
#1
Dec109, 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???




#2
Dec109, 10:06 PM

P: 257

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.) 



#3
Dec209, 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. 



#4
Dec209, 07:56 AM

P: 257

isomorphic groups
Try proving that the group on the right is cyclic.




#5
Dec209, 08:15 AM

P: 3

Jeez i cant even do that.
Im having a terrible day with this. 



#6
Dec209, 08:53 AM

P: 257

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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