Permutation A permutation of the numbers 1,2, and 3 is a rearrangement of these numbers in a definite order. Thus the six possibilities are 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1 可將每一個permutation看成{1,2,3}{1,2,3}的mapping. 11 11 12 12 13 13 22 23 21 23 21 22 33 32 33 31 32 31

Permutation n為正整數且 Xn={1,2,…,n}. 。 一個permutation為XnXn的one-to-one and onto mapping。 The set of all permutations of Xn is denoted Sn and is called the symmetric group of degree n.

Example S3 = {  |  is a permutation of X3 ={1,2,3} } = { 1 , 2 , 3 , 4 , 5 , 6 }, where 1 2 3 4 5 6 11 11 12 12 13 13 22 23 21 23 21 22 33 32 33 31 32 31

Permutation If the permutations  : X4X4 is defined by (1)=3, (2)=1, (3)=4, and (4)=2, we write it as .

Permutation In general, given ∈Sn write it in matrix form as .

Example1 Denote the elements of S3 in matrix notation. Sol:

Theorem The set Sn of permutations of Xn has |Sn|=n! elements.

Let  and  be permutations in Sn Let  and  be permutations in Sn . Both are mappings from Xn to Xn, and we write them as follows: The composite  : XnXn with ( )(k)= ((k)) for all k∈Xn . is again a permutation in Sn.

Example Compute  and  if and . Sol: Note that ≠ in general. 

Identity permutation The identity permutation  in Sn is defined as In other words, (k)=k holds for every k∈Xn. It is easy to verify that == holds for all ∈Sn.

inverse If ∈Sn, the fact that  : XnXn is one-to-one and onto implies that a uniquely determined permutation -1 : XnXn exists (called the inverse of ), which satisfies (-1(k))=k and -1((k))=k, for all k∈Xn.

Example Find the inverse of in S8. Sol:

Theorem Let ,, and  denote permutations in Sn. Then (1)  is in Sn. (2) ==. (3) ()=(). (4) -1==-1.

Example The set Sn of all permutations of {1,2,···,n} is a group under composition, called the symmetric group of degree n.

Consider the permutation in S6. The action of  is described graphically as: 1 4 6 2 3

Let k1, k2, ···,kr be distinct elements of Xn Let k1, k2, ···,kr be distinct elements of Xn. Then the cycle =(k1, k2,···, kr ) is the permutation in Sn defined as (ki) = ki+1 if 1  i  r-1 (kr) = k1 (k) = k if k{k1, k2, ···,kr } We say that  has length r and refer to  as an r-cycle.

Example Write in cycle notation. =(1, 4, 6, 2, 7, 3). Sol: =(1, 4, 6, 2, 7, 3). 因為5對到5，所以沒出現在cycle中。

Theorem If  is an r-cycle, then -1 is also an r-cycle. In fact, if  = (k1,k2,···, kr-1, kr) , then -1= (kr, kr-1, ···, k2, k1) .

Example Consider in S10 . 1 2 3 7 5 10 8 4 6 9

 is the product of these cycles =(1, 3, 7, 2)(4, 6)(5, 10, 8). The four cycles are (1, 3, 7, 2),(4, 6),(5, 10 , 8), and (9) =  .These are pairwise disjoint.  is the product of these cycles =(1, 3, 7, 2)(4, 6)(5, 10, 8). 4 6 9

Example Factor as a product of (pairwise) disjoint cycles. =(1, 5, 9, 7, 4)(2, 12, 8, 3)(6, 11, 13)

Cycle Decomposition 若≠為Sn中的permutation，則一定可以表示成length至少為2的之disjoint cycle的composition 。 若不考慮的排序，則此表示法為唯一。

List all the elements of S4, each factored into disjoint cycles. (1,2) (1,2,3) (1,3,2) (1,2)(3,4) (1,2,3,4) (1,3) (1,2,4) (1,4,2) (1,3)(2,4) (1,2,4,3) (1,4) (1,3,4) (1,4,3) (1,4)(2,3) (1,3,2,4) (2,3) (2,3,4) (2,4,3) (1,3,4,2) (2,4) (1,4,2,3) (3,4) (1,4,3,2)

Transposition A cycle of length 2 is called a transposition. Thus each transposition  has the form =(m, n), where m≠n.

Theorem Every cycle of length r > 1 is a product of r –1 transpositions: (k1, k2,···, kr)=(k1,k2)(k2,k3) ··· (kr-2,kr-1)(kr-1,kr). Hence every permutation is a product of transpositions.

Example (1, 2, 3) = (1, 2)(2, 3) (1, 2, 3, 4) = (1, 2)(2, 3)(3, 4) (1, 2, 3, 4, 5) = (1, 2)(2, 3)(3, 4)(4, 5) (1, 2, 3, 4, 5, 6) = (1, 2)(2, 3)(3, 4)(4, 5)(5, 6)

Remark Factorizations into transpositions are not unique. For example, (2,3)(1,2)(2,5)(1,3)(2,4)=(1,2,4,5)=(1,5)(1,4)(1,2).

Parity Theorem 若n ··· 21與m ··· 21為permutation 的兩種transposition的composition，則m與n必同為even或同為odd。

A permutation  is called even or odd according as it can be written in some way as the product of an even or odd number of transpositions. The set of all even permutations in Sn is denoted An and is called the alternating group of degree n.

Example Determine the parity of Solution: We have  = (1, 5, 7, 2, 4)(3, 6, 8, 9) =(1,5)(5,7)(7,2)(2,4)(3,6)(6,8)(8,9) So  is odd because it has a product of 7 transpositions.

Remark Since every cycle of length r > 1 is a product of r –1 transpositions: (k1 k2 ··· kr)=(k1 k2)(k2 k3) ··· (kr-2 kr-1)(kr-1 kr). Hence permutation =12···r, where i is a cycle of length ai, has a product of a1+a2+···+ar -r transpositions.

Example Determine the parity of Solution: We have  = (1 5 7 2 4)(3 6 8 9) So  is odd because it has a product of 5+4-2=7 transpositions.

List all the elements of S4, each factored into disjoint cycles. (1,2) (1,2,3) (1,3,2) (1,2)(3,4) (1,2,3,4) (1,3) (1,2,4) (1,4,2) (1,3)(2,4) (1,2,4,3) (1,4) (1,3,4) (1,4,3) (1,4)(2,3) (1,3,2,4) (2,3) (2,3,4) (2,4,3) (1,3,4,2) (2,4) (1,4,2,3) (3,4) (1,4,3,2)

List all the elements of A4. (1,2) (1,2,3) (1,3,2) (1,2)(3,4) (1,2,3,4) (1,3) (1,2,4) (1,4,2) (1,3)(2,4) (1,2,4,3) (1,4) (1,3,4) (1,4,3) (1,4)(2,3) (1,3,2,4) (2,3) (2,3,4) (2,4,3) (1,3,4,2) (2,4) (1,4,2,3) (3,4) (1,4,3,2)

List all the elements of A4.  (1, 2, 3) (1, 3, 2) (1, 2)(3, 4) (1, 2, 4) (1, 4, 2) (1, 3)(2, 4) (1, 3, 4) (1, 4, 3) (1, 4)(2, 3) (2, 3, 4) (2, 4, 3)

Theorem Let n> 1. The set An has following properties. If , An, then -1,  An.  An. |An|=n!/2.

Example 由所有even permutations所成集合An會是Sn的subgroup，稱為 alternating group of degree n. Not that: 由所有odd permutations所成集合不會是group。

Example Find the order of Solution:  = (1, 2, 3)(4, 5) || = 6 = lcm(2, 3)

Theorem Let  be a permutation in Sn with factorization  = 1 2… r into disjoint cycles. Then || = lcm(|1|, |2|, …, |r|).

Example Find the order of Solution:  = (1,5,10,13,4,14)(2,7,12,6,11)(3,9) || = lcm(6,5,2) = 30.

Example    = (1, 2)(3, 4) and  = (1, 4)(2, 3) 2 = , 2 = , 1 2  = =(13)(24) 1 2 4 3  

Example Find a group of motions of an equilateral triangle(正三角形). Solution:  =(123) —旋轉120° 2 =(132) —旋轉240°  =(12) —對L3翻轉   =(23) —對L1翻轉  2 =(13) —對L2翻轉 The group of motion is S3  D3

Theorem If n3, the group of motions of a regular n-gon(正n邊形) is isomorphic to Dn.

Cayley’s Theorem Every group G of order n is isomorphic to a subgroup of Sn.

Exercise 4.1 True or False 1,2,3,4,6,7,8,11 Exercise 7,8,10,14,15,18

Coset Let H be a subgroup of a group G and a∈G. Ha = { ha | h∈H } --- right coset of H generated by a aH = { ah | h∈H } --- left coset of H generated by a

Klein group K4 = {1,a,b,ab}, |a| = |b| = 2, ab = ab Let H = {1,a} right cosets H1 =H Ha ={a,a2}={a,1}=H Hb ={b,ab} H(ab) ={ab,a2b}={ab,b}=Hb

Theorem Let H be a subgroup of G, a,b∈G. (1) Ha = H iff a∈H (2) If a∈Hb ,then Ha = Hb (3) Either Ha  Hb =  or Ha = Hb. Note that Ha may not be equal to aH.

Example G = a with |a| = 6, find coset of H = a3  and K = a2 Solution: H = {1, a3}= Ha3 Ha = {a, a4}= Ha4 Ha2 = {a2, a5}= Ha5 K = {1, a2, a4} =Ka2=Ka4 Ka = {a, a3, a5} =Ka3=Ka5

Lemma Let H be a finite subgroup of a group G. Then |H| = |Ha| = |aH| for all a∈G. Proof: Let f: H  Ha with f(h) = ha. Then f is a bijection. So |H| = |Ha|. Similarly, |H| = |aH|.

index 若H為G的subgroup，則H對G的相異的left coset的數量稱為H在G上的index，記為[G:H]。 If H is a subgroup of a finite group G, then .

Lagrange’s Theorem 若H為finite groupG的subgroup，則 |G| = |H|[G:H]. Proof: Suppose Ha1,………, Hak are distinct cosets of H in G. Then k=[G:H] and |Ha1|+………+|Hak| = |G|. This implies k|H| = |G| since |H| = |Hai| for each i. So |G| = |H|[G:H].

Corollary 若H為G的subgroup，則|G|被|H|整除。也就是說 |H|為|G|的因子。

Normal subgroup Let H be a subgroup of G. Note that Hg may not be equal to gH. H is normal if gH=Hg for all g∈G. We write H  G. Example: {1}  G

Theorem Let K be a normal subgroup of G and let G/K={Ka|a∈K}.Then G/K is a group under Ka‧Kb=Kab. The group G/K of all cosets of K in G is called the quotient group (or factor group) of G by K. Example: Z/nZ  Zn

Theorem Let K be a normal subgroup of a finite group G. Then |G/K|=|G|/|K|=[G:K]

Basic facts Let K be a normal subgroup of a group G, and let a,bG. Then (1) Ka=Kb if and only if ab-1K. (2) Ka=K if and only if aK. (3) Ka  Kb = Kab. (4) K=K1 is the identity of G/K. (5) (Ka)-1=Ka-1. (6) (Ka)k=Kak for all kZ.

Example G=<a>,|a|=12, K=<a4> The cosets are K={1,a4,a8} Ka={a,a5,a9}=Ka5=Ka9 Ka2={a2,a6,a10}=Ka6=Ka10 Ka3={a3,a7,a11}=Ka7=Ka11 G/K=<Ka> The Cayley table G/K K Ka Ka2 Ka3 C4

Example Let K={,(12)(34),(13)(24),(14)(23)}. Show that KA4 . Find A4/K and write down the Cayley table. Solution: A4 = K  {(123),(132),(124),(142),(134),(143),(234),(243)} K=K=K (123)K=K(123)={(123),(243),(142),(134)} (132)K=K(132)={(132),(143),(234),(124)} =[K(123)]2 A4/K=<K(123)> A4/K K K(123) K(132)

Image and Kernel  : G  G1 homomorphism Image im() = (G) = { (g) | g∈G} Kernel ker() = { k∈G | (k) = 1} (G) G1 G 1 ker()

Theorem  : G  G1 homonorphism (1) (G) is a subgroup of G1 (2) ker() is a normal subgroup of G

Remark im() may not normal in G. Let H be a subgroup of G and H  G. Then  : H  G with (h) = h is homomorphism and (H) = H  G

Fundamental Theorem of Homomorphisms  : G  G1 homomorphism and onto. Then G/ker()  G1. Let K = ker(). Then  : G/K  G1 with (Ka) = (a) is an isomorphism.

Exercise 4.4 True or False 1,3,5,7,8 Exercise 5,11,20,21,24,27

Exercise 4.5 True or False 2,3 Exercise 2,8

Exercise 4.6 True or False 1,2 Exercise 5,13a,24