What is the group $\langle U, * \rangle$ where $U$ is the set of roots of unity and * is normal multiplication? I'm having trouble understanding my textbook in this regard. Everything else seems to make sense, such as the group $\langle \mathbb{Z}_{1000}, + \rangle$ to list an example. 
In the text, it is stated that $\langle U, * \rangle$ contains two elements that are their own inverse (namely, $-1$ and $1$), but what does that mean in this scope? I am having a tough time picturing it. Sure, $1*1 = 1$ and $-1 * -1 = 1$, but is that what they're getting at? Doesn't $U$ have some relation to the unit circle, that I should be factoring in? I understand $\langle U_4, * \rangle$ having four solutions to $z^4 = 1$, but where does the $U$ versus $U_4$ come in similarly? I'm just having a tough time comprehending $U$ and would love for some insight into it. 
 A: Guide: The main crux of the answer to your question begins with the bold words "The main point is outlined ...". Nevertheless, I recommend carefully reading the material up to this point, in particular Exercises 1 and 2 below, because if you understand the solutions to those Exercises, then I think you will have an answer to your question.
Let $(U,\ast)$ denote the multiplicative group consisting of all roots of unity in $(\mathbb{C}^{\times},\cdot)$ (the multiplicative group of non-zero complex numbers). In words, $U$ consists of all (necessarily non-zero) complex numbers $z$ such that $z^n=1$ for some natural number $n\geq 1$. 
Exercise 1: Prove that $(U,\ast)$ is a subgroup of $(\mathbb{C}^{\times},\cdot)$. (Note that I have used $\ast$ and $\cdot$ to indicate the multiplications on $U$ and $\mathbb{C}^{\times}$, respectively. The reason is that $\ast:U\times U\to U$ and $\cdot:\mathbb{C}^{\times}\times \mathbb{C}^{\times}\to \mathbb{C}^{\times}$ are different functions because they have different domains. Essentially, one component of this exercise asks you to verify that the function $\ast$ is the restriction of the function $\cdot$; but this should be obvious.)
Of course, we also have groups $(U_n,\ast)$ for all natural numbers $n\geq 1$. (I denote the multiplications on all of these groups by the same symbol; technically, this is an abuse of notation as I explained above but this should not cause any confusion.) In words, $U_n$ consists of all (necessarily non-zero) complex numbers $z$ such that $z^n=1$. 
Exercise 2: Prove that $(U_n,\ast)$ is a subgroup of $(U,\ast)$ for all natural numbers $n\geq 1$. By transitivity of the relation "is a subgroup of", it follows from Exercise 1 that $(U_n,\ast)$ is a subgroup of $(\mathbb{C}^{\times},\cdot)$.
The main point is outlined in the form of the following Exercise:
Exercise 3: Prove that as sets $U=\bigcup_{n=1}^{\infty} U_n$ but that this is not an ascending union, i.e., $U_n\not\subseteq U_{n+1}$ for all $n\geq 1$. For which $n,m\geq 1$ is it true that $U_n\subseteq U_m$?
In fact, we have more information than is strictly contained in Exercise 3; the $U_n$'s are all groups and their union is also a group. What does this mean/imply? The main thing relevant to your question is that every element $z\in U$ is an element of $U_n$ for some $n$ and, in this case, the element $z^{-1}\in U$ will also be an element of $U_n$. (If this is not obvious, prove it! It's formally a consequence of Exercise 2; that $U_n$ is a subgroup of $U$.)
Similarly, if $z\in U$, then the inverse of $z$ in $\mathbb{C}^{\times}$ is also an element of $U$. (Again, because formally $U$ is a subgroup of $\mathbb{C}^{\times}$!) 
Exercise 4: Let $S$ and $T$ be the sets of elements in $\mathbb{C}^{\times}$ and $U$, respectively, consisting of those elements equal to their own inverses. Prove that $S\cap U=T$.
Exercise 5: What is $S$? Does that answer your question in conjunction with Exercise 4?
The role of the group $(\mathbb{C}^{\times},\cdot)$ in this situation can also be played by the circle group $(\mathbb{S}^{1},\cdot)$ consisting of all complex numbers $z$ with $\left|z\right|=1$. $U$ is a subgroup of the circle group and the circle group is, in turn, a subgroup of $\mathbb{C}^{\times}$. 
I hope this helps! 
A: The set $U$ of unit roots is the set of all numbers $z$ so that $z^n=1$ for some positive $n\in \mathbb N$. Equivalently, it is the set of all numbers $\mathrm e^{2\pi\mathrm iq}$ where $q$ is a rational number, Proof: Be $q=m/n$, then $(\mathrm e^{2\pi\mathrm iq})^n = \mathrm e^{2\pi\mathrm inq} = \mathrm e^{2\pi\mathrm im}=1$.
From that, we can see that $U$ is a dense subset of the unit circle.
Also we can see that $\langle U,\cdot\rangle$ is closed under multiplication by noting that $\mathrm e^{2\pi\mathrm iq}\cdot\mathrm e^{2\pi\mathrm ir} = \mathrm e^{2\pi\mathrm i(q+r)}$. Clearly $q+r$ is also rational.
Similarly we can prove that it is a group by noting that the multiplicative inverse on $U$ maps on the additive inverse (negative) on $\mathbb Q$. Associativity and neutral element are obvious. Indeed, it is an abelian group.
It is also not hard to see this way that $U$ is isomorphic to the additive group $\mathbb Q/\mathbb Z$.
Clearly $U_n\subset U$ for all $n$. Since $U_n$ is a group by itself, it is a subgroup of $U$. Indeed, $U$ is the union of all $U_n$.
