Hello, I have a basic (I think) question about Roots of Unity. First I feel like I should mention that I have only the most basic understanding of complex numbers. Here is what I have been asked to prove:
An th root of unity is a complex number such that . Prove that the th roots of unity form a cyclic subgroup of of order .
This is my first exposure to roots of unity, but after looking through a number theory book, I feel like the first sentence should include "...where . Am I right about this?
If I am correct about this then how do I prove that each th root of unity has an inverse?
Then, how do I express the identity element of ?
I am feeling pretty lost here. Thanks in advance.
January 22nd 2011, 01:36 PM
The roots of unity are .
Now is the identity.
Also is the generator.
January 22nd 2011, 01:38 PM
Well, if you can assume that is a group, then you only have two things you have to prove: closure and inverses.
Yes, I would definitely say that
For inverses, think about a root of unity, where What do you have to multiply by to get
For closure, the product of two roots of unity must be a root of unity. So, take and The product is What happens when you raise this product to the th power?
January 22nd 2011, 01:53 PM
The problem isn't just showing that each root has an inverse (all roots are nonzero and so have inverses in C), but that each inverse is also a root of unity. But that's really not too hard if you know enough about complex numbers. Specifically, I'm thinking that for any complex number with absolute value 1, , combined with the fact that and so ...
If this is more than you know, then it's not hard to list all nth roots of unity explicitly and then figure out the inverses that way. Also, if you know how the nth roots of unity are located in the complex plane and how multiplication of complex numbers adds angles and multiplies absolute values, then finding at least one generator shouldn't be too bad.
If you're not sure about the geometric view of complex numbers and their operations I'd suggest learning that because it makes problems like this so much easier to visualize.
January 22nd 2011, 03:19 PM
How is this? We are still in the first week of this class (abstract algebra) so I'm not sure how rigorous he expects us to be.
Let be the set of th roots of unity, with elements and . Then and . If we multiply both sides of on the right by we have
But since we have and therefore is an th root of unity. Hence, .
Since 1 is an identity element in and , the identity element in is also in .
Finally, let the inverse of be . Since , we have that . Hence, .
January 22nd 2011, 04:34 PM
You need to show that not
January 23rd 2011, 07:07 AM
Ah yes, well that is a glaring mistake! Thanks for your help. I've got a feeling that I'll have several more questions throughout the semester.
Finally, let the inverse of be . Since , we have that is an th root of unity. Hence, .
January 24th 2011, 05:24 AM
Right. That looks better.
January 26th 2011, 06:41 AM
I believe that I have found a problem with my proof. While I have shown that a set consisting of the th roots of unity form a subgroup of , I have not shown that is cyclic, or that it is of order .
I am kinda stumped as to how I should go about showing this. While it seems obvious that my proof holds for any powers of , the issue seems that the order of could be infinite.
January 26th 2011, 06:47 AM
Take a closer look at Plato's post # 2. There's a generator there.
January 26th 2011, 07:28 AM
As for possibly being infinite, there are several ways to show that it isn't. You can argue that there are only finitely many places the root can lie on the unit circle. For example, one 6th root of unity makes an angle of 360/6 degrees with the real axis. This will "go around" the circle once on its trip to 1. But a point can also go around twice. This would correspond to an angle of 360/3. Sooner or later you start to repeat points.
Also, you can show that an infinite cyclic group is isomorphic to . In that group the equation would be written using additive notation. This equation only has one solution in , whereas the corresponding equation in the group of roots of unity has at least n.
Of course the easiest way is just to invoke the fact that an nth degree polynomial in any field can have at most n zeros. If you're allowed to assume this then it's the easiest way to go.