CHAPTER 1

Groups

1.1. Definitions and Easy Facts

Definition 1.1. A group is a nonempty set G equipped with a binary opera-tion · satisfying the following axioms.

(i) (associativity) (a · b) · c = a · (b · c) for all a, b, c ∈ G;(ii) (existence of identity) there exists e ∈ G such that ea = ae = a for all

a ∈ G;(iii) (existence of inverse) for each a ∈ G there exists a−1 ∈ G such that

aa−1 = a−1a = e.If ab = ba for all a, b ∈ G, G is called abelian. (Note. It is more appropriate totreat abelian groups as Z-modules.)

Facts. Let G be a group.(i) The identity e and the inverse of a ∈ G are unique. (Let e′ be another

identity of G and b another inverse of a. Then e′ = e′e = e and b = be =b(aa−1) = (ba)a−1 = a−1.)

(ii) (Cancelation) Let a, b, c ∈ G. Then ab = ac ⇒ b = c; ba = ca ⇒ b = c;ab = e⇒ a = b−1.

(iii) If a1, . . . , an ∈ G, parentheses are not needed to define a1 · · · an becauseof the associativity. Moreover, (a1 · · · an)−1 = a−1n · · · a−11 .

Examples. Abelian groups: (Z,+), (R,+), (C,+), (R×, ·), (R× = R \ {0}),(C×, ·), Zn (integers modulo n).

Automorphism groups (groups of bijections which preserve a certain structure):SX = the group of all permutations of a set X; GL(V ) = the group of all invertiblelinear transformations of a vector space V ; GL(n, F ) = the group of n×n invertiblematrices over a field F , GL(n, F ) ∼= GL(Fn); the group of isometries of a metricspace; groups of automorphisms of groups, rings, modules, fields, graphs, etc.

Groups as invariants: the fundamental group and homology groups of a topo-logical space.

The law of exponents. Let G be a group, a ∈ G, m,n ∈ Z. Then aman =am+n, (am)n = amn. (Note. a0 := e; a−n := (a−1)n for n ∈ Z+.)

Additive notation (usually for abelian groups). We write a+ b for ab, 0 fore, −a for a−1 and na for an.

Cartesian product. If G and H are groups, the G × H is a group withoperation defined component wise.

Homomorphism. Let G and H be groups. A map f : G → H is called ahomomorphism if f(ab) = f(a)f(b) for all a, b ∈ G. It follows that f(eG) = eH and

1

2 1. GROUPS

f(a−1) = f(a)−1 for a ∈ G. ker f := f−1(eH) = {a ∈ G : f(a) = eH}. f is 1-1⇔ ker f = {eG}. If f : G → H is a homomorphism and a bijection, f is called anisomorphism. It follows that f−1 : H → G is also an isomorphism. If there existsan isomorphism f : G→ H, G and H are called isomorphic (G ∼= H). Isomorphicgroups have the same structure.

Aut(G) := the set of all automorphisms of G (isomorphisms from G to G).

(Aut(G), ◦) is the automorphism group of G.

Example. det : GL(n,R)→ R×, A 7→ detA, is a homomorphism.

Example. G = the set of all fractional linear transformations of C (functions

of the form f(z) =az + bcz + d

, where ad− bc 6= 0). (G, ◦) is a group.

φ : GL(2,C) −→ G[ a bc d

]7−→ f,

where f(z) =az + bcz + d

, is a homomorphism with kerφ = {aI2 : a ∈ C×}.

Subgroups. Let (G, ·) be a group. H ⊂ G is called a subgroup of G, denotedas H < G, if (H, ·) is a group. If X ⊂ G,

〈X〉 :=⋂

X⊂H

1.2. NORMAL SUBGROUPS AND QUOTIENT GROUPS 3

Proof. (i) α : G/H → H\\G, aH 7→ (aH)−1 = Ha−1, is a bijection.(ii) f : G → G/H, a 7→ aH, is onto. For each aH ∈ G/H, f−1(aH) = aH;

hence |f−1(aH)| = |aH| = |H|. So |G| = |G/H| |H|. �

Corollary 1.4 (Lagrange’s theorem). If H < G and |G|

4 1. GROUPS

Easy facts.(i) If H < G and K < G, then HK < G⇔ HK = KH.(ii) If H < G and K C G, then HK < G.(iii) If H C G and K C G, then HK C G.

Theorem 1.6 (The correspondence theorem). Let N C G and let A = {K <G : N ⊂ K}, B = the set of all subgroups of G/N . Then

f : A −→ BK 7−→ K/N

is a bijection. Moreover, K/N C G/N ⇔ K C G.

Universal mapping property of the quotient group. Let N C G andf : G→ H a group homomorphism such that ker f ⊃ N . Then ∃! homomorphismf̄ : G/N → H such that the following diagram commutes.

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f

πf̄

G H

G/N

Proof. Let f̄(aN) = f(a). �

Three isomorphism theorems.

First isomorphism theorem. Let f : G → H be a homomorphism. ThenG/ ker f ∼= f(G).

Proof. By the universal mapping property, f : G → H induces a homo-morphism f̄ : G/ ker f → f(G). f̄ is onto and ker f̄ = {ker f}. Hence f̄ is anisomorphism. �

Second isomorphism theorem. Let H < G and N C G. Then HN/N ∼=H/(N ∩H).

Proof. f : H → HN/N , h 7→ hN , is an onto homomorphism with ker f =N ∩H. �

Third isomorphism theorem. Let N C G, H C G and N ⊂ H. Then(G/N)

/(H/N) ∼= G/H.

Proof. f : G/N → G/H, aN 7→ aH, is a well-defined onto homomorphismwith ker f = H/N . �

Example. If G is a cyclic group, then G ∼= Z or Zn.

Proof. Let G = 〈a〉. Then f : Z→ G, n 7→ an, is an onto homomorphism. Ifker f = {0}, G ∼= Z. If ker f 6= {0}, let n be the smallest positive integer in ker f .Then ker f = nZ. So G ∼= Z/nZ = Zn. �

Fact. If (m,n) = 1, then Zm × Zn ∼= Zmn.

Fact. If H,K C G, HK = G and H ∩K = {e}, then G ∼= H ×K.

1.2. NORMAL SUBGROUPS AND QUOTIENT GROUPS 5

Proof. For all h ∈ H, k ∈ K, we have hkh−1k−1 ∈ H∩K; hence hkh−1k−1 =e, i.e., hk = kh. Define f : H ×K → HK, (h, k) 7→ hk. Then f is an isomorphism.

�

Conjugation. For a ∈ G,

φa : G −→ Gx 7−→ axa−1

is an automorphism of G (conjugation by a). A subgroup H < G is normal iffφa(H) = H for all a ∈ G. H is called a characteristic subgroup of G if f(H) = Hfor all f ∈ Aut(G). “Characteristic”⇒ “normal”; the converse is false, e.g., Z2×{0}is a normal but not characteristic subgroup of Z2 × Z2.

Inn(G) := {φa : a ∈ G} is a normal subgroup of Aut(G) and is called the innerautomorphism group of G. (In fact, if f ∈ Aut(G), fφaf−1 = φf(a).) Moreover,Inn(G) ∼= G/Z(G). (Proof. g : G → Inn(G), a 7→ φa, is an onto homomorphismwith ker g = Z(G).)

Example. Let G be a group with |G| > 2. Then |Aut(G)| > 1.

Proof. If G is nonabelian, |Inn(G)| = |G/Z(G)| > 1. So, assume that G isabelian. If a2 6= e for some a ∈ G, x 7→ x−1 is a nontrivial automorphism of G. Ifa2 = e for all a ∈ G, G is a vector space over Z2 of dimension > 1. In this case,Aut(G) = GL(G). Let E be a basis of G and choose �1, �2 ∈ E distinct. Then∃f ∈ GL(G) such that f(�1) = �2. �

The commutator subgroup and abelianization. An element of the formx−1y−1xy (x, y ∈ G), denoted by [x, y], is called a commutator of G. G′ =〈{x−1y−1xy : x, y ∈ G}〉 is the commutator subgroup of G.

Proposition 1.7.(i) G′ is a characteristic subgroup of G and G/G′ is abelian.(ii) Let H < G. Then H C G with G/H abelian ⇔ H ⊃ G′.

Note. G′ is the smallest normal subgroup H of G such that G/H is abelian. G/G′

is called the abelianization of G.

Proof. (i) {x−1y−1xy : x, y ∈ G} is invariant under any automorphism of G.(ii) (⇒) ∀x, y ∈ G, since G/H is abelian, (xH)(yH) = (yH)(xH); so x−1y−1xy

∈ H.(⇐) H/G′ C G/G′ ⇒ H C G. G/H ∼= (G/G′)/(H/G′) is abelian. �

Normal subgroups at top and bottom. Let H < G. If H ⊂ Z(G) orH ⊃ G′, then H C G.

If G is abelian, every subgroup of G is normal. The converse is false, as shownby the next example.

Example. (The quaternion group) Q8 = {±1,±i,±j,±k}. The multiplicationin Q8 is defined by rules i2 = j2 = k2 = −1, ij = k, jk = i, ki = j, ik = −j,kj = −i, ji = −k; see Figure 1.1. Z(Q8) = {±1}. Every subgroup of Q8 isnormal, yet Q8 is nonabelian. Q8 = 〈{i, j}〉. Let A =

[0 1−1 0

], B = [ 0 ii 0 ], C =[

i 00 −i

]∈ GL(2,C). Then G = {±I,±A,±B,±C} < GL(2,C) and Q8 ∼= G. (The

isomorphism is given by i 7→ A, j 7→ B.)

6 1. GROUPS

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i

kj

ij = kjk = iki = j

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i

kj

ik = −ikj = −iji = −k

Figure 1.1. Multiplication in Q8

Advanced Reading

Hamiltonian groups. A nonabelian group G is called Hamiltonian if all itssubgroups are normal.

Theorem 1.8 (Dedekind, Baer). G is Hamiltonian ⇔G ∼= Q8 ×A×B,

where A is an elementary abelian 2-group and B is an abelian group whose elementsall have odd order.

Lemma 1.9. Let G be a group and x, y ∈ G such that [x, y] commutes with xand y. Then

(i) [xi, yj ] = [x, y]ij, i, j ∈ Z;(ii) (xy)i = xiyi[y, x](

i2), i ∈ N.

Proof. (i) First assume i, j ≥ 0. We have [x, y]2 = (x−1y−1xy)x−1y−1xy =x−1(x−1y−1xy)y−1xy = [x2, y]. Induction shows that [x, y]i = [xi, y]. In the sameway, [xi, y]j = [xi, yj ].

If i < 0, note that [x−1, y] = [x, y]−1.(ii) Induction on i.

(xy)i+1 = xiyi[y, x](i2)xy = xiyixy[y, x](

i2) = xixyi[yi, x]y[y, x](

i2)

= xi+1yi+1[y, x]i+(i2) = xi+1yi+1[y, x](

i+12 ).

�

Proof of Theorem 1.8. (⇐) Let H < G. Then H = (H ∩ (Q8×A))× (H ∩B). (Here we abuse the notation by treating (Q8 × A) × B as an internal directproduct.) It suffices to show that H1 := H ∩ (Q8 × A) C Q8 × A. If H1 containsan element of order 4, then H1 ⊃ {h2 : h ∈ H1} = Z(Q) = G′. So H C G. If H1contains no element of order 4, then H1 ⊂ Z(G). Also, H1 C G.

(⇒) 1◦ Let x, y ∈ G such that c := [x, y] 6= 1. Since 〈x〉, 〈y〉 C G⇔ c ∈ 〈x〉∩〈y〉.So, xi = c = yj for some i, j ∈ Z+. Put Q = 〈x, y〉. Then Q′ = 〈c〉 ⊂ Z(Q). ByLemma 1.9 (i), ci = [xi, y] = [c, y] = 1. Hence o(x), o(y) 0 since otherwise [x, y] = 1.

1.3. THE SYMMETRIC GROUP 7

3◦ In 2◦, assume r ≥ s. Since x−pr−sy does not commute with x, by theminimality of o(x) + o(y), o(x−p

r−sy) ≥ o(y) = ps+1. By Lemma 1.9 (ii),

1 6= (x−pr−s

y)ps

= x−pr

yps

[y, x−pr−s

](ps

2 ) = [y, x]−pr−s(ps2 ) = c−

12p

r(ps−1).

So, p - 12pr(ps − 1)⇒ p = 2 and r = 1. So, o(x) = 4, x2 = y2, yxy−1 = x−1 ⇒ Q is

a homomorphic image of Q8. Q is nonabelian ⇒ Q = Q8.4◦ G = QC where C = CG(Q). (Let g ∈ G. Then gxg−1 = x±1 = x(−1)

a

,gyg−1 = y±1 = y(−1)

b

, a, b ∈ {0, 1}. Then yaxbg commutes with x and y, i.e.yaxbg ∈ C.

5◦ C has no element of order 4. (Assume g ∈ C with o(g) = 4. Then g /∈ Q, soo(gx) = 2 or 4. Since [gx, y] 6= 1, ygxy−1 = (gx)−1, i.e. g = g−1. Contradiction.)

6◦ C is abelian. (Otherwise, C would be a Hamiltonian group without elementof order 4, which contradicts 3◦.) For every g ∈ G, gx does not commute withy. By 1◦, o(gx) < ∞ ⇒ o(g) < ∞. So, C is torsion ⇒ C = A × B, where Ais an elementary abelian 2-group and every element of B has odd order. ThenG = QC ∼= QA×B. Since A is a vector space over Z2, A = (Q ∩A)×A′ for someA′ < A. It’s easy to see that QA = Q×A′. �

End Advanced Reading

1.3. The Symmetric Group

The symmetric group of a setX, denoted by SX , is the group of all permutationsof X. If |X| = n, (may assume X = {1, . . . , n}), write SX = Sn. |Sn| = n!.

Cayley’s theorem. Every group G is isomorphic to a subgroup of SG. If|G| = n, G is embedded in Sn.

Proof. Define

f : G → SGa 7→ f(a)

wheref(a) : G → G

x 7→ ax.

f is a homomorphism with ker f = {e}. �

Notation. σ ∈ Sn is denoted by(1 · · · n

σ(1) · · · σ(n)

).

Let i1, . . . , ik ∈ {1, . . . , n} be distinct. (i1, . . . , ik) ∈ Sn is the permutation whichmaps i1 to i2, i2 to i3, ..., ik−1 to ik, ik to i1 and leaves other elements unchanged;it is a called a k-cycle. A transposition is a 2-cycle. A 1-cycle is the identity of Sn.

Facts.(i) Disjoint cycles commute.(ii) (i1, . . . , ik) = (i2, . . . , ik, i1) = · · · .(iii) (i1, . . . , ik)−1 = (ik, . . . , i1).(iv) σ · (i1, . . . , ik) · σ−1 = (σ(i1), . . . , σ(ik)).

Theorem 1.10. Every σ ∈ Sn is a product of disjoint cycles. The cycles areunique except for the order in which they appear in the product.

8 1. GROUPS

Proof. Existence. Induction on n. In the sequence 1, σ(1), σ2(1), . . . , let i beth smallest positive integer such that σi(1) = 1. Then τ := (1, σ(1), . . . , σi−1(1))−1σfixes 1, σ(1), . . . , σi−1(1); hence τ ∈ S{1,...,n}\{σ(1),...,σi−1(1)}. By the induction hy-pothesis, τ is a product of disjoint cycles involving {1, . . . , n} \ {σ(1), . . . , σi−1(1)}.So, σ = (1, σ(1), . . . , σi−1(1))τ is a product of disjoint cycles.

Uniqueness. Assume σ = (1, i2, . . . , ik)α = (1, j2, . . . , jl)β, where α (β) is aproduct of disjoint cycles involving {1, . . . , n} \ {1, i1, . . . , ik} ({1, . . . , n}\{1, j1, . . . , jl}). Then for 2 ≤ s ≤ min{k, l}, is = σs(1) = js. Claim that k = l. (Ifk < l, then 1 = σ(ik) = σ(jk) = jk+1, which is a contradiction.) So α = β. Byinduction, the disjoint cycles in α and β are the same. �

Easy fact. If the disjoint cycles in σ ∈ Sn are of lengths l1, . . . , lk, theno(σ) = lcm(l1, . . . , lk).

Fact. Sn is generated by each of the following subsets.(i) {(1, 2), (1, 3), . . . , (1, n)}.(ii) {(1, 2), (2, 3), . . . , (n− 1, n)}.(iii) {(1, 2), (1, 2, . . . , n)}.

Proof. (i) (i1, . . . , ik) = (i1, i2)(i2, i3) · · · (ik−1, ik); (i, j) = (1, i)(1, j)(1, i).(ii) (1, j) = (j − 1, j)(1, j − 1)(j − 1, j), 2 ≤ j ≤ n.(iii) (i, i+ 1) = (1, 2, . . . , n)(i− 1, i)(1, 2, . . . , n)−1. �

The sign of a permutation. Let f(x1, . . . , xn) be a polynomial in x1, . . . , xnwith coefficients in Z and let σ ∈ Sn. Define σ(f) = f(xσ(1), . . . , xσ(n)).

σ( ∏

1≤i

1.3. THE SYMMETRIC GROUP 9

The dihedral group Dn. Dn is the subgroup of Sn generated by α =(1, . . . , n) and β = (2, n)(3, n− 1) · · · (dn2 e, b

n2 c+ 2).

Facts about Dn.

(i) o(α) = n, o(β) = 2, βαβ−1 = α−1.(ii) For n ≥ 3, |Dn| = 2n. Dn = {αiβj : 0 ≤ i ≤ n− 1, 0 ≤ j ≤ 1}.(iii)

Z(Dn) =

{{id, αn/2} if n is even,{id} if n is odd.

Proof. (i) βαβ−1 = β(. . . , n−1, n, 1, 2, 3, . . . )β−1 = (. . . , 3, 2, 1, n, n−1, . . . ) =α−1.

(ii) By (i), {αiβj : 0 ≤ i ≤ n − 1, 0 ≤ j ≤ 1} is closed under multiplication.So Dn = {αiβj : 0 ≤ i ≤ n − 1, 0 ≤ j ≤ 1}. αiβj , 0 ≤ i ≤ n − 1, 0 ≤ j ≤ 1,are all distinct. (Assume αi1βj1 = αi2βj2 , 0 ≤ i1, i2 ≤ n− 1, 0 ≤ j1, j2 ≤ 1. Thenαi1−i2 = βj2−j1 , 1 = βj2−j1(1) = αi1−i2(1) ≡ 1 + i1 − i2 (mod n). So, i1 = i2;hence j1 = j2.) �

Let Xn be a regular n-gon in R2 with vertices labeled by 1, 2, . . . , n (Figure 1.2).A symmetry of X is a rigid motion in R3 which permutes the vertices of Xn. Asymmetry of X can be treated as a permutation of the vertices of X.

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........1

2

3

4

7

6

5

Figure 1.2. Regular heptagon

Fact. Dn = the group of symmetry of Xn.

Proof. Let σ ∈ Sn be a symmetry of Xn with σ(1) = i. Then α−iσ is asymmetry of Xn which fixes 1. Thus α−iσ is either the identity or the reflection β,i.e. α−iσ = βj , j = 0 or 1. So, σ = αiβj ∈ Dn. �

Definition 1.12. A group G 6= {e} is called simple if the only normal sub-groups of G are {e} and G.

Fact. A finite abelian group G is simple ⇔ G ∼= Zp where p is a prime.

Simplicity of An (n ≥ 5). If n ≥ 5, An is simple.

Proof. 1◦ If n ≥ 5, all 3-cycles in An are conjugate in An.

10 1. GROUPS

Let i1, . . . , i5 ∈ {1, . . . , n} be distinct (arbitrary). Let

α =

(1 2 3 4 5 · · ·i1 i2 i3 i4 i5 · · ·

)or

(1 2 3 4 5 · · ·i1 i2 i3 i5 i4 · · ·

)such that α ∈ An. Then α(1, 2, 3)α−1 = (i1, i2, i3).

2◦ Let {id} 6= N C An. To show N = An, it suffices to show that N containsa 3-cycle. (By the normality of N and 1◦, N contains all 3-cycles. By Proposi-tion 1.11, N = An.) Let id 6= σ ∈ N , written as a product of disjoint cycles.

Case 1. σ has a cycle of length r ≥ 4, say σ = (1, . . . , r) · α. Let δ = (1, 2, 3) ∈An. Then N 3 σ−1(δσδ−1) = (r, . . . , 2, 1)(2, 3, 1, 4, . . . , r) = (1, 3, r).

Case 2. σ has two 3-cycles, say, σ = (1, 2, 3)(4, 5, 6) · α. Let δ = (1, 2, 4) ∈ An.Then, N 3 σ−1(δσδ−1) = (3, 2, 1)(6, 5, 4)(2, 4, 3)(1, 5, 6) = (1, 4, 2, 6, 3). We are incase 1.

Case 3. One cycle of σ has length 3; all others have length ≤ 2. Then σ2 is a3-cycle.

Case 4. All cycles of σ have length ≤ 2. Say, σ = (1, 2)(3, 4)α. Let δ =(1, 2, 3) ∈ An. Then N 3 σ(δσδ−1) = (1, 2)(3, 4)(2, 3)(1, 4) = (1, 3)(2, 4) := τ . Let� = (1, 3, 5). Then N 3 τ(�τ�−1) = (1, 3)(2, 4)(3, 5)(2, 4) = (1, 3, 5). �

Note. A1, A2, A3 are simple. A4 is not simple. K = {id, (1, 2)(3, 4), (1, 3)(2, 4),(1, 4)(2, 3)} C A4.

1.4. Group Actions

SX and XS. Let X be a set and σ a permutation of X. When applying σ tox ∈ X, we can write σ to the left or right of x, i.e. σ(x) or (x)σ. In the first notation,the rule of composition is (τσ)(x) = τ(σ(x)); the symmetric group on X is denotedby SX . In the second notation, the rule of composition is (x)(τσ) = ((x)τ)σ; thesymmetric group on X is denoted by XS. (Note. σ 7→ σ (σ 7→ σ−1) is an antiisomorphism (isomorphism) between SX and XS.)

Definition 1.13. Let G be a group and X a set. A left (right) action of Gon X is a homomorphism φ : G → SX (XS). Usually, for g ∈ G, the image φ(g)is still denoted by g. So, for a left (right) action, gx = φ(g)(x) (xg = (x)φ(g)).Equivalently, a left action on G on X is a map

G×X −→ X(g, x) 7−→ gx

such that ex = x and g1(g2x) = (g1g2)x for all x ∈ X and g1, g2 ∈ G. A rightaction of g on X is a map X ×G→ X with similar properties.

Example. Let V be a vector space over F and HomF (V, F ) the dual of V .

GL(V )× V −→ V(f, x) 7−→ f(x)

is a left action on GL(V ) on V .

HomF (V, F )×GL(V ) −→ HomF (V, F )(l, f) 7−→ l ◦ f

is a right action of GL(V ) on HomF (V, F ).

1.4. GROUP ACTIONS 11

Definition 1.14. Let G act on X (left action). For x ∈ X, [x] := {gx : g ∈ G}is the G-orbit of x, Gx := {g ∈ G : gx = x} < G is the stabilizer of x. The G-orbitsform a partition of X. If X has only one G-orbit, we say that G acts transitivelyon X.

Easy facts. Let G act on X.(i) For x ∈ X, [G : Gx] = |[x]|.(ii) For g ∈ G and x ∈ X, Ggx = gGxg−1.

Proof. (i) Definef : G/Gx −→ [x]

gGx 7−→ gx.Then f is a well defined bijection. �

Conjugation and the class equation. G acts on G by conjugation: G×G → G, (a, x) 7→ axa−1. [x] is the conjugacy class of x in G. Gx = {a ∈ G :axa−1 = x} =: CG(x) is the centralizer of x in G. Note that |[x]| = 1 iff x ∈ Z(G).Let {xi : i ∈ I} be a set of representatives of the conjugacy classes of G notcontained in Z(G). Then [x], x ∈ Z(G), and [xi], i ∈ I, are all the conjugacyclasses of G. Hence

(1.1) |G| =∑

x∈Z(G)

|[x]|+∑i∈I|[xi]| = |Z(G)|+

∑i∈I

[G : CG(xi)].

(1.1) is called the class equation of G.

Normalizer. Let G be a group and S(G) the set of all subgroups of G. Gacts on S(G) by conjugation: G × S(G) → S(G), (a,H) 7→ aHa−1. For H < G,GH = {a ∈ G : aHa−1 = H} =: NG(H) is the normalizer of H in G. (NG(H)is the largest subgroup K of G such that H C K.) [G : NG(H)] is the number ofconjugates of H in G.

p-group actions. Let p be a prime. A group G is called a p-group if and onlyif for each a ∈ G, o(a) is a power of p. A finite group G is a p-group ⇔ |G| is apower of p. (This follows from Sylow’s theorem.)

Lemma 1.15. Let G be a finite p-group acting on a finite set X. Let X0 = {x ∈X : gx = x ∀g ∈ G}. Then |X| ≡ |X0| (mod p).

Proof. For x ∈ X, [x] = {x} ⇔ x ∈ X0. Let [x1], . . . , [xn] be all the distinctG-orbits such that {x1, . . . , xk} = X0. Then

|X| =n∑i=1

|[xi]| = k + |[xk+1]|+ · · ·+ |[xn]| ≡ k (mod p).

�

Proposition 1.16 (Facts about finite p-groups). Let p be a prime and G anontrivial finite p-group.

(i) Z(G) 6= {e}.(ii) If H < G with [G : H] = p, then H C G.(iii) If |G| = p2, then G is abelian.

12 1. GROUPS

Table 1.1. Examples of Lemma 1.15

G X action X0 conclusion in particular

GN

{e} 6= N C Gconjugation N ∩ Z(G) |N ∩ Z(G)| ≡ 0 (mod p) |N ∩ Z(G)| > 1

HG/H

H < G

left

multiplicationNG(H)/H

[NG(H) : H] ≡ [G : H](mod p)

if p | [G : H],NG(H) 6= H

Proof. (i) and (ii) are special cases of Table 1.1.(iii) 1◦ Let G be any group. If G/Z(G) is cyclic, G is abelian.2◦ If |G| = p2, |Z(G)| = p or p2. However, |Z(G)| 6= p. (Otherwise, G/Z(G) is

cyclic ⇒ G is abelian ⇒ |Z(G)| = p2.) �

Proposition 1.17. Let G be a group. If there exists H < G such that [G :H] = n 2. If ∃H < G such that[G : H] = n > 1, then G ↪→ An.

Proof. In the proof of Proposition 1.17, N = {e} and G = G/N ↪→ SG/H =Sn. Assume G ⊂ Sn. We claim that G ⊂ An. (If G 6⊂ An, then [G : G ∩ An] = 2.So, G ∩An C G,→←.) �

Theorem 1.18 (Burnside’s lemma). Let G be a finite group acting on a finiteset X. Then the number of G-orbits in X is

n =1|G|

∑g∈G

Fix(g).

where Fix(g) = |{x ∈ X : gx = x}|.

Proof. Let X = {(g, x) : g ∈ G, x ∈ X, gx = x}. Counting (g, x) ∈ X in theorder of g, x, we have |X | =

∑g∈G Fix(g). Counting (g, x) ∈ X in the order of x, g,

we also have |X | =∑x∈X |Gx|. Let [x1], . . . , [xn] be the G-orbits in X. Then∑

x∈X|Gx| =

n∑i=1

∑x∈[xi]

|Gx| =n∑i=1

∑x∈[xi]

|G||[x]|

=n∑i=1

∑x∈[xi]

|G||[xi]|

=n∑i=1

|G| = n|G|.

So, n|G| =∑g∈G Fix(g). Hence the theorem. �

1.5. Sylow’s Theorem

Theorem 1.19 (Sylow). Let G be a finite group with |G| = pem, where p is aprime, e > 0, and p - m.

(i) Every p-subgroup (including {e}) of G is contained in a subgroup of orderpe. (A subgroup of G of order pe is called a Sylow p-subgroup.)

(ii) All Sylow p-subgroups are conjugate in G.(iii) Let np denote the number of Sylow p-subgroups of G. Then np | m and

np ≡ 1 (mod p).

1.5. SYLOW’S THEOREM 13

Proof. 1◦ Let X = {X ⊂ G : |X| = pe}. Then |X | =(pempe

)≡ m 6≡ 0

(mod p). G acts on X : G× X → X , (g,X) 7→ gX. There exists X ∈ X such that|[X]| 6≡ 0 (mod p), i.e. |G|/|GX | 6≡ 0 (mod p). So, pe

∣∣ |GX |. Choose x0 ∈ X.Then |GX | = |GXx0| ≤ |GXX| = |X| = pe. So, P := GX is a Sylow p-subgroup ofG.

2◦ Let P = {gPg−1 : g ∈ G}. P acts on P by conjugation. {P} is the onlyorbit with one element. (If P1 ∈ P such that P1 6= P and {P1} is an orbit, thenPP1 is a p-subgroup of G with |PP1| > pe, which is a contraction.) So, |P| ≡ 1(mod p) by Lemma 1.15. Since |P|

∣∣ |G|, we must have |P| ∣∣ m.3◦ Let P1 be a p-subgroup of G. P1 acts on P by conjugation. Since |P| ≡

1 (mod p), there exists P2 ∈ P such that {P2} is a P1-orbit. Then P1 ⊂ P2.(Otherwise, P1P2 is a p-subgroup of G with |P1P2| > pe.) This also shows that Pis the set of all Sylow p-subgroups of G, i.e. np = |P|. �

Proposition 1.20. Let P be a Sylow p-subgroup of G. Then NG(NG(P )) =NG(P ).

Proof. Let x ∈ NG(NG(P )). Then xPx−1 ⊂ xNG(P )x−1 = NG(P ). Sinceboth xPx−1 and P are Sylow p-subgroups of NG(P ), by Sylow’s theorem, thereexists y ∈ NG(P ) such that xPx−1 = yPy−1 = P . So, x ∈ NG(P ). �

Proposition 1.21. Let H C G, |H| < ∞ and P a Sylow p-subgroup of H.Then G = H NG(P ).

Proof. Let x ∈ G. Then P and xPx−1 are conjugate in H, i.e. xPx−1 =yPy−1 for some y ∈ H. So, y−1x ∈ NG(P ) and x = y(y−1x) ∈ HNG(P ). �

Note. The method in the above two proofs is called the Frattini argument.

Semidirect product. Let N and H be groups and θ : H → Aut(N) ahomomorphism. The semidirect product NoθH is a group where NoθH = N×Has a set and

(n1, h1)(n2, h2) =(n1[θ(h1)(n2)], h1h2

).

N ′ := N × {eH} C N oθ H, H ′ := {eG} × H < N oθ H; N ′ ∼= N , H ′ ∼= H;N oθ H = N ′H ′, N ′ ∩H ′ = {e}; H ∼= (N oθ H)/N ′.

G ∼= N oθ H for some θ ⇔ G = N ′G′ where G B N ′ ∼= N , G > H ′ ∼= H andN ′ ∩H ′ = {e}.

Proposition 1.22. Let |G| = pq where p < q are primes.(i) If p - q − 1, G ∼= Zpq.(ii) If p | q−1, then G ∼= Zpq or G ∼= ZqoθZp, where θ : Zp → Aut(Zq) (∼= Z×q )

is any 1-1 homomorphism.

Proof. (ii) Assume G is nonabelian. Let Q be a Sylow q-subgroup and P aSylow p-subgroup of G. Then Q C G; hence G = QP . Write Q = 〈a〉, P = 〈b〉.Then bab−1 = ak, where k ∈ Z×q has multiplicative order p. Let [θ(1)](1) = l ∈ Zq.Then [θ(1)](x) = lx∀x ∈ Zq. Since θ(1) ∈ Aut(Zq), l ∈ Z×q . Since θ : Zp → Aut(Zq)is 1-1, o(θ(1)) = o(1) (in Zp) = p. So the multiplicative order of l in Z× is p. Weuse the fact that Z×q is cyclic (Proposition 3.40). Since k, l ∈ Z×q are both of orderp, we can write l = ks, p - s. (Cf. Exercise 1.3 (iv).) Let c = bs. Then P = 〈c〉 and

14 1. GROUPS

cac−1 = al. Thenφ : Zq oθ Zp −→ G

(i, j) 7−→ aicj

is an isomorphism. In fact,

φ((i1, j1)(i2, j2)

)= φ(i1 + [θ(j1)](i2), j1 + j2)

= φ(i1 + [θ(1)j1 ](i2), j1 + j2

)= φ(i1 + lj1i2, j1 + j2)

= ai1+lj1 i2cj1+j2

= ai1(cj1ac−j1)i2cj1+j2 (∵ cj1ac−j1 = alj1 )

= ai1cj1ai2c−j1cj1+j2

= ai1cj1ai2cj2

= φ(i1, j1)φ(i2, j2).

�

Example. Let |G| = pqr, where p < q < r are primes. Then the Sylowr-subgroup of G is normal.

Proof. Let Q < G, R < G such that |Q| = q, |R| = r. Assume to thecontrary that R 6C G. Then nr = pq, i.e. NG(R) = R. If Q C G, then R C QR,NG(R) ⊃ QR, which is a contradiction. If Q 6C G, nq = r or pr. Then G haspq(r − 1) elements of order r and at least r(q − 1) elements of order q. Note thatpq(r − 1) + r(q − 1) > pqr, contradiction. �

Example. If G is a simple group of order 60, then G ∼= A5.

Proof. 1◦ Let P be the set of all Sylow 5-subgroups of G. Then |P| = 6. LetG act on P by conjugation. Then G ↪→ SP = S6. Thus G ↪→ A6. (Otherwise,[G : G ∩A6] = 2⇒ G ∩A6 C G.) So, we assume G ⊂ A6.

2◦ We claim that ∃ θ ∈ Aut(A6) such that θ(G) = {σ ∈ A6 : σ(1) = 1} ∼= A5.Define

φ : A6 −→ SA6/Gσ 7−→ φ(σ),

whereφ(σ) : A6/G −→ A6/G

αG 7−→ σαG.Then φ is a homomorphism with kerφ ⊂ G. Hence kerφ = {id} and φ : A6 ↪→SA6/G. Since A6 is simple, φ : A6

∼=→ AA6/G. Write A6/G = {G1, . . . , G6}, whereG1 = G. Let f : {1, . . . , 6} → A6/G, i 7→ Gi. Define

θ : A6 −→ A6σ 7−→ f−1φ(σ)f.

Then θ ∈ Aut(A6). If σ ∈ G, [θ(σ)](1) = [f−1φ(σ)f ](1) = [f−1φ(σ)](G) =f−1(G) = 1. �

1.6. FINITELY GENERATED ABELIAN GROUPS 15

1.6. Finitely Generated Abelian Groups

If A and B are abelian groups, A×B is also denoted by A⊕B.

Theorem 1.23 (Structure of finitely generated abelian groups). Every finitelygenerated abelian group G is isomorphic to

Zpe11 ⊕ · · · ⊕ Zpess ⊕ Zr,

where p1, . . . , ps are primes (not necessarily distinct), ei > 0, r ≥ 0. The num-bers pe11 , . . . , p

ess (the elementary divisors of G) and r (the rank of G) are uniquely

determined by G.

Proof. This is a special case of Theorem 2.57, the structure theorem of finitelygenerated modules over a PID. �

Facts.

(i) The number of nonisomorphic abelian groups of order pe11 · · · pess isP (e1) · · ·P (es), where P (e) is the number of partitions of e.

(ii) Assume A,B,C are finitely generated abelian groups. Then A ⊕ A ∼=B⊕B ⇒ A ∼= B; A⊕C ∼= B⊕C ⇒ A ∼= B. (Use the structure theorem.)These are false if A,B,C are not finitely generated. Let A = Zℵ0 andB = Z. Then A⊕A ∼= B⊕A but A 6∼= B. Corner’s theorem (next) showsthat there is an abelian group A such that A⊕A⊕A ∼= A but A⊕A 6∼= A.Let B = A⊕A. Then B ⊕B ∼= A⊕A but B 6∼= A.

Advanced Reading

Theorem 1.24 (Corner [4]). Let r ∈ Z+. There exists a countable reducedtorsion-free abelian group G such that for m,n ∈ Z+, Gm ∼= Gn ⇔ m ≡ n (mod r).(An abelian group is called reduced if its only divisible subgroup is {0}.)

Lemma 1.25 (Corner [3]). Let A be a ring such that (A,+) is countable, re-duced, and torsion-free. Then A ∼= End(G) for some countable, reduced, torsion-freeabelian group G.

Lemma 1.26. Let G be an abelian group and ω1, ω2 two idempotents in End(G).Then ω1G ∼= ω2G⇔ there exist x, y ∈ End(G) such that ω1 = xy and ω2 = yx.

Proof. (⇐) Let y∗ : ω1G → ω2G, ω1a 7→ yω1a and x∗ : ω2G → ω1G, ω2a 7→xω2a. Then x∗ and y∗ are inverses of each other.

(⇒) Let α : ω1G → ω2G and β : ω2G → ω1G be inverse isomorphisms. Lety = αω1 and x = βω2. �

Proof of Theorem 1.24. Only have to consider r > 1. Let A be the ringfreely generated by xi, yi, 0 ≤ i ≤ r, subject to the relation

yixj =

{1 if i = j,0 if i 6= j,

r∑i=0

xiyi = 1.

1◦ (A,+) is free abelian of countable rank.We have A ∼= B/C, where B is the ring freely generated by xi, yi, 0 ≤ i ≤ r,

subject to the relations

yixj =

{1 if i = j,0 if i 6= j

16 1. GROUPS

and C is the ideal of B generated by 1−∑ri=0 xiyi.

We claim that B is free abelian with a basis

(1.2) xi1 · · ·ximyjn · · · yj1 , n,m ≥ 0, 0 ≤ ik, jk ≤ r.

Let A be the set of elements in (1.2), B the set of products of elements of{x1, . . . , xr, y1, . . . , yr} containing at least one string yixj (i 6= j) but no stringyixi, and C the set of products of elements in {x1, . . . , xr, y1, . . . , yr} ∪ {yixi − 1 :1 ≤ i ≤ r} containing at least one yixi − 1 but no string yixi. Then A∪ B ∪ C is aZ-basis of the free ring generated by {x1, . . . , xr, y1, . . . , yr}. To see this, it sufficesto show that A ∪ B ∪ C is linearly independent over Z. Assume

(1.3) α1a1 + · · ·+ αsas + β1b1 + · · ·+ βtbt + γ1c1 + · · ·+ γucu = 0,

where ai ∈ A, bj ∈ B, ck ∈ C, αi, βj , γk ∈ Z. Assume that c1 has the largest numberof elements in {x1, . . . , xr, y1, . . . , yr} (counted with multiplicity). In (1.3), expandeach ck as a product of elements in {x1, . . . , xr, y1, . . . , yr}. Let c′1 be obtained fromc1 by replacing each yixi − 1 with yixi. Then in the LHS of (1.3), γ1c′1 is the onlyterm of involving c′1. Therefore, γ1 = 0. In the same way, γ1 = · · · = γu = 0. Itfollows immediately that α1 = · · · = αs = β1 = · · · = βt = 0.

Since B is the free ring on {x1, . . . , xr, y1, . . . , yr} modulo the ideal generatedby B ∪ C, B is free abelian with a basis A.

C is generated by

xi1 · · ·xim(1−

r∑i=0

xiyi

)yjn · · · yj1 , n,m ≥ 0, 0 ≤ ik, jk ≤ r.

It can be seen that B = C ⊕D, where

D := 〈xi1 · · ·ximyjn · · · yj1 , n,m ≥ 0, 0 ≤ ik, jk ≤ r, (im, jn) 6= (r, r)〉.

Thus B/C ∼= D.2◦ By Lemma 1.25, A = End(G) for some countable, reduced, torsion-free

abelian group G. Let ωi = xiyi, 0 ≤ i ≤ r. Then 1 = ω0+· · ·+ωr is a decompositionof 1 into orthogonal idempotents. So,

G =r∑i=0

ωiG.

Since ωi = xiyi and 1 = yixi, by Lemma 1.26, G ∼= ωiG. Thus G ∼= Gr+1; henceGm ∼= Gn if m ≡ n (mod r).

3◦ There exists Z-linear map T : A→ Zr such that(i) T (xy) = T (yx), x, y ∈ A;(ii) T (ωi) = 1, 0 ≤ i ≤ r.

We only have to construct a Z-linear map T ∗ : B → Zr satisfying the analogiesof (i) and (ii) and T ∗|C = 0. Define

T ∗(xi1 · · ·ximyjn · · · yj1) =

{1 if m = n and ik = jk, 1 ≤ k ≤ m,0 otherwise

and extend T ∗ to B by linearity.4◦ Gm 6∼= Gn if m 6≡ n (mod r).

1.7. FREE GROUPS 17

Assume Gm ∼= Gn for some 1 ≤ m,n ≤ r. Then(m−1∑i=0

ωi

)G =

m−1∑i=0

ωiG ∼=n−1∑i=0

ωiG =(n−1∑i=0

ωi

)G.

By Lemma 1.26, there exist x, y ∈ A such that∑m−1i=0 ωi = xy,

∑n−1i=0 ωi = yx.

Then in Zr,

m = T(m−1∑i=0

ωi

)= T (xy) = T (yx) = T

(n−1∑i=0

ωi

)= n.

�

End Advanced Reading

1.7. Free Groups

FX , the free group on X. Let X be a set and

W ={xe11 · · ·xess : s ≥ 0, xi ∈ X, ei ∈ {±1}

}.

(Here, xe11 · · ·xess is a formal product and is called a word in X.) For u, v ∈ W, sayu ∼ v if v can be obtained form u by deleting and inserting strings of the form xx−1or x−1x, x ∈ X. ∼ is an equivalence relation on W; the equivalence class of u ∈ Wis denoted by [u]. Put FX = W/∼ and for [u], [v] ∈ W/∼, define [u][v] = [uv],where uv is the concatenation of u and v. This gives a well defined operation onFX ; FX with this operation is a group, called the free group on X; rankFX := |X|.

Let G be any group and f : X → G. Then there is a unique homomorphismf̄ : FX → G such that

.................................................................................................................. ...........

........

........

........

........

........

........

........

............

...........

................................................................................................. ...........

f

ιf̄

F

GX

commutes, where ι(x) = [x], x ∈ X.

Every group is a homomorphic image of a free group.A word w ∈ W is called reduced if it contains no strings xx−1 or x−1x, x ∈ X.

Fact. Every class in W/∼ contains a unique reduced word.

Proof. (Uniqueness) Let R be the set of all reduced words in W. Define

f : X −→ SRx 7−→ f(x)

wheref(x) : R −→ R

xe11 · · ·xess 7−→

{xxe11 · · ·xess if x

e11 6= x−1,

xe22 · · ·xess if xe11 = x

−1.

18 1. GROUPS

Note that for x ∈ X, we have

f(x)−1 : R −→ R

xe11 · · ·xess 7−→

{x−1xe11 · · ·xess if x

e11 6= x,

xe22 · · ·xess if xe11 = x.

There exists homomorphism f̄ : FX → SR such that f̄([x]) = f(x) for all x ∈X. Assume xe11 · · ·xess , y

f11 · · · y

ftt ∈ R such that [x

e11 · · ·xess ] = [y

f11 · · · y

ftt ]. Then

xe11 · · ·xess =(f(x1)e1 · · · f(xs)es

)(1) = f̄([xe11 · · ·xess ])(1) = f̄([y

f11 · · · y

ftt ])(1) =(

f(y1)f1 · · · f(yt)ft)(1) = yf11 · · · y

ftt . �

So, ι : X → FX is an injection and X can be regarded as a subset of FX .

Presentation. Let X be a set and R a set of words in X. Define

〈X | R〉 = FX/N,

where N is the normal closure of {[r] : r ∈ R} in FX , i.e., the smallest subgroup ofFX containing {[r] : r ∈ R}. If G ∼= 〈X | R〉, 〈X | R〉 is called a presentation of G;elements in X are called generators; elements in R are called relations.

Theorem 1.27 (Van Dyck). Let G be a group generated by X ⊂ G. Let R bea set of words in X such that for every r ∈ R, r = 1 in G, i.e., the relation r = 1is satisfied in G. Then ∃! onto homomorphism 〈X | R〉 → G such that [x] 7→ x forall x ∈ X.

Proof. Let N be the normal closure of {[r] : r ∈ R} in FX . The map f :X → G, x 7→ x, induces a homomorphism f̄ : FX → G such that [x] 7→ x. SinceG = 〈X〉, f̄ is onto. By assumption, {[r] : r ∈ R} ⊂ ker f̄ . So N ⊂ ker f̄ . Thus, f̄induces an onto homomorphism from FX/N = 〈X | R〉 to G.

The uniqueness is obvious since 〈X | R〉 is generated by {[x]N : x ∈ X}. �

Abuse of notation. In a presentation 〈X | R〉 = FX/N , an element [u]N ,where u is word in X, is usually denoted by u.

Examples.

• Dn = 〈α, β | αn = β2 = 1, βαβ−1 = α−1〉.• The infinite dihedral group

D∞ = 〈α, β | β2 = 1, βαβ−1 = α−1〉.

• The quaternion group

Q8 = 〈x, y | x4 = 1, y2 = x2, yxy−1 = x−1〉.

Proof. Q8 = 〈i, j〉, where i, j satisfies relations i4 = 1, j2 = i2, jij−1 =i−1. By Van Dyck’s theorem, ∃ onto homomorphism

(1.4) G := 〈x, y | x4 = 1, y2 = x2, yxy−1 = x−1〉 −→ Q8

such that x 7→ i, y 7→ j. Every element in G can be written as xiyj ,0 ≤ i ≤ 3, 0 ≤ j ≤ 1. So, |G| ≤ 8. Hence the homomorphism in (1.4) isan isomorphiam. �

1.7. FREE GROUPS 19

• The generalized quaternion groupQ4n = 〈x, y | x2n = 1, y2 = xn, yxy−1 = x−1〉.

Let ξ = e2πi/2n, A =[ ξ

ξ−1

], B =

[ −11

]. Then ∃ an isomorphism

Q4n∼=−→ 〈A,B〉 ⊂ GL(2,C) such that x 7→ A, y 7→ B. |Q4n| = 4n.

The braid group on n strings.

Bn = 〈σ1, . . . , σn−1 | σiσi+1σi = σi+1σiσi+1, 1 ≤ i ≤ n−2; σiσj = σjσi, |i−j| > 1〉.

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..1 2 i i+1 n

σi

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..

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..1 2 i i+1 n

σ−1i

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σ1

σ2

σ1

σ2

σ1

σ2........................................................................................................................................................................................

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............. ............. ............. ............. ............. ............. ............. ............. ............. .............

............. ............. ............. ............. ............. ............. ............. ............. ............. .............

σ1σ2σ1 σ2σ1σ2=

=

Figure 1.3. The braid group

Fact. Let X = A�∪ B, H the subgroup of FX generated by A and N the

normal closure of B in FX . Then FX = HN and H ∩ N = {1}. In particular,F/N ∼= H is free on A.

Proof. Only have to show H ∩N = {1}. Define

φ : X −→ FA

x 7−→

{x if x ∈ A,1 if x ∈ B.

φ induces a homomorphism φ̄ : FX → FA such that N ⊂ ker φ̄ and φ̄|H : H → FAis an isomorphism. If w ∈ H ∩N , then φ̄|H(w) = φ̄(w) = 1, hence w = 1. �

Fact. FX1 ∼= FX2 ⇔ |X1| = |X2|.

Proof. (⇒) FXi/(FXi)′ is free abelian of rank |Xi|. �

Theorem 1.28 (The Nielson-Schreier theorem). Let F be a free group on X andG < F . Then G is a free group. If [F : G] = m

20 1. GROUPS

Let F̃ be the free group on X× (G\\F ). φ := α|X×(G\\F ) induces a homomorphismφ̄ : F̃ → G.

2◦ We construct a map ψ : F × (G\\F ) → F̃ such that the following diagramcommutes.

........

........

........

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................................................................................................. ...........α

φ̄ψ

φ=α|X×(G\\F )

G

F̃

X × (G\\F )

F × (G\\F )

Define

ψ : F × (G\\F ) −→ F̃(u,C) 7−→ uC

where uC is defined inductively on the length of u:

uC =

1 if u = 1,(x,C) if u = x ∈ X,(x,Cx−1)−1 if u = x−1, x ∈ X,vCyCv if u = vy is reduced and y ∈ X ∪X−1.

Then for all u, v ∈ F and C ∈ G\\F , (uv)C = uCvCu, (u−1)C = (uCu−1)−1 and

φ̄(uC) = τ(C)u τ(Cu)−1.

3◦ Letβ : G −→ F̃

u 7−→ uG

β is a homomorphism and φ̄ ◦ β = idG. So, φ̄ is onto, β is 1-1, and

G ∼= F̃ / ker φ̄.

4◦ Let N be the normal closure of {τ(C)G : C ∈ G\\F} in F̃ . We claim thatker φ̄ = N .

φ̄(τ(C)G) = τ(G)τ(C)τ(C)−1 = 1. So, N ⊂ ker φ̄. To show that ker φ̄ ⊂N , note that ker φ̄ = ker γ, where γ = β ◦ φ̄. It suffices to show that γ(xC) ≡xC (mod N) for (x,C) ∈ X × (G\\F ). We have γ(xC) =

[τ(C)x τ(Cx)−1

]G =τ(C)GxC

(τ(Cx)−1

)Gτ(C)x = τ(C)GxC(τ(Cx)G)−1 ≡ xC (mod N).5◦ There exists a (right) transversal τ of G in F (τ(G) = 1) such that if a

reduced word uy ∈ im(τ), y ∈ X ∪X−1, then u ∈ im(τ). im(τ) is said to have theSchreier property and τ is called a Schreier transversal. (For each C ∈ G\\F , thelength of C, denoted by |C|, is defined to be the minimum length of words in C.Define τ(C) inductively on |C|. τ(G) = 1. Assume that τ(D) have been defined forall D ∈ G\\F with |D| < |C|. Let uy ∈ C (y ∈ X ∪X−1) be of minimum length.Define τ(C) = τ(Gu)y.)

1.8. NONABELIAN GROUPS OF ORDER ≤ 30 21

6◦ Let τ be a Schreier transversal of G in F . For each G 6= C ∈ G\\F ,τ(C) = uxe (reduced), x ∈ X, e ∈ {±1}. So, τ(C)G = uG(xe)Gu, where

(xe)Gu =

{xGu if e = 1(xGux

−1)−1 if e = −1

}= (xD)e for some D ∈ G\\F.

Since im(τ) has the Schreier property, u ∈ im(τ), so uG ∈ N , hence xD ∈ N .Applying the same argument to u and use induction on the length of u, we see thatτ(C)G is a product of certain elements in N of the form xD (x ∈ X, D ∈ G\\F )and their inverses. So N is the normal closure of N ∩

(X × (G\\F )

). Then

G ∼= F̃ /N (by 3◦, 4◦)∼= the free group on X × (G\\F )

)\N.

7◦ Assume [F : G] = m

22 1. GROUPS

Table 1.2. Nonabelain groups of order ≤ 30

|G| G #6 S3 18 D4, Q8 210 D5 112 A4, D6, Z3 oα Z4 314 D7 1

16D8, Q16, D4 × Z2, Q8 × Z2, Z8 oβ1 Z2, Z8 oβ2 Z2,

(Z4 × Z2) oβ3 Z2, (Z4 × Z2) oβ4 Z2, Z4 oβ5 Z49

18 D9, S3 × Z3, (Z3 × Z3) o Z2 320 D10, Z5 oγ1 Z4, Z5 oγ2 Z4 321 Z7 o Z3 122 D11 1

24D4 × Z3, Q8 × Z3, Z3 oδ1 Z8, Z3 oδ2 (Z4 × Z2),

Z3 oδ3 (Z4 × Z2), Z2 ×D6, Z3 oδ4 (Z2 × Z2 × Z2),Q8 oδ5 Z3, Z3 oδ6 Q8, Z3 oδ7 D4, Z3 oδ8 D4, S4

12

26 D13 127 Z9 o Z3, (Z3 × Z3) o Z3 328 D14, Z7 o Z4 230 D15, D5 × Z3, S3 × Z5 3

Order 12. G ∼= A4 or D6 or Z3 oα Z4, where α : Z4 → Aut(Z3), [α(1)](1) =−1.

Proof. Let P be a Sylow 3-subgroup of G and Q a Sylow 2-subgroup of G.G acts on G/P by left multiplication which gives a homomorphism f : G→ SG/Pwith ker f ⊂ P . If ker f = {e}, then G ↪→ S4, so G ∼= A4. If ker f = P , then P C Ghence G = P oQ. In this case it is easy to show that G ∼= D6 if Q ∼= Z2 × Z2 andG ∼= Z3 oα Z4 if Q ∼= Z4. �

We have

〈x, y | x3 = y4 = 1, yxy−1 = x−1〉∼=−→ Z3 oα Z4 (x 7→ (1, 0), y 7→ (0, 1)).

Let a = ((1, 2, 3), 2), b = ((1, 2), 1) ∈ S3 × Z4 and T = 〈a, b〉 < S3 × Z4. Then|T | = 12 and

T = 〈a, b | a6 = 1, a3 = b2, bab−1 = a−1〉.Also

〈x, y | x3 = y4 = 1, yxy−1 = x−1〉∼=−→ T (x 7→ a2, y 7→ b).

Order 16. Table 1.3. (Cf. [2, §118].)

Order 20. G ∼= D10 or Z5 oγ1 Z4 or Z5 oγ2 Z4, where γ1, γ2 : Z4 → Aut(Z5),[γ1(1)](1) = 2, [γ2(1)](1) = −1.

1.8. NONABELIAN GROUPS OF ORDER ≤ 30 23

Table 1.3. Nonabelian groups of order 16

group presentation

D8 〈x, y | x8 = y2 = 1, yxy−1 = x−1〉Q16 〈x, y | x8 = 1, y2 = x4, yxy−1 = x−1〉

D4 × Z2〈x, y, z | x4 = y2 = z2 = 1, y−1xy = x−1,

xz = zx, yz = zy〉

Q8 × Z2〈x, y, z | x4 = z2 = 1, y2 = x2,y−1xy = x−1, xz = zx, yz = zy〉

Z8 oβ1 Z2,β1 : Z2 → Aut(Z8), [β1(1)](1) = 5.

〈x, y | x8 = y2 = 1, y−1xy = x5〉

Z8 oβ2 Z2β1 : Z2 → Aut(Z8), [β1(1)](1) = −1.

〈x, y | x8 = y2 = 1, y−1xy = x−1〉

(Z4 × Z2) oβ3 Z2,β3 : Z2 → Aut(Z4 × Z2),

β3(1) :

{(1, 0) 7→ (1, 0),(0, 1) 7→ (2, 1).

〈x, y, z | x4 = y2 = z2 = 1, xy = yx,z−1xz = x, z−1yz = x2y〉

(Z4 × Z2) oβ4 Z2,β4 : Z2 → Aut(Z4 × Z2),

β4(1) :

{(1, 0) 7→ (1, 1),(0, 1) 7→ (0, 1).

〈x, y, z | x4 = y2 = z2 = 1, xy = yx,z−1xz = xy, z−1yz = y〉

Z4 oβ5 Z4β5 : Z4 → Aut(Z4), [β1(1)](1) = −1.

〈x, y | x4 = y4 = 1, y−1xy = x−1〉

Proof. Let 〈a〉 C G be the Sylow 5-subgroup and P a Sylow 2-subgroup. IfP = 〈b〉〈c〉, where o(b) = o(c) = 2, we may assume bab−1 = a. (If bab−1 = a−1and cac−1 = a−1, then (bc)a(bc)−1 = a.) Then G = 〈ab〉〈c〉, where Z10 ∼= 〈ab〉 CG. Thus G ∼= Z10 o Z2 ∼= D10. If P = 〈b〉 ∼= Z4, then G ∼= Z5 oγ Z4, whereγ : Z4 → Aut(Z5) is a homomorphism such that [γ(1)](1) 6= 1. If [γ(1)](1) = 2 or3, G ∼= Z5 oγ1 Z4. If [γ(1)](1) = −1, G ∼= Z5 oγ2 Z4.

Z5 oγ1 Z4 6∼= Z5 oγ2 Z4 since Z(Z5 oγ1 Z4) = {(0, 0)} but Z(Z5 oγ2 Z4) =〈(0, 2)〉 �

Order 24. Table 1.4. (Cf. [2, §126].)

Order 27. Z9 o�1 Z3, �1 : Z3 → Aut(Z9), [�1(1)](1) = 4; (Z3 × Z3) o�2 Z3,

�2 : Z3 → Aut(Z3 × Z3), �2(1) :

{(1, 0) 7→ (1, 0),(0, 1) 7→ (1, 1).

Order 28. G ∼= D14 or Z7 o Z4.

24 1. GROUPS

Table 1.4. Nonabelian groups of order 24

group presentation

D4 × Z3Q8 × Z3

Z3 oδ1 Z8,δ1 : Z8 → Aut(Z3), [δ1(1)](1) = −1

〈a, b | a8 = b3 = 1, a−1ba = b−1〉

Z3 oδ2 (Z4 × Z2),δ2 : Z4 × Z2 → Aut(Z3),

[δ2(1, 0)](1) = 1, [δ2(0, 1)](1) = −1

〈a, b, c | a3 = b4 = c2 = 1, bc = cb,b−1ab = a, cac = a−1〉

Z3 oδ3 (Z4 × Z2),δ3 : Z4 × Z2 → Aut(Z3),

[δ3(1, 0)](1) = −1, [δ3(0, 1)](1) = 1

〈a, b, c | a3 = b4 = c2 = 1, bc = cb,b−1ab = a−1, cac = a〉

Z2 ×D6Z3 oδ4 (Z2 × Z2 × Z2)

δ4 : Z2 × Z2 × Z2 → Aut(Z3),[δ4(1, 0, 0)](1) = −1,[δ4(0, 1, 0)](1) = 1,[δ4(0, 0, 1)](1) = 1

〈a, b, c, d | a3 = b2 = c2 = d2 = 1,bc = cb, cd = dc, db = bd,

bab = a−1, cac = a, dad = a〉

Q8 oδ5 Z3,δ5 : Z3 → Aut(Q8),

δ5(1) :

{i 7→ jj 7→ k

〈a, b, c | a4 = c3 = 1, a2 = b2,b−1ab = a−1, c−1ac = b, c−1bc = ab〉

Z3 oδ6 Q8,δ6 : Q8 → Aut(Z3),

[δ6(i)](1) = 1, [δ6(j)](1) = −1

〈a, b, c | a3 = b4 = 1, b2 = c2,c−1bc = b−1, b−1ab = a, c−1ac = a−1〉

Z3 oδ7 D4,δ7 : D4 → Aut(Z3),

[δ7(α)](1) = 1, [δ7(β)](1) = −1

〈a, b, c | a3 = b4 = c2 = 1, cbc = b−1,b−1ab = a, cac = a−1〉

Z3 oδ8 D4,δ8 : D4 → Aut(Z3),

[δ8(α)](1) = −1, [δ8(β)](1) = 1

〈a, b, c | a3 = b4 = c2 = 1, cbc = b−1,b−1ab = a−1, cac = a〉

S4 〈a, b | a4 = b3 = (ab)2 = 1〉

Proof. Let P be the Sylow 7-subgroup of G and Q a Sylow 2-subgroup ofG. Then P ∼= Z7 and G ∼= P o Q. If Q ∼= Z4, G ∼= Z7 o Z4. If Q ∼= Z2 × Z2,G ∼= D14. �

Order 30. G ∼= D15 or D5 × Z3 or S3 × Z5.

Proof. Let P be the Sylow 5-subgroup of G and Q a Sylow 3-subgroup ofG. Since P C G, PQ < G and PQ ∼= Z5 × Z3. So, G ∼= (Z5 × Z3) oθ Z2, where

1.9. GROUP EXTENSIONS 25

θ : Z2 → Aut(Z5 × Z3),

θ(1) :

{(1, 0) 7→ (−1, 0),(0, 1) 7→ (0, 1),

or

{(1, 0) 7→ (1, 0),(0, 1) 7→ (0,−1),

or

{(1, 0) 7→ (−1, 0),(0, 1) 7→ (0, 1).

In the three cases, G ∼= D5 × Z3, S3 × Z5 and D15 respectively. �

Advanced Reading

1.9. Group Extensions

If 1 → K i→ G p→ Q → 1 is an exact sequence of groups, G is called anextension of K by Q. If there is a homomorphism j : Q→ G such that p ◦ j = idQ,the exact sequence (or the extension) is called split. Given K,G,Q there exists split

sequence 1→ K i→ Gp

�jQ→ 1 iff G ∼= K oQ.

Q-modules induced by extensions. Let 0 → K i→ G p→ Q → 1 be anextension, where K is abelian and G is written additively. There exists a homo-morphism φ : Q→ Aut(K). Let j : Q→ G be any lifting. Then for x ∈ Q,

φ(x) : K −→ Ka 7−→ i−1

(j(x) + i(a)− j(x)

).

φ is independent of j. φ defines an action of Q on K; hence K is a (left) Q-module.To recap, an extension 0→ K i→ G p→ Q→ 1 with abelian K induces a Q-moduleK.

Given a Q-module K, an extension 0→ K i→ G p→ Q→ 1 is said to realize theQ-module K if the Q-module K coincides with the one induced by the extension.

Let K be a Q-module and f : Q×Q→ K a function. Define a binary operation+ on K ×Q:

(a, x) + (b, y) =(a+ xb+ f(x, y), xy

).

The (K ×Q,+) is a group ⇔(1.5) xf(y, z)− f(xy, z) + f(x, yz)− f(x, y) = 0 ∀x, y, z ∈ Q.f is called a 2-cocycle if (1.5) holds. (K ×Q,+) is denoted by G(K,Q, f) if f is a2-cocycle.

Note.(i) (1.5) ⇔ the associativity of +.(ii) (1.5) implies that f(1, y) = f(1, 1), f(x, 1) = xf(1, 1) ∀x, y ∈ Q. (1.5) also

implies the existence of identity and the inverse inG(K,Q, f). (0G(K,Q,f) =(−f(1, 1), 1), −(a, x) = (−x−1a− f(1, 1)− f(x−1, x), x−1).)

(iii) If f is a 2-cocycle, then there exists an exact sequence

(1.6) 0 −→ K i−→ G(K,Q, f) p−→ Q −→ 1where

(1.7)i : K −→ G(K,Q, f)

a 7−→ (a− f(1, 1), 1),

(1.8)p : G(K,Q, f) −→ Q

(a, x) 7−→ x.

26 1. GROUPS

Moreover, extension (1.6) realizes the Q-module K.

Let 0 → K i→ G p→ Q → 1 be an extension making K a Q-module. Letj : Q→ G be a lifting. Then(1.9) j∗(x, y) := i−1

(j(x) + j(y)− j(xy)

)is a 2-cocycle. Moreover,

φ : G(K,Q, j∗) −→ G(a, x) 7−→ i(a) + j(x)

is an isomorphism and the diagram

0 −→ K i′

−→ G(K,Q, j∗) p′

−→ Q −→ 1yid yφ yid0 −→ K i−→ G p−→ Q −→ 1

commutes, where i′ and p′ are defined by (1.7) and (1.8). If j′ : Q→ G is anotherlifting. Then j′(x)− j(x) = h(x) for some function h : Q→ K. Moreover,

j′∗(x, y)− j∗(x, y) = xh(y)− h(xy) + h(x).

The second cohomology group. Let K be a Q-module. Z2(Q,K) :=the abelian group of all 2-cocycles f : Q × Q → K. A function f : Q × Q → Kis called a 2-coboundary if f(x, y) = xh(y) − h(xy) + h(x) for some h : Q →K. B2(Q,K) := the set of all 2-coboundaries. Then B2(Q,K) < Z2(Q,K).H2(Q,K) := Z2(Q,K)/B2(Q,K) is the 2nd cohomology group of Q with coeffi-cients in K.

Equivalence of extensions. Let K be a Q-module and let

0 −→ K i−→ G p−→ Q −→ 1(1.10)

0 −→ K i′

−→ G′ p′

−→ Q −→ 1(1.11)be two extensions realizing the Q-module K. The two extensions are called equiv-alent if there exists an isomorphism φ : G → G′ such that the following diagramcommutes.

0 −→ K i−→ G p−→ Q −→ 1yid yφ yid0 −→ K i

′

−→ G′ p′

−→ Q −→ 1

Theorem 1.29 (The meaning ofH2(Q,K)). Let K be a Q-module. Let E(Q,K)be the family of all equivalence classes of extensions of K by Q realizing the Q-module K. There is a bijection α : H2(Q,K)→ E(Q,K); α(0) = the class of splitextensions.

Proof. α : f + B2(Q,K) 7→ [G(K,Q, f)], f ∈ Z2(Q,K); α−1 : [G] 7→ j∗ +B2(Q,K), where j : Q→ G is a lifting and j∗ ∈ Z2(Q,K) is given by (1.9). �

Corollary 1.30. Let K be a Q-module. If H2(Q,K) = 0, then every exten-sion of K by Q realizing the Q-module K is a semidirect product K oQ.

Proposition 1.31. Let K be a Q-module where |K| = m, |Q| = n and (m,n) =1. Then H2(Q,K) = 0.

1.10. NORMAL AND SUBNORMAL SERIES 27

Proof. Let f ∈ Z2(Q,K). Then

xf(y, z)− f(xy, z) + f(x, yz)− f(x, y) = 0 ∀x, y, z ∈ Q.

Sum over z ∈ Q⇒xh(y)− h(xy) + h(x) = nf(x, y),

where h(x) =∑z∈Q f(x, z). So, nf ∈ B2(Q,K). Since (m,n) = 1, f ∈ B2(Q,K).

�

Complement. Two subgroups H,K of G are called complements of each otherif G = HK and H ∩K = {0}.

Corollary 1.32. If |G| = mn, (m,n) = 1 and G has an abelian normalsubgroup K of order m. Then K has a complement.

Theorem 1.33 (The Schur-Zassenhaus theorem). If |G| = mn, (m,n) = 1 andG has a normal subgroup K of order m, then K has a complement.

Proof. Induction on m. If K has a nontrivial subgroup N such that N C G,K/N has a complement H/N in G/N . Since N C H, |H| = |N |n, (|N |, n) = 1 and|N | < m, H has a subgroup of order n.

So, we may assume that K is a minimal normal subgroup of G. Let P 6= 1be a Sylow subgroup of K. Since G = KNG(P ) (Frattini argument), G/K =KNG(P )/K ∼= NG(P )/K ∩ NG(P ) = NG(P )/NK(P ). If NG(P ) 6= G, then|NK(P )| < |K| = m. The induction hypothesis applied to NK(P ) C NG(P )implies that NG(P ) has a subgroup of order n.

So, assume NG(P ) = G, i.e. P C G. Then 1 6= Z(P ) C G. Since Z(P ) ⊂ K,the minimality of K ⇒ K = Z(P ), which is abelian. Corollary 1.32 applies. �

End Advanced Reading

1.10. Normal and Subnormal Series

Definition 1.34. Let G be a group.

(1.12) G = G0 B G1 B · · · B Gm = {e}

is called a subnormal series of G. Gi/Gi+1, 0 ≤ i ≤ m − 1, are the factor groupsof the series. The length of (1.12) is |{0 ≤ i ≤ m − 1 : Gi 6= Gi+1}|. If Gi C G,1 ≤ i ≤ m, (1.12) is called a normal series. IfGi/Gi+1 is simple for all 0 ≤ i ≤ m−1,(1.12) is called a composition series. If Gi/Gi+1 is abelian for all 0 ≤ i ≤ m − 1,(1.12) is called a solvable series.

A (sub)normal series

(1.13) G = H0 B H1 B · · · B Hn = {e}

is called a refinement of (1.12) if {G0, . . . , Gm} ⊂ {H0, . . . ,Hn}. Two subnormalseries S : G = G0 B · · · B Gm = {e} and T : G = H0 B · · · B Hn = {e} are calledequivalent if there is a bijection between the nontrivial factors of S and those of Tsuch that the corresponding factors are isomorphic.

Theorem 1.35 (The Schreier refinement theorem). Any two (sub)normal seriesof G have equivalent refinements.

28 1. GROUPS

Lemma 1.36 (Zassenhaus). Let A∗, A,B∗, B be subgroups of G such that A∗ CA and B∗ C B. Then A∗(A ∩B∗) C A∗(A ∩B), B∗(A∗ ∩B) C B∗(A ∩B) and

A∗(A ∩B)A∗(A ∩B∗)

∼=B∗(A ∩B)B∗(A∗ ∩B)

.

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A

A∗(A ∩B)

A∗(A ∩B∗)

A∗

A∗ ∩B

B

B∗(A ∩B)

B∗(A∗ ∩B)

B∗

A ∩B∗

A ∩B

(A ∩B∗)(A∗ ∩B)

Proof. Let D = (A∗ ∩ B)(A ∩ B∗). We claim that D C A ∩ B. SinceA∗ C A, we have A∗ ∩ B C A ∩ B; since B∗ C B, we have A ∩ B∗ C A ∩ B. So,D = (A∗ ∩B)(A ∩B∗) C A ∩B.

It suffices to showA∗(A ∩B)A∗(A ∩B∗)

∼=A ∩BD

.

(By symmetry, we have B∗(A∩B)

B∗(A∗∩B)∼= A∩BD .) Define

φ : A∗(A ∩B) −→ A ∩BD

ax 7−→ Dx, a ∈ A∗, x ∈ A ∩B.

1◦ φ is well defined. If ax = a′x′, where a, a′ ∈ A∗ and x, x′ ∈ A ∩ B, thenx′x−1 = (a′)−1a ∈ A∗ ∩ (A ∩B) = A∗ ∩B ⊂ D. So, Dx = Dx′.

2◦ φ is onto. Obvious.3◦ φ is a homomorphism. ∀ ax, a′x′ ∈ A∗(A ∩ B), we have φ(axa′x′) =

φ(axa′x−1xx′) = Dxx′ = φ(ax)φ(a′x′) since xa′x−1 ∈ A∗.4◦ kerφ = A∗D = A∗(A∗ ∩B)(A ∩B∗) = A∗(A ∩B∗).By the 1st isomorphism theorem, A

∗(A∩B)A∗(A∩B∗)

∼= A∩BD . �

Proof of Theorem 1.35. Let S : G = G0 B · · · B Gm = {e} and T : G =H0 B · · · B Hn = {e} be two (sub)normal series. Let

Gij = Gi+1(Gi ∩Hj), Hij = Hj+1(Gi ∩Hj), 0 ≤ i ≤ m, 0 ≤ j ≤ n.

1.10. NORMAL AND SUBNORMAL SERIES 29

(Gm+1 := {e}, Hn+1 := {e}.) Then

Gi−1,n = Gi,0 = Gi, 1 ≤ i ≤ m− 1,Hm,j−1 = H0,j = Hj , 1 ≤ j ≤ n− 1,Gm−1,n−1 = Hm−1,n−1 = Gm−1 ∩Hn−1.

We have a refinement S ′ for S

S ′ :

G = G00 B G01 B · · · B G0,n−1 BG10 B G11 B · · · B G1,n−1 B

......

...Gm−1,0 B Gm−1,1 B · · · B Gm−1,n−1 B {e}

and a refinement T ′ for T

T ′ :

G

qH00 H01 · · · H0,n−15 5 5H10 H11 · · · H1,n−15 5 5...

......

5 5 5Hm−1,0 Hm−1,1 · · · Hm−1,n−15 5 5

{e}

We claim that S ′ and T ′ are equivalent. It suffices to showGij/Gi,j+1 ∼= Hij/Hi+1,j ,0 ≤ i ≤ m− 1, 0 ≤ j ≤ n− 1. By the Zassenhaus lemma,

Gij/Gi,j+1 =Gi+1(Gi ∩Hj)Gi+1(Gi ∩Hj+1)

∼=Hj+1(Gi ∩Hj)Hj+1(Gi+1 ∩Hj)

= Hij/Hi+1,j .

�

Fact. Every finite group has a composition series.

Theorem 1.37 (Jordan-Hölder). Any two composition series of a group G areequivalent.

Proof. Let S and T be two composition series of G. By Theorem 1.35, S andT have refinements S ′ and T ′ such that S ′ and T ′ are equivalent. But S ′ = S andT ′ = T . �

Examples of composition series. Sn B An B {id}, where n ≥ 5. S4 BA4 B K B 〈(1, 2)(3, 4)〉 B {id}.

Derived groups and solvable groups. Let G be a group. G(0) := G,G(i+1) := (G(i))′. G(i) is called the ith derived group of G. If G(n) = {e} for somen ∈ Z+, G is called solvable.

30 1. GROUPS

Example.

S′n =

{An if n ≥ 3,{id} if n ≤ 2,

A′n =

An if n ≥ 5,K if n = 4,{id} if n ≤ 3.

Proof. Assume n ≥ 3. Since Sn/An is abelian, S′n ⊂ An. Since (i, j, k) =(i, j)(i, k)(i, j)−1(i, k)−1 ∈ S′n for all distinct i, j, k ∈ {1, . . . , n}, we have An ⊂ S′n.

If n ≥ 5, {id} 6= A′n C An and An is simple. So A′n = An. A4/K is abelian, soA′4 ⊂ K. Since A′4 C S4 and A′4 6= {id}, we have A′4 = K. �

Proposition 1.38. G is solvable ⇔ G has a solvable series.

Proof. (⇒) G = G(0) B G(1) B · · · B G(n) = {e} is a solvable series.(⇐) Let G = G0 B G1 B · · · B Gn = {e} be a solvable series. Then Gi ⊃ G′i−1,

1 ≤ i ≤ n. So, {e} = Gn ⊃ G′n−1 ⊃ G′′n−2 ⊃ · · · ⊃ G(n)0 = G

(n). �

Fact. Let N C G. Then G is solvable ⇔ N and G/N are both solvable.

Proof. (⇒) Assume G(n) = {e}. Then N (n) = {e} and (G/N)(n) = {N}.(⇐) A solvable series of G/N and a solvable series of N give rise to a solvable

series of G. �

Examples. Finite p-groups are solvable. Sn (n ≥ 5) is not solvable.

Theorem 1.39 (The Burnside p-q theorem). If |G| = paqb, where p, q areprimes, then G is solvable.

The proof needs representation theory.

Theorem 1.40 (The Feit-Thompson theorem). Every finite group of odd orderis solvable.

The proof is over 250 pages [7].

Central series and nilpotent groups.

{e} = G0 C G1 C · · · C Gn = Gsuch that Gi+1/Gi ⊂ Z(G/Gi) is called a central series of G. A central series is anascending series. If G has a central series, G is called nilpotent.

Easy fact. Let G be a group. Z0(G) := {e}. Zi+1(G) is defined byZ(G/Zi(G)) = Zi+1(G)/Zi(G). Then G is nilpotent ⇔ Zn(G) = G for somen ∈ Z+.

Proof. (⇒) Let {e} = G0 C G1 C · · · C Gn = G be a central series. Useinduction to show Gi ⊂ Zi(G). Assume Gi−1 ⊂ Zi−1(G). Since f : G/Gi−1 →G/Zi−1(G), gGi−1 7→ gZi−1(G), is an onto homomorphism and since Gi/Gi−1 ⊂Z(G/Gi−1), we have GiZi−1(G)/Zi−1(G) = f(Gi/Gi−1) ⊂ Z(G/Zi−1(G)) =Zi(G)/Zi−1(G). So Gi ⊂ Zi(G). �

Easy fact. G nilpotent ⇒ G solvable.Example. Let 3 ≤ n = 2tm, 2 - m. Consider the dihedral group

Dn = 〈α, β | αn = β2 = 1, βαβ−1 = α−1〉.

1.11. EXAMPLES OF AUTOMORPHISM GROUPS 31

First assume m ≥ 3. We claim that

(1.14) Zi(Dn) =

{〈αn/2i〉 if 0 ≤ i ≤ t,〈αm〉 if i > t.

To see (1.14), use induction on i. Assume that i < t and Zi(Dn) = 〈αn/2i〉. Then

Dn/Zi(Dn) = 〈ᾱ, β̄〉 ∼= Dn/2i , where ᾱ = αZi(Dn), β̄ = βZi(Dn), o(ᾱ) = n/2i,o(β̄) = 2, β̄ᾱβ̄−1 = ᾱ−1. Hence, Z(Dn/Zi(Dn)) = 〈ᾱn/2

i+1〉 = 〈αn/2i+1〉/Zi(Dn).So Zi+1(Dn) = 〈αn/2

i+1〉. Since Dn/Zt(Dn) = Dn/〈αm〉 ∼= Dm, Z(Dn/Zt(Dn)) =Zt(Dn)/Zt(Dn), so Zt+1(Dn) = Zt(Dn). In the same way, Zi(Dn) = Zt(Dn) forall i > t.

Now assume m = 1, i.e., n = 2t. The same argument shows that

Zi(Dn) =

{〈αn/2i〉 if 0 ≤ i < t,Dn if i ≥ t.

So, Dn is nilpotent iff n is a power of 2. (Note. Dn is always solvable.)

Fact 1.41. If G is nilpotent and H � G, then H 6= NG(H).

Proof. Let {e} = G0 C G1 C · · · C Gn = G be a central series. Choosei such that Gi ⊂ H but Gi+1 6⊂ H. Since Gi+1/Gi ⊂ Z(G/Gi). Gi+1/Gi ⊂NG/Gi(H/Gi)⇒ Gi+1 ⊂ NG(H). �

Fact. Finite p-groups are nilpotent.

Proof. Let |G| = pk. If Zi(G) 6= G, then Zi+1(G)/Zi(G) = Z(G/Zi(G)) isnontrivial. So Zi(G) ( Zi+1(G). Thus Zn(G) = G for some n > 0. �

Proposition 1.42. Let G be a finite group. Then G is nilpotent ⇔ G ∼= thedirect product of its Sylow subgroups.

Proof. (⇒) Only have to show that every Sylow subgroup P of G is nor-mal. By Proposition 1.20, NG(NG(P )) = NG(P ). If NG(P ) 6= G, by Fact 1.41,NG(NG(P )) 6= NG(P ), which is a contradiction. So, NG(P ) = G, i.e. P C G.

(⇐) H ×K is nilpotent ⇔ H and K are both nilpotent. �

1.11. Examples of Automorphism Groups

Automorphism groups of Sn and An.

Lemma 1.43. If f ∈ Aut(An) maps a 3-cycle to a 3-cycle, then ∃α ∈ Sn suchthat f(σ) = ασα−1 ∀σ ∈ An.

Proof. When n ≤ 3, the claim is obviously true. So assume n ≥ 4.First note that f maps all 3-cycles to 3-cycles. When n = 4, every 3-cycle is

conjugate to (1, 2, 3) or (1, 2, 3)−1 in A4. When n ≥ 5, all 3-cycles are conjugate inAn.

Let f((1, 2, 3)

)= (a1, a2, a3). Since (a1, a2, a3)f

((1, 2, 4)

)= f

((1, 2, 3)(1, 2, 4)

)= f

((1, 3)(2, 4)

)has order 2, after a cyclic shift of (a1, a2, a3), we have f

((1, 2, 4)

)=

(a1, a2, a4). For i ≥ 5, since (a1, a2, a3)f((1, 2, i)

)= f

((1, 3)(2, i)

)has order 2,

we must have f((1, 2, i)

)= (a1, a2, ai) or (a2, a3, c) or (a3, a1, d), where ai /∈

32 1. GROUPS

{a1, . . . , a4} and c, d /∈ {a1, a2, a3}. In the last two cases, o[f((1, 2, 4)

)f((1, 2, i)

)]6=

2, which is a contradiction. So, we must have f((1, 2, i)

)= (a1, a2, ai). Therefore,

f((1, 2, i)

)= (a1, a2, ai) = α(1, 2, i)α−1, i ≥ 3,

where α(i) = ai, 1 ≤ i ≤ n. Since An is generated by (1, 2, i), 3 ≤ i ≤ n, we havef(σ) = ασα−1 ∀σ ∈ An. �

Lemma 1.44. Let G < Sn with [Sn : G] = n. Then ∃θ ∈ Aut(Sn) such thatθ(G) = {σ ∈ Sn : σ(1) = 1} ∼= Sn−1.

Proof. Let φ : Sn → SSn/G be the action of Sn on Sn/G by left multiplication.Then kerφ ⊂ G. Since the only normal subgroup of Sn with an order dividing(n − 1)! is {id}, we have kerφ = {id}. So φ is an isomorphism. Write Sn/G ={G1, . . . , Gn}, where G1 = G. Let f : {1, . . . , n} → Sn/G, i 7→ Gi. Define

θ : Sn −→ Snσ 7−→ f−1φ(σ)f.

Then θ ∈ Aut(Sn). If σ ∈ G,[θ(σ)

](1) =

[f−1φ(σ)f

](1) =

[f−1φ(σ)

](G) = 1. �

Theorem 1.45. (i) For n ≥ 4, ρ : Aut(Sn) → Aut(An), f 7→ f |An , isan isomorphism.

(ii) Assume n 6= 6. Then

Aut(Sn) = Inn(Sn) ∼=

{1 if n ≤ 2,Sn if n ≥ 3, n 6= 6,

Aut(An) ∼=

1 if n ≤ 2,Z2 if n = 3,Sn if n ≥ 4, n 6= 6.

(iii) Aut(S6) = Aut(A6) and [Aut(S6) : Inn(S6)] = 2.

Proof. 1◦ For n ≥ 4, ρ is 1-1. (Thus we may write Aut(Sn) ⊂ Aut(An).)Assume f ∈ Aut(Sn) such that f |An = idAn . To prove f = id, it suffices

to show that f((1, 2)

)= (1, 2). Assume to the contrary that f

((1, 2)

)6= (1, 2).

f((1, 2)

)is a product of k disjoint transpositions. Thus

2(n− 2)! =∣∣CSn((1, 2))∣∣ = ∣∣CSn[f((1, 2))]∣∣ = 2kk!(n− 2k)!,

which implies that k = 1 or n = 6 and k = 3. In either cases, we have CAn((1, 2)

)6=

CAn[f((1, 2)

)], which is a contradiction.

2◦ For n ≥ 4, n 6= 6, Aut(Sn) = Inn(Sn) and ρ : Aut(Sn) → Aut(An) is anisomorphism.

Let g ∈ Aut(An) and let σ ∈ An be a 3-cycle. The f(σ) is a product of kdisjoint 3-cycles. We have

(1.15)12

3(n− 3)! = |CAn(σ)| = |CAn(g(σ)

)| = 1

23kk!(n− 3k)!.

The integer solutions of (1.15) are (n, k) = (n, 1) and (n, k) = (6, 2). Since n 6= 6, wehave k = 1. By Lemma 1.43, g = ρ(f) for some f ∈ Inn(Sn). Therefore, ρ

∣∣Inn(Sn)

:Inn(Sn) → Aut(An) is onto. By 1◦, Aut(Sn) = Inn(Sn) and ρ : Aut(Sn) →Aut(An) is an isomorphism.

1.11. EXAMPLES OF AUTOMORPHISM GROUPS 33

3◦ [Aut(A6) : Inn(S6)] ≤ 2.Let f, g ∈ Aut(A6). If f maps a 3-cycle to a 3-cycle, by Lemma 1.43, f ∈

Inn(S6). Assume that both f and g maps every 3-cycle to a product of two disjoint3-cycles. Then f maps every product of two disjoint 3-cycles to a 3-cycle. (Notethat in A6, the number of 3-cycles equals the number of products of two disjoint3-cycles.) Hence fg maps all 3-cycles to 3-cycles. By Lemma 1.43, fg ∈ Inn(S6).So, [Aut(A6) : Inn(S6)] ≤ 2.

4◦ Aut(S6) 6= Inn(S6).Let P be the set of 6 Sylow 5-subgroups of S5. Let φ : S5 → SP (= S6) be the

action of S5 on P by conjugation. Then kerφ = {id}. (Since S5 acts transitivelyon P, 6

∣∣ |S5/ kerφ|. Since A5 is only one nontrivial normal subgroup of S5, wehave kerφ = {id}.) Thus φ(S5) is a subgroup of SP of index 6. By Lemma 1.44,∃f ∈ Aut(SP) such that f

(φ(S5)

)is the stabilizer of some P ∈ P. Since φ(S5) acts

transitively on P, we must have f /∈ Inn(SP). (If f = α( )α−1 for some α ∈ SP ,then φ(S5) stabilizes α−1(P ), →←.)

5◦ By 1◦ , 3◦ and 4◦, we have

Inn(S6) ( Aut(S6) ⊂ Aut(A6)

and [Aut(A6) : Inn(S6)] ≤ 2. So, Aut(S6) = Aut(A6) and [Aut(S6) : Inn(S6)] = 2.

6◦ The remaining claims in (ii) about Aut(Sn) and Aut(An) for n ≤ 3 areobvious. (To see that Aut(S3) = Inn(S3), note that ∀f ∈ Aut(S3), f

((1, 2)

)is a

2-cycle and f((1, 2, 3)

)is a 3-cycle; hence |Aut(S3)| ≤ 3 · 2 = |Inn(S3)|.) �

Note.

(i) Assume f ∈ Aut(S6) \ Inn(S6). By Lemma 1.43, f maps every 3-cycle toa product of two disjoint 3-cycles and vice versa. By a similar argument,f maps every transposition to a product of 3 disjoint transpositions andvice versa.

(ii) By Lemma 1.44 and Theorem 1.45, if n 6= 6, every subgroup G < Sn with|G| = (n− 1)! must fix one of 1, . . . , n. This is false for n = 6.

The Automorphism group of GL(n, F ).Let F be a field and n ≥ 2. For each P ∈ GL(n, F ), σ ∈ Aut(F ) and represen-

tation χ : GL(n, F )→ F×, define

gP,σ,χ : GL(n, F ) −→ GL(n, F )A 7−→ χ(A)PAσP−1.

Then gP,σ,χ ∈ Aut(GL(n, F )

)and

G = {gP,σ,χ : P ∈ GL(n, F ), σ ∈ Aut(F ), χ : GL(n, F )→ F× is a representation}

is a subgroup of Aut(GL(n, F )

). The automorphism

τ : GL(n, F ) −→ GL(n, F )A 7−→ (A−1)T

is an involution. τ /∈ G unless n = 2; see Exercise 1.17. Aut(GL(n, F )

)is generated

by G and τ ; see [5].

34 1. GROUPS

Exercises

1.1. Let f : G → H be a homomorphism of groups and X ⊂ G. Prove thatf(〈X〉) = 〈f(X)〉.

1.2. (Double cosets) Let H and K be subgroups of G. For a ∈ G, HaK := {hak :h ∈ H, k ∈ K} is called an (H,K)-double coset in G.(i) Prove that the set of all (H,K)-cosets form a partition of G.(ii) Assume H and K are finite. Prove that |HaK| = |H||K||H∩aKa−1| .

1.3. (Cyclic groups)(i) Prove that every subgroup of a cyclic group is cyclic.(ii) Prove that every quotient group of a cyclic group is cyclic.(iii) Let G be an infinite group. Prove that G ∼= Z ⇔ G is isomorphic to

every nontrivial subgroup of G.(iv) If G is a cyclic group of order n. Then for every m | n, G has a unique

subgroup of order m. (If G = 〈a〉, then 〈a nm 〉 is the unique subgroup oforder m.)

1.4. (Diagram chasing) A sequence of groups and homomorphisms

· · · −→ Gi−1fi−1−→ Gi

fi−→ Gi+1 −→ · · ·

is called exact at Gi if im fi−1 = ker fi; the sequence is called exact if it is

exact at Gi for all i. An exact sequence 1 → Kf→ G h→ Q → 1 is called a

short exact sequence. Let

1 1 1y y y1 −→ A f−→ B g−→ C −→ 1

α

y yβ yγ1 −→ A′ f

′

−→ B′ g′

−→ C ′ −→ 1α′y yβ′ yγ′

1 −→ A′′ f′′

−→ B′′ g′′

−→ C ′′ −→ 1y y y1 1 1

be a commutative diagram of groups such that all columns are exact and the2nd and 3rd rows are exact. Prove that the 1st row is also exact.

1.5. Determine all subgroups of S4 and identify the ones that are normal.

1.6. (The converse of Lagrange’s theorem is false.) Prove that An does not havea subgroup of index 2.

1.7. Assume H < Sn such that [Sn : H] = 2. Prove that H = An.

1.8. (Application of Burnside’s lemma) Each vertex of a regular n-gon is to becolored with any of c colors. Two colored regular n-gons are considered thesame if one can be obtained from the other through a rotation or a reflection.Find the total number of different ways to color the regular n-gon. (Let X

EXERCISES 35

be the set of all colored regular n-gons. Then Dn acts on X and the numberto be found is the number of Dn-orbits in X .)

1.9. Prove that there are no simple groups of order 120 and 300.

1.10. (Groups of order 2pq) Let G be a nonabelian group of order 2pq, where2 < p < q are primes.(i) Assume q 6≡ 1 (mod p). Then

G ∼= (Zq × Zp) oαi Z2for some i = 1, 2, 3, where αi : Z2 → Aut(Zq × Zp),

α1(1) :

{(1, 0) 7→ (−1, 0),(0, 1) 7→ (0, 1);

α2(1) :

{(1, 0) 7→ (1, 0),(0, 1) 7→ (0,−1);

α3(1) :

{(1, 0) 7→ (−1, 0),(0, 1) 7→ (0,−1).

Moreover, (Zq × Zp) oαi Z2, i = 1, 2, 3, are pairwise nonisomorphic.(ii) Assume q ≡ 1 (mod p). In addition to the three groups in (i), there is

a fourth group(Zq o Zp) oβ Z2,

where Zq o Zp = 〈a, b | aq = bp = 1, bab−1 = ak〉, o(k) = p in Z×p , andβ : Z2 → Aut(Zq o Zp),

β(1) :

{a 7→ a−1,b 7→ b.

Note that

(Zq o Zp) oβ Z2 = 〈a, b, c | aq = bp = c2 = 1, bab−1 = ak, cac−1 = a−1, cbc−1 = b〉.

1.11. (Generalized quaternion groups) Let n ≥ 1 and Q4n = 〈x, y | x2n = 1, xn =y2, yxy−1 = x−1〉. Also let

A =

[ξ

ξ−1

], A =

[−1

1

]∈ GL(2,C),

where ξ = e2πi/2n, and let G = 〈A,B〉 < GL(2,C).(i) Prove that ∃ an onto homomorphism f : Q4n → G such that f(x) = A

and f(y) = B.(ii) Prove that |Q4n| ≤ 4n and |G| ≥ 4n. Thus, |Q4n| = |G| = 4n and

f : Q4n → G is an isomorphism.

1.12. Prove that S4 ∼= 〈a, b | a4 = b3 = (ab)2 = 1〉.

1.13. Prove that Q8 cannot be embedded in S7.

1.14. Prove that if n is odd, D2n ∼= Dn × Z2.

1.15. Let |X| =∞. Prove that Aut(SX) 6= Inn(SX).

1.16. Let n ≥ 4, α = (1, 2, . . . , n), β = (1, 4)(2, 3) ∈ Sn. Then

〈α, β〉 =

Dn if n = 4, 5,Sn if n ≥ 6 and n is even,An if n ≥ 6 and n is odd.

36 1. GROUPS

1.17. Let F be a field and n ≥ 3. Then there do not exist P ∈ GL(n, F ), σ ∈Aut(F ) and a representation χ : GL(n, F )→ F× such that

(1.16) (A−1)T = χ(A)PAσP−1 for all A ∈ GL(n, F ).Note. If n = 2, then

(A−1)T =1

detA

[0 1−1 0

]A

[0 1−1 0

]−1for all A ∈ GL(2, F ).

1.18. Let F be a field and f : GL(n, F )→ F× a homomorphism. Prove that ∃! ahomomorphism g : F× → F× such that f = g ◦ det.

1.19. Let G = 〈{A,B}〉 < GL(2,Q), where

A =

[2 00 1

], B =

[1 10 1

].

Let U = {[ 1 a0 1 ] : a ∈ Q} < GL(2,Q)}. Prove that G ∩ U is not finitelygenerated.

1.20. (Dixon [6, Problem 6.39]) Let G be a finite p-group such that Z(G′) is cyclic.Prove that G′ is abelian.

1.21. Let H and K be normal subgroups of G such that G/H and G/K are bothsolvable. Prove that G/H ∩K is solvable. (Hint. Consider H/H ∩K and(G/H ∩K)/(H/H ∩K). Use the 2nd and 3rd isomorphism theorems.)

1.22. Let G be group. A normal subgroup N C G is called minimal if N 6= {e} and6 ∃ K C G such that {e} 6= K ( N . Assume that G is a finite solvable groupand N a minimal normal subgroup of G. Prove that N is an elementaryabelian p-group, i.e., N ∼= Zp × · · · × Zp for some prime p.

CHAPTER 2

Rings and Modules

2.1. Rings, Basic Definitions

Definition 2.1. A ring is a nonempty set R equipped with two operations +and · such that

(i) (R,+) is an abelian group;(ii) (ab)c = a(bc) ∀a, b, c ∈ R;(iii) a(b+ c) = ab+ ac, (a+ b)c = ac+ bc ∀a, b, c ∈ R.

If ab = ba for all a, b ∈ R, R is called commutative. If ∃1R ∈ R such that 1Ra =a1R = a ∀a ∈ R, 1R is called the identity of R.

Subring. Let (R,+, ·) be a ring. S ⊂ R is called a subring of R if (S,+, ·) isa ring.

Homomorphism. Let R and S be rings. A map f : R → S is called ahomomorphism if f(a + b) = f(a) + f(b), f(ab) = f(a)f(b) for all a, b ∈ R. Anisomorphism is a bijective homomorphism.

Note. In general, a ring may not have an identity, e.g. 2Z. If S is a subringof R, any of the following could happen: (i) R has identity, S does not (R = Z,S = 2Z); (ii) S has identity, R does not (R = Z × 2Z, S = Z × {0}); (iii) R andS both have identity but 1R 6= 1S (R = Z× Z, S = Z× {0}). If R and S are tworings with identity, a homomorphism f : R → S does not necessarily map 1R to1S . However, we make the following declaration.

Declaration. In these notes, unless specified otherwise, it is assumed that aring has identity; if S is a subring of R, 1S = 1R; a homomorphism maps identityto identity.

Basic properties of rings.(i) 0R · a = a · 0R = 0R, a ∈ R.(ii) (na)b = a(nb) = n(ab), m(na) = (mn)a, a, b ∈ R, m,n ∈ Z.(iii) ( n∑

i=1

ai

)( m∑j=1

bj

)=

n∑i=1

m∑j=1

aibj .

(iv) Assume a1, . . . , as ∈ R are pairwise commutative. Then

(a1 + · · ·+ as)n =∑

i1+···+is=n

n!i1! · · · is!

ai11 · · · aiss .

The multiplicative group. a ∈ R is call a unit (or invertible) if ∃b ∈ R suchthat ab = ba = 1R. R× := the set of all units of R. (R×, ·) is the multiplicativegroup of R.

37

38 2. RINGS AND MODULES

Types of rings.Integral domain. R: commutative, 1R 6= 0, no zero divisors (i.e., ab = 0⇒ a =

0 or b = 0).Division ring (skew field). R: 1R 6= 0, R× = R \ {0}.Field. Commutative division ring.

Examples.Fields: Q, R, C, Zp (p prime).Integral domains (not fields): Z, D[x] (the polynomial ring over an integral

domain D).Noncommutative rings: Mn×n(R) = the ring of n× n matrices over a ring R.Endomorphism ring. Let A be an abelian group, End(A) = Hom(A,A).

(End(A),+, ◦) is the endomorphism ring of A.Fact. Every ring R is a subring of End

((R,+)

).

Proof. We havef : R ↪→ End

((R,+)

)r 7−→ f(r)

wheref(r) : (R,+) −→ (R,+)

x 7−→ rx.�

Example (Real quaternions, a division ring which is not a field).

H = {a1 + a2i+ a3j + a4k : a1, . . . , a4 ∈ R}.Addition: coordinate wise; multiplication: defined by the distributive laws and therules i2 = j2 = k2 = −1, ij = k, jk = i, ki = j, ik = −j, kj = −i, ji = −k. Ifz = a1 + a2i+ a3j + a4k, define z̄ = a1 − a2i− a3j − a4k. zz̄ = a21 + a22 + a23 + a24.If z 6= 0, z−1 = 1zz̄ z.

Group rings. Let G be a group (written multiplicatively) and R a ring. Thegroup ring R[G] := the set of all formal sums

∑g∈G rgg, where rg ∈ R and rg = 0

except for finitely many g ∈ G.∑g∈G

rgg +∑g∈G

sgg :=∑g∈G

(rg + sg)g,(∑h∈G

rhh)(∑

k∈G

skk)

=∑g∈G

( ∑h,k∈Ghk=g

rhsk

)g.

If X ⊂ G is closed under multiplication and e ∈ X, then R[X] = {∑g∈X rgg ∈

R[G]} is a subring of R[G].Characteristic. The characteristic of a ring R (charR) is the smallest n ∈

Z+ such that na = 0 for all a ∈ R. If no such n exists, charR = 0. ( char Zn = n,char Q = 0.)

Fact. If D is an integral domain, charD = 0 or a prime.

Ideals. Let R be a ring. I ⊂ R is called a left (right ) ideal of R if I is asubgroup of (R,+) and ax ∈ R for all a ∈ R, x ∈ I. An ideal is a two-sided ideal.

2.1. RINGS, BASIC DEFINITIONS 39

If X ⊂ R, the ideal of R generated by X (the smallest ideal containing X) is

〈X〉 (or (X)) ={ n∑i=1

aixibi : n ≥ 0, ai, bi ∈ R, xi ∈ X}.

An ideal generated by one element is called a principal ideal.

Sum and product of ideals. Let I, J be left (right) ideals of R. Define

I + J = {a+ b : a ∈ I, b ∈ J}.

I + J is the smallest left (right) ideal of R containing I ∪ J .If I and J are ideals of R, define

IJ ={ n∑i=1

aibi : n ≥ 0, ai ∈ I, bi ∈ J}.

IJ is an ideal of R and IJ ⊂ I ∩ J .

The quotient ring. Let I be an ideal of R. Then R/I is an abelian group.For a + I, b + I ∈ R/I, define (a + I)(b + I) = ab + I. The multiplication is welldefined and (R/I,+, ·) is a ring, called the quotient ring of R by I.

π : R −→ R/Ir 7−→ r + I

is an onto homomorphism (canonical homomorphism).

Fact. I is an ideal of R ⇔ I = ker f for some homomorphism f : R→ S.

Proposition 2.2 (Universal mapping property). Let f : R → S be a homo-morphism of rings and let I be an ideal of R such that I ⊂ ker f . Then there existsa unique homomorphism f̄ : R/I → S such that the following diagram commutes.

................................................................................................. ...........

...............................................................................

.............................................................................................................................

f

πf̄

R S

R/I

Isomorphism theorems.(i) Let f : R→ S be a homomorphism of rings. Then R/ ker f ∼= f(R).(ii) Let I ⊂ J be ideals of R. Then (R/I)/(J/I) ∼= R/J .

The correspondence theorem. Let I be an ideal of R. Let A = the setof all ideals of R containing I, B = the set of all ideals of R/I. Then A → B:J 7→ J/I, is a bijection.

m-adic topology. Let R be a ring and m an ideal of R. For each x ∈ R,{x + mn : n ∈ N} form a neighborhood base of x. The topology on R defined bythis neighborhood base is called the m-adic topology. The following mappings arecontinuous in the m-adic topology.

(i) R×R→ R, (x, y) 7→ x+ y;(ii) R→ R, x 7→ −x;(iii) R×R→ R, (x, y) 7→ xy.

40 2. RINGS AND MODULES

(A ring R endowed with a topology such that mappings (i) – (iii) are continuous iscalled a topological ring. Thus R with the m-adic topology is a topological ring.)

Proof. (i) (x+ mn) + (y + mn) ⊂ x+ y + mn.(ii) −(x+ mn) ⊂ −x+ mn.(iii) (x+ mn)(y + mn) ⊂ x+ y + mn. �

The ideal mn is both open and closed. (For every x ∈ mn, x+ mn ⊂ mn; hencemn is open. R \ mn =

⋃x∈R\mn(x + m

n) is open. So mn is closed.) The m-adictopology is Hausdorff ⇔

⋂∞n=0 m

n = {0}. The m-adic topology is discrete ⇔ m isnilpotent (i.e., mn = 0 for some n > 0).

2.2. Prime Ideals and Maximal Ideals

Definition 2.3. An ideal P of R is called a prime ideal if (i) P 6= R and (ii)if A,B are ideals of R such that AB ⊂ P , then A ⊂ P or B ⊂ P .

An ideal M of R is called maximal if M 6= R and there is no ideal strictlybetween M and R. Maximal left (right) ideal