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Isomorphic groups

by razorg425
Tags: groups, isomorphic
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razorg425
#1
Dec1-09, 08:46 AM
P: 3
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???
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rochfor1
#2
Dec1-09, 10:06 PM
P: 256
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.)
razorg425
#3
Dec2-09, 04:49 AM
P: 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.

rochfor1
#4
Dec2-09, 07:56 AM
P: 256
Isomorphic groups

Try proving that the group on the right is cyclic.
razorg425
#5
Dec2-09, 08:15 AM
P: 3
Jeez i cant even do that.
Im having a terrible day with this.
rochfor1
#6
Dec2-09, 08:53 AM
P: 256
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.


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