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Prove that the 12-th roots of unity in C form a cyclic group

AutGuy98

New member
Sep 11, 2019
20
Hey guys,

Sorry that it's been a decent amount of time since my last posting on here. Just want to say upfront that I am extremely appreciative of all the support that you all have given me over my last three or four posts. Words cannot express it and I am more than grateful for it all. But, in light of that, I actually have some more questions for an exercise set that I have to do for one of my classes and I'm really unsure how to begin doing them. There are four of them and they all require proofs to some degree. Anyway, I was going to make one post and put all four parts of the same question in it (i.e. 2(a),2(b),2(c), and 2(d)), but was unsure whether or not it would be allowed here. So, for those reasons and to play it safe rather than try to do so, here is the second part that I've been having trouble with. Any help here is, once again, greatly appreciated and will leave me forever further in your gratitude.

Question: 2(b): "Prove that the 12-th roots of unity in C form a cyclic group under multiplication (we denote it by C12). List the generators of this group."

Again, I have no idea where to start with this, so any help is extremely gracious and appreciated.

P.S. If possible at all, I'd need help on these by tomorrow, Wednesday, at 12:30 E.S.T., so please try to look this over at your earliest conveniences. Thank you all again for your help with everything already.
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
Hi AutGuy98 ,

The twelfth roots of unity are the solutions to the complex equation $z^{12} = 1$, which are $\exp(2\pi i k/12)$ for $k = 0, 1,\ldots, 11$. In particular they are of the form $\omega^k$ $(k\in \Bbb Z)$ where $\omega = \exp(2\pi i/12)$. Thus, $C_{12} = \{\omega^k : k\in \Bbb Z\}$ is a cyclic group (under multiplication) generated by $\omega$. Since there are 12 distinct elements of $C_{12}$ (namely, $1, \omega, \omega^2,\ldots, \omega^{11}$), the order of the group $C_{12}$ is $12$.

To find generators of $C_{12}$, think of the following. Each integer $k\pmod{12}$ identifies a unique element $\omega^{k}$ in $C_{12}$ (for $\omega^a = \omega^b \iff a \equiv b \pmod{12}$). So if you know the generators of the group of integers modulo 12, $\Bbb Z_{12}$, then you can find the corresponding generators for $C_{12}$.
 

AutGuy98

New member
Sep 11, 2019
20
Hi AutGuy98 ,

The twelfth roots of unity are the solutions to the complex equation $z^{12} = 1$, which are $\exp(2\pi i k/12)$ for $k = 0, 1,\ldots, 11$. In particular they are of the form $\omega^k$ $(k\in \Bbb Z)$ where $\omega = \exp(2\pi i/12)$. Thus, $C_{12} = \{\omega^k : k\in \Bbb Z\}$ is a cyclic group (under multiplication) generated by $\omega$. Since there are 12 distinct elements of $C_{12}$ (namely, $1, \omega, \omega^2,\ldots, \omega^{11}$), the order of the group $C_{12}$ is $12$.

To find generators of $C_{12}$, think of the following. Each integer $k\pmod{12}$ identifies a unique element $\omega^{k}$ in $C_{12}$ (for $\omega^a = \omega^b \iff a \equiv b \pmod{12}$). So if you know the generators of the group of integers modulo 12, $\Bbb Z_{12}$, then you can find the corresponding generators for $C_{12}$.
So would the generators of C12 then be (e1,e2,e3,...,e12)? I am unsure of how exactly to finish this off. Could you please either give me another hint or solve it for me, because I am limited on time and honestly have only a minute sense of what is going on in my Elementary Abstract Algebra class? I don't mean to sound rude or anything, it's just that, like I said, there is a lot of pressure on me right now about things for my other classes as well and getting this done before class tomorrow. Again, I apologize if that is how I seem because that is not how I want to sound. Please let me know when you can and thanks for everything so far Euge.
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,892
A generator of a group $G$ is an element $x\in G$ such that the cyclic subgroup generated by $x$ is all of $G$: $\langle x\rangle = G$. In the integers modulo $n$, the generators are given by positive integers not exceeding $n$ and relatively prime to $n$. If $n = 12$, these would be $1$, $5$, $7$, and $11$, right? So the corresponding generators of $C_{12}$ are $\omega, \omega^5, \omega^7$, and $\omega^{11}$. They are known as the primitive twelfth roots of unity.