Abstract Algebra: List elements of Subgroup

In summary, U(20) is the group of units mod 20, which includes the elements 1, 3, 7, 9, 11, 13, 17, and 19. The cyclic subgroup generated by 3 is <3>={3,9,7,...} and the cyclic subgroup generated by 7 is <7>={7,9,3,...}. These subgroups are not equal to U(20), as some elements are missing. The correct elements for <3> and <7> are {3,9,7,1}, which can be found by continuing to multiply and mod out by 20. The mistaken answer in the book may be due to using addition
  • #1
srfriggen
307
6

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?
 
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  • #2
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? =)
 
  • #3
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!
 
  • #4
No problem. =)
 

Related to Abstract Algebra: List elements of Subgroup

1. What is a subgroup in abstract algebra?

A subgroup is a subset of a larger algebraic structure, such as a group, that also forms a group under the same operation. This means that the elements in the subgroup have closure, associativity, identity, and inverse properties.

2. How are subgroups determined in abstract algebra?

Subgroups are determined by finding a subset of the original group that satisfies the four properties of closure, associativity, identity, and inverse. These properties must hold for all elements in the subset in order for it to be considered a subgroup.

3. Can a subgroup have different operations than the original group?

No, a subgroup must have the same operation as the original group in order to maintain the group structure. However, the elements in the subgroup may be a different set of elements from the original group.

4. What is the significance of subgroups in abstract algebra?

Subgroups allow us to break down larger algebraic structures into smaller, more manageable groups. They also help us understand the structure and properties of the original group by studying the properties of the subgroup.

5. How do you list the elements of a subgroup in abstract algebra?

To list the elements of a subgroup, we need to first identify the subset that satisfies the four properties of closure, associativity, identity, and inverse. Then, we can list the elements in the subset, which will also be the elements in the subgroup.

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