Abstract Algebra: List elements of Subgroup

Click For Summary

Homework Help Overview

The problem involves identifying the elements of the subgroups generated by 3 and 7 in the group of units U(20), specifically under the operation of multiplication modulo 20.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to list the elements of the subgroups <3> and <7> and questions the accuracy of the answer key, suggesting a potential typo. Some participants clarify the operation involved and correct the original poster's assumption about the operation being addition instead of multiplication.

Discussion Status

Participants are exploring the definitions and properties of the group U(20) and its subgroups. Clarification has been provided regarding the operation used in the group, and there is an ongoing discussion about the elements of the subgroups generated by 3 and 7.

Contextual Notes

The original poster's confusion stems from the assumption that the operation is addition rather than multiplication, which is critical in determining the subgroup elements.

srfriggen
Messages
304
Reaction score
7

Homework Statement



List the elements of the subgroups <3> and <7> in U(20).



Homework Equations





The Attempt at a Solution



U(20)= {1, 3, 7, 9, 11, 13, 17, 19} = <3> = <7>.



So basically I have that the common elements of, <3> and <7> and U(20), under + modulo 20, are all the elements in U(20). Other than calculating this all out, I reasoned it to be true since 3 and 7 are both coprime to 20.

The answer key in the book has <3>=<7>= {3,9,7,1}.

I believe this is a typo, and that either the answer key is incorrect, or U(20) should be U(10).

Can someone please confirm my suspicion or let me know if I am missing something?
 
Physics news on Phys.org
Remember, U(n) is the group of units mod n and the operation is multiplication mod n. You are right that U(20)={1,3,7,9,11,13,17,19}. But, the cyclic subgroup generated by 3 is certainly not U(20). We note that 3(3) mod 20 = 9. Then, 3(9) mod 20 = 7 mod 20. [Remember, 3(9) = 27 mod 20 = 7]. Now, so far we see that <3>={3,9,7,...}. Next, multiply 7 by 3 and mod out by 20. What do we get? =)
 
jmjlt88 said:
Remember, U(n) is the group of units mod n and the operation is multiplication mod n. You are right that U(20)={1,3,7,9,11,13,17,19}. But, the cyclic subgroup generated by 3 is certainly not U(20). We note that 3(3) mod 20 = 9. Then, 3(9) mod 20 = 7 mod 20. [Remember, 3(9) = 27 mod 20 = 7]. Now, so far we see that <3>={3,9,7,...}. Next, multiply 7 by 3 and mod out by 20. What do we get? =)

I see. I was using addition as the operation. Thank you!
 
No problem. =)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 28 ·
Replies
28
Views
5K