}and[G, G] =G.

By Lagrange’s Theorem, any group of prime order is simple. All other simple groups are nonabelian; they are extremely rare and interesting. Of the49 910 529 484groups of order at most2000, exactly six are nonabelian simple groups, namely

simple groupG |G| A5 'P SL2(5) 60

PSL2(7) 'GL3(2) 168

A6 'PSL2(9) 360

PSL2(8) 504

PSL2(11) 660

PSL2(13) 1092

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These small simple groups belong to two families An and P SL2(q) for n ≥ 5 and q ≥ 5 a prime

power. We prove that the groups in these families are simple.

3.8.1 Simplicity of alternating groups

Theorem 3.14 Forn ≥5the alternating groupAnis simple.

Proof: Every element of Anis a product of an even number of transpositions. HenceAnis generated

by elements of the form(a b)(c d)and(a b)(b c). Now

(a b)(c d) = (a c b)(a c d), and (a b)(b c) = (a b c),

soAnis also generated by3-cycles.

Since n ≥ 5, the centralizer of a 3-cycle in Sn contains a transposition, so this centralizer is not

contained inAn. It follows that the3-cycles form a single conjugacy class inAn.

LetN _EAnbe a nontrivial normal subgroup ofAn. We must show thatN = An. For allσ ∈N and

α∈ Anthe commutatorσ−1ασα−1 belongs toN, sinceN E An. We use this procedure to show that

N contains a3-cycle. SinceN is a union of conjugacy classes ofAn, it will follow thatN contain all

3-cycles and thereforeN =Anby our previous remarks.

We write elements ofN as products of disjoint cycles. Case 1: SupposeN contains a disjoint product of the form

σ =τ·(a1 a2 . . . ar), r ≥4.

Letα= (a1 a2a3). Then we compute

σ−1ασα−1 = (a1 a3ar)∈N,

so thatN contains a3-cycle in this case.

Case 2: SupposeN contains a disjoint product of the form

σ =τ ·(a b c)(d e f).

Letα= (a b d). Then we compute

σ−1ασα−1 = (a d b f c)∈N.

Then case 1 applies, and shows thatN contains a3-cycle. Case 3: SupposeN contains a disjoint product of the form

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We have already observed that(a b)(c d) = (a c b)(a c d), so that

σ =τ ·(a c b)(a c d).

Then case 2 applies, and shows thatN contains a3-cycle.

Every nonidentity element ofAncan be written in one of these three forms. HenceN must contain a

3-cycle.

3.8.2 Simplicity ofPSL2(F)

We next prove that the groupPSL2(F)is simple for any field with at least four elements. The proof

depends a series of lemmas, each interesting in its own right. We work in the groupG= SL2(F), with

the following subgroups and element:

B =

a b

0 a−1

: a ∈F×, b∈F

, U =

1 b

0 1

: b ∈F

, U =

1 0

c 1

: c∈F

,

T =

a 0 0 a−1

: a∈F×

, Z =

1 0 0 1 ,

−1 0 0 −1

=Z(G), w=

0 −1 1 0

.

Lemma 3.15 (Bruhat Decomposition) We haveG=B∪BwB, a disjoint union.

Proof: A matrix

a b c d

∈Glies outside ofB exactly ifc6= 0. In this case,

a b c d =

c−1 a

0 c

·

0 −1 1 0

·

1 dc−1

0 1

∈BwB.

Lemma 3.16 The subgroupB is a maximal proper subgroup ofG.

Proof: SupposeH is a subgroup ofGproperly containingB. Then there existsh ∈ H withh /∈ B. By Lemma3.15, we can writeh = b1wb2, withbi ∈B. It follows that w ∈H, henceBwB ⊂H, so

H =G.

Lemma 3.17 The groupGis generated byU andU.

Proof: LetHbe the subgroup ofGgenerated byU andU. Ifa∈F×, we have

a 0 0 a−1

= 1 1 0 1 · 1 0

a−1 1

·

1 −a−1

0 1

·

1 0

a−a2 1

Notes on Group Theory