# Thread: Show that this group is cyclic.

1. ## Show that this group is cyclic.

Ok, I have this:
$\displaystyle U_n = \sqrt[n]{1}$ where n $\displaystyle \in [1, \infty]$.
Show that $\displaystyle U_n$ is cyclic.
It is clearly cyclic as all it's elements are equal to 1, but how do I prove it.

2. ## Re: Show that this group is cyclic.

Originally Posted by Legjendat
Ok, I have this:
$\displaystyle U_n = \sqrt[n]{1}$ where n $\displaystyle \in [1, \infty]$.
Show that $\displaystyle U_n$ is cyclic.
It is clearly cyclic as all it's elements are equal to 1, but how do I prove it.
Are we to assume that $\displaystyle n\in\mathbb{Z}^+$, i.e. a positive integer?

3. ## Re: Show that this group is cyclic.

Originally Posted by Plato
Are we to assume that $\displaystyle n\in\mathbb{Z}^+$, i.e. a positive integer?
Yes. I mean n is an integer, we don't assume it as one, I didn't mention that.

4. ## Re: Show that this group is cyclic.

Originally Posted by Legjendat
Ok, I have this:
$\displaystyle U_n = \sqrt[n]{1}$ where n $\displaystyle \in [1, \infty]$.
Show that $\displaystyle U_n$ is cyclic.
OK, then $\displaystyle U_n\text{ is simply the multiplicative group of the }n, ~~n^{th}\text{-complex roots of unity.}$.

So $\displaystyle U_n$ is generated by $\displaystyle \rho=\exp\left(\frac{2\pi i}{n}\right)$

5. ## Re: Show that this group is cyclic.

Originally Posted by Legjendat
Ok, I have this:
$\displaystyle U_n = \sqrt[n]{1}$ where n $\displaystyle \in [1, \infty]$.
Show that $\displaystyle U_n$ is cyclic.
It is clearly cyclic as all it's elements are equal to 1, but how do I prove it.
IF all its elements were "equal to 1" this would be trivial. But it's not true. What is true is that the absolute value of all elements of $\displaystyle U_n$ is 1.

6. ## Re: Show that this group is cyclic.

Actually, given the notation used by the OP, $\displaystyle U_n = 1$ for all $\displaystyle n$. Even if the OP wrote $\displaystyle U_n = \{x \in \mathbb{C} \mid x = \sqrt[n]{1}\}$, then $\displaystyle U_n = \{1\}$ since $\displaystyle \sqrt[n]{1} = 1$ is distinct in $\displaystyle \mathbb{C}$. Now, if $\displaystyle U_n = \{x\in \mathbb{C} \mid x^n = 1\}$, then we still would only have $\displaystyle U_n$ being the $\displaystyle n$-th roots of unity in the case that $\displaystyle n$ is rational. But the OP is allowing $\displaystyle n \in [1,\infty]$, which is an interval of the extended reals. We must assume that $\displaystyle \sqrt[\infty]{x}$ is defined and $\displaystyle \sqrt[\infty]{1} = 1$.

7. ## Re: Show that this group is cyclic.

Originally Posted by SlipEternal
Now, if $\displaystyle U_n = \{x\in \mathbb{C} \mid x^n = 1\}$, then we still would only have $\displaystyle U_n$ being the $\displaystyle n$-th roots of unity in the case that $\displaystyle n$ is rational. But the OP is allowing $\displaystyle n \in [1,\infty]$, which is an interval of the extended reals. We must assume that $\displaystyle \sqrt[\infty]{x}$ is defined and $\displaystyle \sqrt[\infty]{1} = 1$.
That issue is addressed in replies #2 & #3.
He clearly states that $\displaystyle n\in\mathbb{Z}^+$.

8. ## Re: Show that this group is cyclic.

Originally Posted by Plato
That issue is addressed in replies #2 & #3.
He clearly states that $\displaystyle n\in\mathbb{Z}^+$.
Sorry, I skimmed the beginning. Still, $\displaystyle U_n$ is not a set as it is written and $\displaystyle \sqrt[n]{1} = 1$ remain valid issues.

9. ## Re: Show that this group is cyclic.

Originally Posted by SlipEternal
Still, $\displaystyle U_n$ is not a set as it is written and $\displaystyle \sqrt[n]{1} = 1$ remain valid issues.
Well, we are dealing with beginning students.

And this is such a well known result for the n complex nth roots of unity.

I am confident that is what $\displaystyle U_n$ is meant to be.