| Thread Closed |
isomorphic groups |
Share Thread | Thread Tools |
| Dec1-09, 08:46 AM | #1 |
|
|
isomorphic groups
Hi guys just a quick question on how I would go about showing the units of Z21 is isomorphic toC2*C6(cyclic groups).I have done out the multiplicative table but they seem to be different to me. What else can I do???
|
| Dec1-09, 10:06 PM | #2 |
|
|
Start by proving that [tex]\mathbb{Z}_{21} \simeq \mathbb{Z}_3 \times \mathbb{Z}_7[/tex]. What are the groups of units of [tex]\mathbb{Z}_3[/tex] and [tex]\mathbb{Z}_7[/tex]?
(Hint: it's important that GCD(3,7)=1.) |
| Dec2-09, 04:49 AM | #3 |
|
|
Sorry Rochfor,
I actually cant prove that, I know it should be true as gcd(3,7)=1. A little more help please? Thanks. |
| Dec2-09, 07:56 AM | #4 |
|
|
isomorphic groups
Try proving that the group on the right is cyclic.
|
| Dec2-09, 08:15 AM | #5 |
|
|
Jeez i cant even do that.
Im having a terrible day with this. |
| Dec2-09, 08:53 AM | #6 |
|
|
So we want to show that every element of [tex]\mathbb{Z}_3 \times \mathbb{Z}_7[/tex] is of the form [tex]n \cdot ( [1]_3, [1]_7 )[/tex]. So for [tex]x, y \in \mathbb{Z}[/tex], we want [tex]( [x]_3, [y]_7 ) = n \cdot ( [1]_3, [1]_7 ) = ( [n]_3, [n]_7 )[/tex]. So we need to find a number n so that [tex]x \equiv n \mod 3[/tex] and [tex]y \equiv n \mod 7[/tex]. The Chinese Remainder Theorem is your friend.
|
| Thread Closed |
| Thread Tools | |
Similar Threads for: isomorphic groups
|
||||
| Thread | Forum | Replies | ||
| Isomorphic Quotient Groups in Z4 x Z4 | Calculus & Beyond Homework | 7 | ||
| Isomorphic Groups | Set Theory, Logic, Probability, Statistics | 0 | ||
| Isomorphic groups? | Linear & Abstract Algebra | 3 | ||
| isomorphic groups that have different properties? | Linear & Abstract Algebra | 3 | ||
| Showing Two Groups are Isomorphic | Linear & Abstract Algebra | 32 | ||