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Math 343 Intro to Algebraic Structures Spring 2010 Homework 7 Solutions p. 110 14 a The cyclic subgroup of Z24 generated by 18 has order 4. b Z3 Z4 is of order 12. c The element 4 2 of Z12 Z8 has order 12. d The Klein 4-group is isomorphic to Z2 Z2. e Z2 Z Z4 has eight elements of nite order. p* 112 39 Let G be an abelian group* Show that the elements of nite order in G form a subgroup* This subgroup is called the torsion subgroup of G* Proof* Let T g G g has nite order. We will prove T is a subgroup of G* We rst note that the identity element e of G has nite order the order of e is 1. So e T. Now suppose a b T. This means there exist positive integers m and n such that am e and bn e. We will prove ab T. Since G is abelian we have ab mn amn bmn am n bn m e n e m e. The last equation implies ab has nite order that is ab T. Finally we show a 1 T. Consider a 1 m a m am 1 e 1 We conclude that T is a subgroup of G* p* 133 6 Let R R where R is additive and R is multiplicative be given by x 2....

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In a cyclic group of infinite order, identity has order 1 and all other elements have order . In a cyclic group of order , order of is n gcd ( n , k ) . Furthermore, the (distinct) elements which have order are { a d i : i ∈ Z n d ∗ } .

Group G is cyclic if there exists a∈G such that the cyclic subgroup generated by a, ⟨a⟩, equals all of G. That is, G={na|n∈Z}, in which case a is called a generator of G. The reader should note that additive notation is used for G. Z12=[Z12;+12], where +12 is addition modulo 12, is a cyclic group.

Only groups that have the same order and same properties (either both abelian or both nonabelian; either both cyclic or both noncyclic) can be isomorphic. Thus among the above groups, only Z12 and Z3 × Z4 be isomorphic. This two groups are indeed isomorphic by Proposition 3.4.

Z4 × Z6 is also isomorphic to the second group in the list. But then Z2 × Z12 and Z4 × Z6 are isomorphic.

A cyclic group is a group which is equal to one of its cyclic subgroups: G = ⟨g⟩ for some element g, called a generator of G. For a finite cyclic group G of order n we have G = {e, g, g2, ... , gn−1}, where e is the identity element and gi = gj whenever i ≡ j (mod n); in particular gn = g0 = e, and g−1 = gn−1.

Is Z10 ⊕ Z12 ⊕ Z6 ≈ Z15 ⊕ Z4 ⊕ Z12? No. Z15 ⊕ Z4 ⊕ Z12 ≈ Z3 ⊕ Z5 ⊕ Z4 ⊕ Z3 ⊕ Z4, and Z4 is not isomorphic to Z2 ⊕ Z2; one is cyclic the other isn't. Chapter 8: #62 Solution: U(165) ≈ U(3) ⊕ U(5) ⊕ U(11) ≈ Z2 ⊕ Z4 ⊕ Z10.

Since isomorphisms preserve orders of elements, we get that ϕ(1) ∈ Z3 ×Z9 has order 27. However, by Theorem 8.1, the maximum order of an element of Z3 × Z9 is 9, a contradiction. Hence, Z3 × Z9 is not isomorphic to Z27.

The order of Z4×Z6 is 24, and the order of Z24 is 24. However, the two are not isomorphic because 4 and 6 are not relatively prime.

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Living Trust
Name Change
Personal Planning
Small Business
Wills & Estates
Packages A-Z
Form Categories
Affidavits
Bankruptcy
Bill of Sale
Corporate - LLC
Divorce
Employment
Identity Theft
Internet Technology
Landlord Tenant
Living Wills
Name Change
Power of Attorney
Real Estate
Small Estates
Wills
All Forms
Forms A-Z
Form Library
Customer Service
Your Privacy Choices
Terms of Service
Privacy Notice
Legal Hub
Content Takedown Policy
Bug Bounty Program
About Us
Help Portal
Legal Resources
Blog
Affiliates
Contact Us
Delete My Account
Site Map
Industries
Forms in Spanish
Localized Forms
State-specific Forms
Forms Kit
Legal Guides
Real Estate Handbook
All Guides
Prepared for You
Notarize
Incorporation services
Our Customers
For Consumers
For Small Business
For Attorneys
Our Sites
US Legal Forms
USLegal
FormsPass
pdfFiller
signNow
altaFlow
DocHub
Instapage
Social Media
Call us now toll free:
+1 833 426 79 33
As seen in:
  • USA Today logo picture
  • CBC News logo picture
  • LA Times logo picture
  • The Washington Post logo picture
  • AP logo picture
  • Forbes logo picture
© Copyright 1997-2026
airSlate Legal Forms, Inc.
3720 Flowood Dr, Flowood, Mississippi 39232