After reviewing groups from number systems, we now study a finite group that lives inside the complex numbers. The roots of unity are complex numbers whose fixed positive power equals $1$. They form a standard bridge between complex numbers, finite groups, cyclic groups, and geometry on the unit circle. This example is important because it is finite, concrete, and rich enough to show closure, inverses, order of elements, and cyclic generation. In this lesson, students will prove that the set of all $n$-th roots of unity is a finite group under multiplication. :::definition[Roots of Unity] Let $n\in\mathbb{N}$. A complex number $z\in\mathbb{C}$ is called an $n-th root of unity$ if $$\begin{equation*} z^n=1. \end{equation*}$$ The set of all $n$-th roots of unity is $$\begin{equation*} U_n=\{z\in\mathbb{C}:z^n=1\}. \end{equation*}$$ ::: The roots of unity are located on the unit circle in the complex plane. They are $$\begin{equation*} 1,\; e^{\frac{2\pi i}{n}},\; e^{\frac{4\pi i}{n}},\;\ldots,\; e^{\frac{2\pi i(n-1)}{n}}. \end{equation*}$$ Equivalently, $$\begin{equation*} U_n=\left\{e^{\frac{2\pi ik}{n}}:k=0,1,2,\ldots,n-1\right\}. \end{equation*}$$ Thus $U_n$ has exactly $n$ elements. :::theorem Let $n\in\mathbb{N}$ and let $$\begin{equation*} U_n=\{z\in\mathbb{C}:z^n=1\}. \end{equation*}$$ Then $(U_n,\cdot)$ is a finite group under ordinary complex multiplication. ::: :::proof Given that $n\in\mathbb{N}$ and $$\begin{equation*} U_n=\{z\in\mathbb{C}:z^n=1\}. \end{equation*}$$ To prove that $(U_n,\cdot)$ is a finite group. [1] To prove closure. Let $z_1,z_2\in U_n$. Then $$\begin{equation*} z_1^n=1,\quad z_2^n=1. \end{equation*}$$ Since complex multiplication is commutative and associative, $$\begin{align*} (z_1z_2)^n &=z_1^n z_2^n\\ &=1\cdot1\\ &=1. \end{align*}$$ Therefore $z_1z_2\in U_n$. [2] To prove associativity. Let $z_1,z_2,z_3\in U_n$. Since complex multiplication is associative, $$\begin{equation*} (z_1z_2)z_3=z_1(z_2z_3). \end{equation*}$$ [3] To prove the existence of identity element. Since $$\begin{equation*} 1^n=1, \end{equation*}$$ we get $1\in U_n$. Also $$\begin{equation*} z\cdot1=1\cdot z=z \quad \forall\, z\in U_n. \end{equation*}$$ Therefore $1$ is the identity element. [4] To prove the existence of inverse elements. Let $z\in U_n$. Then $$\begin{equation*} z^n=1. \end{equation*}$$ Since $z^n=1$, we have $z\neq0$. Also $$\begin{align*} z\cdot z^{n-1} &=z^n\\ &=1. \end{align*}$$ Similarly, $$\begin{equation*} z^{n-1}\cdot z=1. \end{equation*}$$ Thus $z^{n-1}$ is the inverse of $z$. Since $$\begin{align*} (z^{n-1})^n &=(z^n)^{n-1}\\ &=1, \end{align*}$$ we get $z^{n-1}\in U_n$. Therefore every element of $U_n$ has an inverse in $U_n$. Finally, $$\begin{equation*} U_n=\left\{e^{\frac{2\pi ik}{n}}:k=0,1,2,\ldots,n-1\right\}, \end{equation*}$$ so $U_n$ has $n$ elements. Hence, $(U_n,\cdot)$ is a finite group. $\square$ ::: The proof corrects a subtle point that students sometimes skip. It is not enough to show that $z\cdot z^{n-1}=1$. One must also show that the proposed inverse $z^{n-1}$ belongs to the same set $U_n$. This is verified by checking $(z^{n-1})^n=1$. Membership in the original set is part of the inverse axiom. :::example For $n=4$, $$\begin{equation*} U_4=\{1,i,-1,-i\}. \end{equation*}$$ Under multiplication, this is a group. The identity element is $1$, the inverse of $i$ is $-i$, and the inverse of $-1$ is $-1$. The element $i$ has order $4$, because $$\begin{equation*} i^4=1 \end{equation*}$$ and no smaller positive power of $i$ equals $1$. ::: :::example For $n=3$, $$\begin{equation*} U_3=\left\{1,\omega,\omega^2\right\}, \end{equation*}$$ where $$\begin{equation*} \omega=e^{\frac{2\pi i}{3}}. \end{equation*}$$ Then $$\begin{equation*} \omega^3=1 \end{equation*}$$ and $$\begin{equation*} 1+\omega+\omega^2=0. \end{equation*}$$ The group operation is ordinary multiplication, and $\omega$ is a generator of the group. ::: :::solved-problem Find the inverse of $e^{\frac{2\pi ik}{n}}$ in $U_n$. ::: :::solution Let $$\begin{equation*} z=e^{\frac{2\pi ik}{n}}\in U_n. \end{equation*}$$ To find the inverse of $z$ in $U_n$. Since $z^n=1$, we get $$\begin{equation*} z^{-1}=z^{n-1}. \end{equation*}$$ Also, $$\begin{align*} z^{-1} &=e^{-\frac{2\pi ik}{n}}\\ &=e^{\frac{2\pi i(n-k)}{n}}. \end{align*}$$ Therefore the inverse of $e^{\frac{2\pi ik}{n}}$ is $$\begin{equation*} e^{-\frac{2\pi ik}{n}}. \end{equation*}$$ Hence, $$\begin{equation*} \left(e^{\frac{2\pi ik}{n}}\right)^{-1}=e^{-\frac{2\pi ik}{n}}. \end{equation*}$$ $\square$ ::: :::solved-problem Determine the order of $e^{\frac{2\pi i}{6}}$ in $U_6$. ::: :::solution Let $$\begin{equation*} \zeta=e^{\frac{2\pi i}{6}}. \end{equation*}$$ To determine the order of $\zeta$ in $U_6$. We compute $$\begin{equation*} \zeta^6=e^{2\pi i}=1. \end{equation*}$$ If $1\leq r<6$, then $$\begin{equation*} \zeta^r=e^{\frac{2\pi ir}{6}}\neq1. \end{equation*}$$ Therefore the least positive integer $r$ such that $\zeta^r=1$ is $6$. Hence, $$\begin{equation*} o(\zeta)=6. \end{equation*}$$ $\square$ ::: :::remark The group $U_n$ is not just finite; it is cyclic. The element $$\begin{equation*} e^{\frac{2\pi i}{n}} \end{equation*}$$ generates all elements of $U_n$. This fact will become central when cyclic groups are studied in detail. For now, it gives a concrete example where powers of one element produce the entire group. ::: The roots of unity form equally spaced points on the complex unit circle. We observe the point $e^{2\pi ik/n}$ and its inverse $e^{-2\pi ik/n}$. Change $n$ and $k$ to see how multiplication by roots corresponds to adding exponents modulo $n$. This geometric picture explains why the set is closed under multiplication and why each root has an inverse inside the same set. :::scientific-preview[Roots of Unity Circle Explorer] @libraries: plotly ::: Use the calculator to list the exponents of the $n$-th roots of unity and find inverse exponents. Enter $n$ and $k$. The calculator reports the element $e^{2\pi ik/n}$, its inverse exponent, and the order of the selected root. This connects the geometric circle picture with modular arithmetic on exponents. :::calculator[Roots of Unity List] ::: :::exercise [1] Define $U_n$. [2] Prove that $U_n$ is closed under complex multiplication. [3] Find $U_2$. [4] Find $U_4$. [5] Find the inverse of $e^{\frac{2\pi i\cdot 3}{7}}$ in $U_7$. ::: :::answer [1] $U_n=\{z\in\mathbb{C}:z^n=1\}$. [2] If $z_1^n=1$ and $z_2^n=1$, then $(z_1z_2)^n=z_1^nz_2^n=1$. [3] $U_2=\{1,-1\}$. [4] $U_4=\{1,i,-1,-i\}$. [5] The inverse is $e^{-\frac{2\pi i\cdot 3}{7}}=e^{\frac{2\pi i\cdot 4}{7}}$. ::: :::faq Q: What is an $n$-th root of unity? A: It is a complex number $z$ satisfying $z^n=1$. Q: How many elements does $U_n$ have? A: It has exactly $n$ elements. Q: What is the identity element of $U_n$? A: The identity element is $1$. Q: Why is $U_n$ a group? A: It is closed under multiplication, multiplication is associative, $1$ is the identity, and every element has an inverse in $U_n$. ::: :::call-to-action[Additive Groups Modulo n] subtitle: Continue to residue classes modulo n, another fundamental finite group example. button: Next Lesson | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/additive-groups-modulo-n button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/groups-from-number-systems :::