# Abstract Algebra: List elements of Subgroup

srfriggen

## Homework Statement

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

## 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?

jmjlt88
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? =)

srfriggen
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!

jmjlt88
No problem. =)