Are Z21 and C2*C6 Isomorphic?

[razorg425]

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?

 

[rochfor1]

Start by proving that $$\mathbb{Z}_{21} \simeq \mathbb{Z}_3 \times \mathbb{Z}_7$$. What are the groups of units of $$\mathbb{Z}_3$$ and $$\mathbb{Z}_7$$?

(Hint: it's important that GCD(3,7)=1.)

 

[razorg425]

Sorry Rochfor,
I actually can't prove that, I know it should be true as gcd(3,7)=1.
A little more help please?
Thanks.

 

[rochfor1]

Try proving that the group on the right is cyclic.

 

[razorg425]

Jeez i can't even do that.
Im having a terrible day with this.

 

[rochfor1]

So we want to show that every element of $$\mathbb{Z}_3 \times \mathbb{Z}_7$$ is of the form $$n \cdot ( [1]_3, [1]_7 )$$. So for $$x, y \in \mathbb{Z}$$, we want $$( [x]_3, [y]_7 ) = n \cdot ( [1]_3, [1]_7 ) = ( [n]_3, [n]_7 )$$. So we need to find a number n so that $$x \equiv n \mod 3$$ and $$y \equiv n \mod 7$$. The http://mathworld.wolfram.com/ChineseRemainderTheorem.html" is your friend.