1 Monoids and Groups

Home , Abelian group, Homomorphism, Identity element, Identity function, Isomorphism, Monoid

1 Monoids and groups

1.1 Deﬁnition. A monoid is a set M together with a map

M × M → M, (x, y) 7→ x · y such that

(i) (x · y) · z = x · (y · z) ∀x, y, z ∈ M (associativity);

(ii) ∃e ∈ M such that x · e = e · x = x for all x ∈ M (e = the identity element of M).

1.2 Examples.

1) Z with addition of integers (e = 0)

2) Z with multiplication of integers (e = 1)

3) Mn(R) = {the set of all n × n matrices with coeﬃcients in R} with matrix multiplication (e = I = the identity matrix)

4) U = any set P (U) := {the set of all subsets of U} P (U) is a monoid with A · B := A ∪ B and e = ∅. 5) Let U = any set

F (U) := {the set of all functions f : U → U}

F (U) is a monoid with multiplication given by composition of functions (e = idU = the identity function).

1.3 Deﬁnition. A monoid is commutative if x · y = y · x for all x, y ∈ M.

1.4 Example. Monoids 1), 2), 4) in 1.2 are commutative; 3), 5) are not.

1 1.5 Note. Associativity implies that for x1, . . . , xk ∈ M the expression

x1 · x2 ····· xk has the same value regardless how we place parentheses within it; e.g.:

(x1 · x2) · (x3 · x4) = ((x1 · x2) · x3) · x4 = x1 · ((x2 · x3) · x4) etc.

1.6 Note. A monoid has only one identity element: if e, e0 ∈ M are identity elements then e = e · e0 = e0

1.7 Deﬁnition. A group is a monoid G such that for any x ∈ G there is y ∈ G satistying x · y = e = y · x.

The element y is called the inverse of x and it is denoted by x−1 (or by −x in the additive notation).

A group G is commutative (or abelian) if x · y = y · x for all x, y ∈ G.

1.8 Examples.

1) Z, Q, R, C with addition 2) Q∗ = Q − {0}, R∗ = R − {0}, C∗ = C − {0} with multiplication

3) GLn(R) = {A ∈ Mn(R) | det(A) 6= 0} with matrix multiplication (the n × n general linear group)

4) SLn(R) = {A ∈ Mn(R) | det(A) = 1} with matrix multiplication (the n × n special linear group) 5) Let U = be any set and let Perm(U) := {f : U → U | f is a bijection} Perm(U) with composition of functions is a group (the group of permu- tations of U) Note. If U = {1, 2, . . . , n} then Perm(U) is called the symmetric group on n letters and it is denoted by Sn.

2 7) Let T = an equilateral triangle

GT = {I,R1,R2,S1,S2,S3}

I R2 R1

S S 1 S2 3

GT = the group of symmetries of T .

1.9 Proposition (Cancellation Law). If G is a group, x, y, x ∈ G and

xy = xz then y = z.

Proof.

xy = xz x−1xy = x−1xz y = z

1.10 Note. The cancellation law does not hold for monoids. E.g. in M2(R) take 1 0 0 0 0 0 A = ,B = ,C = 0 0 0 1 0 0 Then AB = AC but A 6= C.

3 2 Subgroups

2.1 Deﬁnition. If G is a group then a subgroup of G is a subset H ⊆ G such that

(i) e ∈ H; (ii) if x, y ∈ H then xy ∈ H; (iii) if x ∈ H then x−1 ∈ H.

2.2 Note. A subgroup of a group is by itself a group.

2.3 Examples.

1) If G is a group then G, {e} are subgroups of G

2) Z is a subgroup of Q, which is a subgroup of R, which is a subgroup of C.

3) SLn(R) is a subgroup of GLn(R)

4) H = {I,R1,R2} is a subgroup of GT

T 2.4 Note. If {Hi}i∈I is a family of subgroups of G then i∈I Hi is also a subgroup of G.

2.5 Deﬁnition. If G is a group and S is a subset of G then denote

hSi = the smallest subgroup of G that contains S hSi is the subgroup of G generated by the set S.

2.6 Proposition. If S ⊆ G then hSi consists of all elements of the form

±1 ±1 ±1 x1 x2 ····· xk where x1, . . . , xk ∈ S.

4 Proof. Exercise.

2.7 Deﬁnition. A set S ⊆ G generates G if hSi = G.

2.8 Example. S = {S1,S2} generates GT .

2.9 Definition. A group G is finitely generated if it is generated by some finite subset S ⊆ G.

2.10 Note.

• Every finite group is finitely generated. • Some infinite groups are finitely generated; e.g. Z = h1i.

2.11 Deﬁnition. A group G is cyclic if G = hai for some a ∈ G

2.12 Note. If G is cyclic, G = hai then every element g ∈ G is of the form

g = an for some n ∈ Z (where a−n := (a−1)n, a0 = e).

2.13 Examples.

1) Z = h1i is cyclic.

2) H := {I,R1,R2} ⊆ GT is cyclic:

H = hR1i and H = hR2i

5 3 Homomorphisms of groups

3.1 Deﬁnition. Let G, H be groups. A function f : G → H is a group homomorphism if for any a, b ∈ G we have

f(ab) = f(a)f(b)

3.2 Proposition. If f : G → H is a homomorphism of groups and eG, eH denote identity elements in, respectively, G and H then

(i) f(eG) = eH (ii) f(a−1) = f(a)−1 for any a ∈ G.

Proof. (i) We have

f(eG) = f(eG · eG) = f(eG) · f(eG)

−1 Multiplying this equation by f(eG) we obtain eH = f(eG).

(ii) Since by (i) we have f(eG) = eH therefore

−1 −1 f(a) · f(a ) = f(a · a ) = f(eG) = eH It is now enough to multiply this equation from the left by f(a)−1.

3.3 Deﬁnition. A homomorphism f : G → H is an isomorphism if there is a homomorphism g : H → G such that g ◦ f = idG and f ◦ g = idH .

3.4 Proposition. A map f : G → H is an isomorphism of groups iﬀ f is a homomorphism and a bijection.

Proof. Exercise.

3.5 Deﬁnition. If there exists an isomorphism f : G → H then we say that the groups G and H are isomorphic and we write G ∼= H.

6 3.6 Deﬁnition. A homomorphism f : G → G is called an endomorphism of G. An isomorphism f : G → G is called an automorphism of G.

3.7 Examples.

1) idG : G → G is an automorphism of G.

2) f : G → G, f(g) = e ∀g∈G is an endomorphism of G. 3) If f : G → H, g : H → K are homomorphisms then so is g ◦ f : G → K.

4) For g ∈ G deﬁne −1 cg : G → G, cg(a) := gag

Check: cg is an automorphism of G. Automorphisms of this form are called inner automorphisms of G.

Note. If G is an abelian group then cg = idG for all g ∈ G.

∗ 5) Recall: GLn(R) = {A ∈ Mn | det(A) 6= 0}, R = R − {0} We have the determinant function:

∗ det: GLn(R) → R Since det(AB) = det(A) · det(B) this function is a homomorphism.

6) Let G ⊆ GL2(R) 1 r G := r ∈ R 0 1

G is a subgroup of GL2(R): 1 r 1 s 1 r + s · = 0 1 0 1 0 1

1 r−1 1 −r = 0 1 0 1

We have homomorphisms:

f : R → G and g : R → G

7 where

1 r 1 r f(r) = , g = r 0 1 0 1 ∼ Since g ◦ f = idG, f ◦ g = R we get G = R.

3.8 Deﬁnition. If G is a group then

|G| := the number of elements of G

|G| is called the order of G.

3.9 Example. |GT | = 6, |Z| = ∞.

3.10 Note. If G ∼= H then |G| = |H|.

8 4 The kernel and the image of a homomorphism

4.1 Proposition. Let f : G → H be a homomorphism.

1) If G0 is a subgroup of G then f(G0) is a subgroup of H. 2) If H0 is a subgroup of H then f −1(H0) is a subgroup of G.

Proof. Exercise.

4.2 Deﬁnition. If f : G → H is a homomorphism then

• the image of f is the subgroup

Im(f) := f(G) ⊆ H

• the kernel of f is the subgroup

−1 Ker(f) := f (eH ) ⊆ G

4.3 Note. f : G → H is an epimorphism (onto) iﬀ Im(f) = H.

4.4 Proposition. f : G → H is a monomorphism (1-1) iﬀ Ker(f) = {eG}

Proof. (⇒) We have f(eG) = eH . Thus if f is 1-1 then f(g) = eH only if g = eH . In other words we have then Ker(f) = {eH }.

(⇐) Assume that Ker(f) = {eG} and let f(a) = f(b) for some a, b ∈ G. We have: −1 −1 f(ab ) = f(a)f(b) = eH −1 −1 so ab ∈ Ker(f). Therefore ab = eG, and so a = b.

9 4.5 Problem. Let G be a group, and let H be a subgroup of G. Is there a homomorphism f : G → K such that Ker(f) = H?

4.6 Note. The dual problem is trivial: if H is a subgroup of G then we have the inclusion homomorphism i: H,→ G and Im(i) = H. It follows that any subgroup of G is an image of some homomorphism.

4.7 Deﬁnition. A subgroup H ⊆ G is a normal subgroup if for every h ∈ H we have aha−1 ∈ H ∀a ∈ G

4.8 Notation. If H is a normal subgroup of G then we write H C G

4.9 Proposition. If f : G → H is a homomorphism then Ker(f) is a normal subgroup of G.

Proof. If a ∈ G, h ∈ Ker(f) then

f(aha−1) = f(a)f(h)f(a)−1 = f(a) · e · f(a)−1 = f(a)f(a)−1 = e so aha−1 ∈ Ker(f).

4.10 Examples.

1) Any subgroup of an abelian group is normal.

2) H := {I,R1,R2} is a normal subgroup of GT (check!).

3) K := {I,S1} is not a normal subgroup of GT (check!). As a consequence K cannot be the kernel of any homomorphism GT → G.

10 5 Normal subgroups, cosets and quotient groups

Recall. If f : G → K is a homomorphism then Ker(f) is a normal subgroup of G.

Next goal: If H is a normal subgroup of G then there is a homomorphism f : G → K such that H = Ker(f).

5.1 Deﬁnition. If H is a subgroup of G then a left coset of H in G is a subset of G of the form aH := {ah | h ∈ H} for some a ∈ G.

A right coset of H in G is a subset of G of the form

Ha := {ha | h ∈ H} for some a ∈ G.

5.2 Example.

Recall: GT = {I,R1,R2,S1,S2,S3}. Take H := {I,S1}. We have:

IH = {I · I,I · S1} = {I,S1} = H S1H = {S1 · I,S1 · S1} = {S1,I} S2H = {S2 · I,S2 · S1} = {S2,R2} S3H = {S3 · I,S3 · S1} = {S3,R1} R1H = {R1 · I,R1 · S1} = {R1,S3} R2H = {R2 · I,R2 · S1} = {R2,S2}

Note: IH = S1H, S2H = R2H, S3H = R1H

5.3 Lemma. If G is a group and H is a subgroup of G then

aH = bH iﬀ a−1b ∈ H

11 Proof. (⇒) Let aH = bH. Since e ∈ H thus

b = be ∈ bH = aH so b = ah for some h ∈ H. Therefore a−1b = h ∈ H.

(⇐) Assume that a−1b ∈ H. For any h ∈ H we have

ah = a(a−1b)(a−1b)−1h = b((a−1b)−1)h ∈ bH

This gives: aH ⊆ bH. Also for any h ∈ H we have:

bh = (aa−1)bh = a(a−1b)h ∈ aH so bH ⊆ aH. Therefore aH = bH.

5.4 Proposition. If H is a subgroup of G then for any a, b ∈ G either

aH = bH or aH ∩ bH = ∅

Proof. Let aH ∩ bH 6= ∅ and let c ∈ aH ∩ bH. Then

ah1 = c = bh2 for some h1, h2 ∈ H. This gives

−1 −1 a b = h1h2 ∈ H and so aH = bH by (5.3).

5.5 Corollary. If H is a subgroup of G then every element of G belongs to one and only left coset of H.

5.6 Note. In general aH 6= Ha. For example, If H ⊆ GT , H = {I,S1} then

S2H = {S2,R2},HS2 = {S2,R1}

12 5.7 Proposition. A subgroup H of G is normal iﬀ

aH = Ha ∀a ∈ G

Proof. Exercise.

5.8 Notation. If H is a subgroup of G then

G/H := the set of all left cosets of H in G

5.9. Multiplication of cosets. Let H ⊆ G, aH, bH ∈ G/H. Deﬁne

aH · bH := (ab)H

5.10 Note. In general this is not well deﬁned, i.e. we may have aH = a0H, bH = b0H but (ab)H 6= (a0b0)H.

For example, take H = {I,S1} ⊆ GT . Recall:

S2H = R2H = {S2,R2},S3H = R1H = {S3,R1}

However: (S2S3)H = R1H = {R1,S2}

(R2R1)H = IH = {I,S1}

5.11 Proposition. If H is a normal subgroup of G then the multiplication of cosets given in (5.9) is well deﬁned.

Proof. If H C G then by (5.7) we have aH = Ha ∀a∈G. Let aH = a0H, bH = b0H. Then

(ab)H = a(bH) = a(b0H) = a(Hb0) = (aH)b0 = (a0H)b0 = a0(b0H) = (a0b0)H

13 5.12 Corollary/Deﬁnition. If H C G then G/H is a group with multiplication deﬁned by (5.9). The identity elements in G/H is the coset eH = H ∈ G/H. The inverse of a coset aH is the coset a−1H.

The group G/H is called the quotient group (or the factor group) of G by H.

5.13 Example.

Take Z, the additive group of integers. Since Z is abelian every its subgroup is normal. For n ∈ Z, n ≥ 2 deﬁne

nZ = {na | a ∈ Z} e.g. 2Z = {· · · − 4, −2, 0, 2, 4,... }, 5Z = {..., −10, −5, 0, 5, 10,... }

Note: nZ is a subgroup of Z.

Cosets of nZ in Z: k + nZ = {k + na | a ∈ Z} e.g. 1 + 5Z = {· · · − 9, −4, 1, 6, 11,... }, 3 + 5Z = {..., −7, −2, 3, 8, 13,... }

Note: k + nZ = l + nZ iﬀ (k − l) ∈ nZ i.e. iﬀ k = l + na for some a ∈ Z. E.g.: 1 + 5Z = 6 + 5Z = 11 + 5Z = −4 + 5Z

Recall: if n, k ∈ Z then there is a unique number l ∈ {0, 1, . . . , n − 1} such that k = l + na for some a ∈ Z. Thus every coset of nZ can be uniquely written as l +nZ where l ∈ {0, 1, . . . , n − 1}.

Denote ¯l := l + nZ. Then Z/nZ = {0¯, 1¯,..., n − 1}

The addition table in Z/5Z:

14 + 0¯ 1¯ 2¯ 3¯ 4¯ 0¯ 0¯ 1¯ 2¯ 3¯ 4¯ 1¯ 1¯ 2¯ 3¯ 4¯ 0¯ 2¯ 2¯ 3¯ 4¯ 0¯ 1¯ 3¯ 3¯ 4¯ 0¯ 1¯ 2¯ 4¯ 4¯ 0¯ 1¯ 2¯ 3¯

Recall: A group G is cyclic if it is generated by a single element: G = hai for some a ∈ G.

Note: For every n the group Z/nZ is cyclic: Z/nZ = h1¯i.

5.14 Note. If H C G then we have a homomorphism π : G → G/H, π(a) := aH

This is the canonical epimorphism of G onto G/H.

We have:

Ker(π) = {a ∈ G | π(a) = eH} = {a ∈ G | aH = eH} = {a ∈ G | e−1a ∈ H} = {a ∈ G | a ∈ H} = H

5.15 Corollary. A subgroup H ⊆ G is the kernel of some homomorphism f : G → K iﬀ H is a normal subgroup

15 6 Isomorphism theorems

6.1 Theorem. If f : G → H is a homomorphism then there is a unique homomorphism f¯: G/ Ker(f) → H such that the following diagram commutes:

f G / H x; x x x x π x x f¯ x x x G/ Ker(f)

Moreover, f¯ is a monomorphism and Im(f¯) = Im(f).

Proof. Denote: K := Ker(f). Deﬁne

f¯: G/K → H, f¯(aK) := f(a)

We have:

1) f¯ is well deﬁned:

If aK = bK then a−1b ∈ K, so f(a−1b) = e. Thus

f(b) = f(aa−1b) = f(a)f(a−1b) = f(a)

2) f¯ is a homomorphism (check).

3) f¯ is a unique homomorphism satisfying f¯◦ π = f

Indeed, if g : G/K → H is some other homomorphism and g ◦ π = f then

f(a) = g ◦ π(a) = g(aK) and so g(aK) = f¯(aK) for all aK ∈ G/K.

16 4) f¯ is 1-1:

We need: Ker(f¯) = {eK}. We have: if f¯(aK) = e then f(a) = e, so a ∈ K and so aK = eK.

5) Im(f) = Im(f¯) (obvious).

6.1 First Isomorphism Theorem. If f : G → H is an epimorphism then

G/ Ker(f) ∼= H

Proof. Take the map f¯: G/ Ker(f) → H. Then Im(f¯) = Im(f) = H, so f¯ is an epimorphism. Also, f¯ is 1-1. Therefore f¯ is a bijective homomorphism and thus it is an isomorphism.

6.2 Example.

Recall: GLn(R) = {A ∈ Mn(R) | det(A) 6= 0}

SLn(R) = {A ∈ Mn(R) | det(A) = 1}

SLn(R) is a normal subgroup of GLn(R). We have the homomorphism

∗ det: GLn(R) → R

Since this is an epimorphism and Ker(det) = SLn(R) we get ∼ ∗ GLn(R)/SLn(R) = R

∼ ∼ ∼ 6.3 Theorem. If G is a cyclic group then G = {e} or G = Z or G = Z/nZ for some n ≥ 2.

17 Proof. Let G = hai for some a ∈ G. Deﬁne

n f : Z → G, f(n) := a Notice that

1) f is a homomorphism 2) f is onto.

∼ Thus by the First Isomorphism Theorem G = Z/ Ker(f).

Check: all subgroups H ⊆ Z are of the form H = nZ for some n ≥ 0. ∼ It follows that G = Z/nZ for some n ≥ 0. Also: ∼ ∼ • if n = 0 then nZ = 0Z = {0} and G = Z/{0} = Z ∼ ∼ • if n = 1 then nZ = 1Z = Z and G = Z/Z = {e} ∼ • if n ≥ 2 then G = Z/nZ.

6.4 Notation. If H,K are subgroups of G then

HK := {hk ∈ G | h ∈ H, k ∈ K}

6.5 Lemma. If H,K are subgroups of G then HK is a subgroup of G iﬀ

HK = KH

Proof. Exercise.

6.1 Second Isomorphism Theorem. If H,K are subgroups of G and H C G then KH is a subgroup of G, (H ∩ K) C K and K/(H ∩ K) ∼= KH/H

18 Proof. Exercise.

6.1 Third Isomorphism Theorem. Let K ⊆ H ⊆ G. If K,H are normal subgroups of G then K C H, H/K C G/K and (G/K)/(H/K) ∼= G/H

Proof. Exercise.