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Introduction to Abstract Algebra

by: Kavon Feest

Introduction to Abstract Algebra MATH 113

Marketplace > University of California - Berkeley > Mathematics (M) > MATH 113 > Introduction to Abstract Algebra
Kavon Feest

GPA 3.93

A. Ogus

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A. Ogus
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This 4 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 113 at University of California - Berkeley taught by A. Ogus in Fall. Since its upload, it has received 21 views. For similar materials see /class/226582/math-113-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.

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Date Created: 10/22/15
Even and odd permutations March 7 2008 Let S be a nite set Recall that any permutation a E SymS can be written as a product of disjoint cycles a 7172quot3977 Furthermore this expression is unique up to reorderingl Here we donlt allow any 71 to be the identity permuationl Recall also that if 7 is a cycle of length Z gt 0 then 7 can be written as a product of E E l transpositions De nition 1 Let a be a permuation of a nite set S and write a as a product of disjoint cycles a 7172quot3977 Then Na Z171Z271Z71Z1Z7r where i is the length of 7139 Thus if i is the length of the cycle 71 above then a can be written as a product of Na transpositionsl However such an expression is not unique and in fact even the number of transpositions in such an expression is not unique However the following is true Theorem 1 Suppose that a is written as a product ofm transpositions a Tlfgnfm Then m E Na mod 2 Since congruence is an equivalence relation it follows that if a is also written as a product of m transpositions a Tif 41 then m E m mod 2 Theorem 1 follows from the following more suggestive result Theorem 2 If a and B are permuations of S then No E Na mod 2 lndeed note that if 739 is a transposition then NT 1 Hence it follows from Theorem 2 that NaN73917392TmEll1 m mod2 Note that in fact we only needed to apply Theorem 2 when 6 was a transposition but in fact the general case of Theorem 2 follows by induction from this case anyway Hint write 6 as a product of transpositions and use the associative law Let us prepare for the proof of Theorem 2 by means of some calculations Lemma 1 If 71 and 72 are two cycles with exactly one element in common then 7172 is a cycle of length 1 5 7 l where i is the length of 7139 Proof Actually I think it is convincing enough to compute a typical example 123445671234567 1 Lemma 2 If 739 is a transposition a b and y is a cycle containing both a and b then 7739 is a product of disjoint cycles 7172 where 1 Z2 Z the length of 7 Proof Again an example should be convincing 123456782512678345 D Proof of Theorem 2 It suf ces to prove the theorem when 6 is a transposition 739 Since NT l we have to prove that Na7 E Na 1 mod Write a as a product of disjoint cycles a 7172 4T so by de nition Na 171nZT71 Case 1 739 is disjoint from all the 7139 Then 1739 7172 7T7 is a product of disjoint cycles and so by de nition NaTZ1139Zr 12 1Na1 Case 2 739 meets just one of the y s in just one element We might as well assume that 739 meets yr and no other Then by Lemma 1 WT is a cycle 7 of length of T 1 Then 1739 7172 7T717 as a product of disjoint cycles so Nar Z171Z271mh471471 1712 1 l quot39Zrilil l zr l lil Nal Case 3 739 meets two of the ns Again we may as well assume that it meets 71 and VT necessarily it meets each in exactly one element Then 7 2 WT is a cycle of length Z T 1 which now contains 7quot Hence 71 I 717 is a cycle of length T71 Z T 71 T71 ZTi Hence a7 71 41 as a product of disjoint cycles7 and Na7 lilJperTilil 171m pl 71 Z171mZP171ZTNa1i Case 4 739 meets one of the ms in two elements Thus we assume that 739 I a b where a and I both occur in 7139 and we may as well assume that i Ti Then 7T7 is a product of two disjoint cycles 771 and Z NJ Ti Hence Na7 Z171Z71Z171 Z171ZT72Na71i Since Na 7 1 E Na 1 mod 2 the result holds in this case tool 1 Cyclicity Theorem Let G be a nite group Then the following conditions are equiv alent 1 G is cyclic 2 For each d E Z7 the number of g E G such that gd e is less than or equal to d 3 For each d E Z4 7 G has at most one subgroup of order d 4 For each d E Z42 G has at most d elements of order d Proof Suppose that G is cyclic of order n lfd E Z4 7 let d gcdd7 n and write d do and n d m Clearly if gd 6 then also 9 1 e Moreover7 since there exist integers Ly such that d zdyn and g 6 9d gmd so gd 6 implies also that gd 6 Thus gd 6 iff gd 6 Now if go generates g the set of all such 9 is just the subgroup of G generated by 96quot which has d elernents Thus 1 implies Suppose that 2 holds and d E Z Let H be a subgroup of G of order d Then 9 1 e for every 9 E G According to 27 there are at most d such elements But then H g E G 9 1 e and hence H is unique Suppose 3 holds If there are no elements of order d7 then there is nothing to check If g is an element of order d7 then 9 is a subgroup of order d7 and by 37 it is the unique such subgroup Hence if g is any element of order d7 9 E Since 9 contains exactly d elements of order d7 we see that G has exactly d elements of order d Suppose that 4 holds For each divisor d of the order of G7 let md denote the number of elements of G of order d Looking at the partition of the group G obtained by grouping together elements of the same order7 we see that the sum of all md is equal to the order of G For exarnple7 if G Zn7 md d if dln and md 0 otherwise Thus Ed n d 71 If G is a group of order n and satis es 3 we nd that n Zmltdgt Z gtltdgt n dln dln Since each 0 S md S d for each d7 we see that the equality Ed nmw Ed n d implies that each md d for every d In particular n 31 0 This means that G has at least one elrneent of order 717 and hence is cyclic 1


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