1.1 Definition. A monoid is a set M together with a map. 1.3 Definition. A monoid is commutative if x y = y x for all x, y M.


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1 1 Monoids and groups 1.1 Definition. A monoid is a set M together with a map M M M, (x, y) 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) M n (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 = id U = 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
2 1.5 Note. Associativity implies that for x 1,..., x k M the expression x 1 x 2 x k has the same value regardless how we place parentheses within it; e.g.: (x 1 x 2 ) (x 3 x 4 ) = ((x 1 x 2 ) x 3 ) x 4 = x 1 ((x 2 x 3 ) x 4 ) etc. 1.6 Note. A monoid has only one identity element: if e, e M are identity elements then e = e e = e 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) GL n (R) = {A M n (R) det(a) 0} with matrix multiplication (the n n general linear group) 4) SL n (R) = {A M n (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 permutations of U) Note. If U = {1, 2,..., n} then Perm(U) is called the symmetric group on n letters and it is denoted by S n. 2
3 7) Let T = an equilateral triangle G T = {I, R 1, R 2, S 1, S 2, S 3 } I R 1 R 2 S 1 S S 2 3 G T = the group of symmetries of T. 1.9 Proposition (Cancellation Law). If G is a group, x, y, x G and then y = z. xy = xz Proof. xy = xz x 1 xy = x 1 xz y = z 1.10 Note. The cancellation law does not hold for monoids. E.g. in M 2 (R) take ( ) ( ) ( ) A =, B =, C = Then AB = AC but A C. 3
4 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) SL n (R) is a subgroup of GL n (R) 4) H = {I, R 1, R 2 } is a subgroup of G T 2.4 Note. If {H i } i I is a family of subgroups of G then i I H i is also a subgroup of G. 2.5 Definition. If G is a group and S is a subset of G then denote S = the smallest subgroup of G that contains S S is the subgroup of G generated by the set S. 2.6 Proposition. If S G then S consists of all elements of the form where x 1,..., x k S. x ±1 1 x ±1 2 x ±1 k 4
5 Proof. Exercise. 2.7 Definition. A set S G generates G if S = G. 2.8 Example. S = {S 1, S 2 } generates G T. 2.9 Definition. A group G is finitely generated if it is generated by some finite subset S G Note. Every finite group is finitely generated. Some infinite groups are finitely generated; e.g. Z = Definition. A group G is cyclic if G = a for some a G 2.12 Note. If G is cyclic, G = a then every element g G is of the form g = a n for some n Z (where a n := (a 1 ) n, a 0 = e) Examples. 1) Z = 1 is cyclic. 2) H := {I, R 1, R 2 } G T is cyclic: H = R 1 and H = R 2 5
6 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 e G, e H denote identity elements in, respectively, G and H then (i) f(e G ) = e H (ii) f(a 1 ) = f(a) 1 for any a G. Proof. (i) We have f(e G ) = f(e G e G ) = f(e G ) f(e G ) Multiplying this equation by f(e G ) 1 we obtain e H = f(e G ). (ii) Since by (i) we have f(e G ) = e H therefore f(a) f(a 1 ) = f(a a 1 ) = f(e G ) = e H It is now enough to multiply this equation from the left by f(a) Definition. A homomorphism f : G H is an isomorphism if there is a homomorphism g : H G such that g f = id G and f g = id H. 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
7 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) id G : 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 c g : G G, c g (a) := gag 1 Check: c g is an automorphism of G. Automorphisms of this form are called inner automorphisms of G. Note. If G is an abelian group then c g = id G for all g G. 5) Recall: GL n (R) = {A M n det(a) 0}, R = R {0} We have the determinant function: det: GL n (R) R Since det(ab) = det(a) det(b) this function is a homomorphism. 6) Let G GL 2 (R) G := G is a subgroup of GL 2 (R): ( ) 1 r 0 1 We have homomorphisms: {( ) 1 r r R} 0 1 ( ) 1 s = 0 1 ( ) 1 1 r = 0 1 ( ) 1 r + s 0 1 ( ) 1 r 0 1 f : R G and g : R G 7
8 where f(r) = ( ) 1 r, g 0 1 Since g f = id G, f g = R we get G = R. (( )) 1 r = r Definition. If G is a group then G is called the order of G. G := the number of elements of G 3.9 Example. G T = 6, Z = Note. If G = H then G = H. 8
9 4 The kernel and the image of a homomorphism 4.1 Proposition. Let f : G H be a homomorphism. 1) If G is a subgroup of G then f(g ) is a subgroup of H. 2) If H is a subgroup of H then f 1 (H ) 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 Ker(f) := f 1 (e H ) G 4.3 Note. f : G H is an epimorphism (onto) iff Im(f) = H. 4.4 Proposition. f : G H is a monomorphism (11) iff Ker(f) = {e G } Proof. ( ) We have f(e G ) = e H. Thus if f is 11 then f(g) = e H g = e H. In other words we have then Ker(f) = {e H }. only if ( ) Assume that Ker(f) = {e G } and let f(a) = f(b) for some a, b G. We have: f(ab 1 ) = f(a)f(b) 1 = e H so ab 1 Ker(f). Therefore ab 1 = e G, and so a = b. 9
10 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 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) Examples. 1) Any subgroup of an abelian group is normal. 2) H := {I, R 1, R 2 } is a normal subgroup of G T (check!). 3) K := {I, S 1 } is not a normal subgroup of G T (check!). As a consequence K cannot be the kernel of any homomorphism G T G. 10
11 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 for some a G. 5.2 Example. Ha := {ha h H} Recall: G T = {I, R 1, R 2, S 1, S 2, S 3 }. Take H := {I, S 1 }. We have: IH = {I I, I S 1 } = {I, S 1 } = H S 1 H = {S 1 I, S 1 S 1 } = {S 1, I} S 2 H = {S 2 I, S 2 S 1 } = {S 2, R 2 } S 3 H = {S 3 I, S 3 S 1 } = {S 3, R 1 } R 1 H = {R 1 I, R 1 S 1 } = {R 1, S 3 } R 2 H = {R 2 I, R 2 S 1 } = {R 2, S 2 } Note: IH = S 1 H, S 2 H = R 2 H, S 3 H = R 1 H 5.3 Lemma. If G is a group and H is a subgroup of G then ah = bh iff a 1 b H 11
12 Proof. ( ) Let ah = bh. Since e H thus b = be bh = ah so b = ah for some h H. Therefore a 1 b = h H. ( ) Assume that a 1 b H. For any h H we have ah = a(a 1 b)(a 1 b) 1 h = b((a 1 b) 1 )h bh This gives: ah bh. Also for any h H we have: bh = (aa 1 )bh = a(a 1 b)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 and let c ah bh. Then for some h 1, h 2 H. This gives and so ah = bh by (5.3). ah 1 = c = bh 2 a 1 b = h 1 h 1 2 H 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 Ha. For example, If H G T, H = {I, S 1 } then S 2 H = {S 2, R 2 }, HS 2 = {S 2, R 1 } 12
13 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 = a H, bh = b H but (ab)h (a b )H. For example, take H = {I, S 1 } G T. Recall: S 2 H = R 2 H = {S 2, R 2 }, S 3 H = R 1 H = {S 3, R 1 } However: (S 2 S 3 )H = R 1 H = {R 1, S 2 } (R 2 R 1 )H = IH = {I, S 1 } 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 G then by (5.7) we have ah = Ha a G. Let ah = a H, bh = b H. Then (ab)h = a(bh) = a(b H) = a(hb ) = (ah)b = (a H)b = a (b H) = (a b )H 13
14 5.12 Corollary/Definition. If H 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 1 H. The group G/H is called the quotient group (or the factor group) of G by H 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 Z = { 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 = Z = 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
15 Recall: A group G is cyclic if it is generated by a single element: G = a for some a G. Note: For every n the group Z/nZ is cyclic: Z/nZ = Note. If H 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 1 a 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
16 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 π f G/ Ker(f) Moreover, f is a monomorphism and Im( f) = Im(f). H 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 1 b K, so f(a 1 b) = e. Thus f(b) = f(aa 1 b) = f(a)f(a 1 b) = 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
17 4) f is 11: 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 11. Therefore f is a bijective homomorphism and thus it is an isomorphism. 6.2 Example. Recall: GL n (R) = {A M n (R) det(a) 0} SL n (R) = {A M n (R) det(a) = 1} SL n (R) is a normal subgroup of GL n (R). We have the homomorphism det: GL n (R) R Since this is an epimorphism and Ker(det) = SL n (R) we get GL n (R)/SL n (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
18 Proof. Let G = a for some a G. Define f : Z G, f(n) := a n 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 G then KH is a subgroup of G, (H K) K and K/(H K) = KH/H 18
19 Proof. Exercise. 6.1 Third Isomorphism Theorem. Let K H G. If K, H are normal subgroups of G then K H, H/K G/K and (G/K)/(H/K) = G/H Proof. Exercise. 19
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