### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Introduction to Abstract Algebra MATH 113

GPA 3.93

### View Full Document

## 21

## 0

## Popular in Course

## Popular in Mathematics (M)

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.

## Reviews for Introduction to Abstract Algebra

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

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

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.