After studying the group of roots of unity, we now turn to a second fundamental finite example: residue classes modulo $n$ under addition. These groups are among the most important examples in algebra and number theory. They show how arithmetic can be performed inside a finite set while still satisfying all group axioms. The case $\mathbb{Z}_3$ is small enough to compute completely, but the same proof works for every positive integer $n$. In this lesson, students will prove that $(\mathbb{Z}_n,+_n)$ is a finite group. :::definition[Residue Class] Let $n\in\mathbb{N}$ with $n\geq2$. For $a\in\mathbb{Z}$, the $residue class$ of $a$ modulo $n$ is denoted by $\overline{a}$ and is defined by $$\begin{equation*} \overline{a}=\{a+kn:k\in\mathbb{Z}\}. \end{equation*}$$ The set of all residue classes modulo $n$ is $$\begin{equation*} \mathbb{Z}_n=\{\overline{0},\overline{1},\ldots,\overline{n-1}\}. \end{equation*}$$ ::: :::definition[Addition Modulo n] Let $n\in\mathbb{N}$ with $n\geq2$. Addition modulo $n$ on $\mathbb{Z}_n$ is defined by $$\begin{equation*} \overline{a}+_n\overline{b}=\overline{a+b}. \end{equation*}$$ This operation is called $addition modulo n$. ::: The operation $+_n$ adds representatives and then takes the residue class modulo $n$. For example, in $\mathbb{Z}_5$, $$\begin{equation*} \overline{3}+_5\overline{4}=\overline{7}=\overline{2}. \end{equation*}$$ The result is again a class in $\mathbb{Z}_5$. This is the finite version of closure under addition. :::theorem Let $n\in\mathbb{N}$ with $n\geq2$. Then $(\mathbb{Z}_n,+_n)$ is a finite group. ::: :::proof Given that $n\in\mathbb{N}$ with $n\geq2$ and $$\begin{equation*} \mathbb{Z}_n=\{\overline{0},\overline{1},\ldots,\overline{n-1}\}. \end{equation*}$$ To prove that $(\mathbb{Z}_n,+_n)$ is a finite group. [1] To prove closure. Let $\overline{a},\overline{b}\in\mathbb{Z}_n$. By definition, $$\begin{equation*} \overline{a}+_n\overline{b}=\overline{a+b}. \end{equation*}$$ Since $\overline{a+b}$ is a residue class modulo $n$, we get $$\begin{equation*} \overline{a}+_n\overline{b}\in\mathbb{Z}_n. \end{equation*}$$ [2] To prove associativity. Let $\overline{a},\overline{b},\overline{c}\in\mathbb{Z}_n$. Then $$\begin{align*} (\overline{a}+_n\overline{b})+_n\overline{c} &=\overline{a+b}+_n\overline{c}\\ &=\overline{a+b+c}\\ &=\overline{a}+_n\overline{b+c}\\ &=\overline{a}+_n(\overline{b}+_n\overline{c}). \end{align*}$$ Therefore addition modulo $n$ is associative. [3] To prove the existence of identity element. The element $\overline{0}\in\mathbb{Z}_n$ satisfies $$\begin{equation*} \overline{a}+_n\overline{0}=\overline{0}+_n\overline{a}=\overline{a} \quad \forall\, \overline{a}\in\mathbb{Z}_n. \end{equation*}$$ Therefore $\overline{0}$ is the identity element. [4] To prove the existence of inverse elements. Let $\overline{a}\in\mathbb{Z}_n$. Then $\overline{-a}\in\mathbb{Z}_n$ and $$\begin{equation*} \overline{a}+_n\overline{-a}=\overline{-a}+_n\overline{a}=\overline{0}. \end{equation*}$$ Therefore every element has an additive inverse in $\mathbb{Z}_n$. Since $\mathbb{Z}_n$ has $n$ elements, the group is finite. Hence, $(\mathbb{Z}_n,+_n)$ is a finite group. $\square$ ::: :::example For $n=3$, $$\begin{equation*} \mathbb{Z}_3=\{\overline{0},\overline{1},\overline{2}\}. \end{equation*}$$ The addition table is $$\begin{equation*} \begin{array}{c|ccc} +_3 & \overline{0} & \overline{1} & \overline{2}\\ \hline \overline{0} & \overline{0} & \overline{1} & \overline{2}\\ \overline{1} & \overline{1} & \overline{2} & \overline{0}\\ \overline{2} & \overline{2} & \overline{0} & \overline{1} \end{array} \end{equation*}$$ The identity is $\overline{0}$, and the inverse of $\overline{1}$ is $\overline{2}$. ::: :::example In $\mathbb{Z}_8$ under addition, the inverse of $\overline{3}$ is $\overline{5}$ because $$\begin{equation*} \overline{3}+_8\overline{5}=\overline{8}=\overline{0}. \end{equation*}$$ The order of $\overline{3}$ is $8$ because $$\begin{equation*} 8\overline{3}=\overline{24}=\overline{0} \end{equation*}$$ and no smaller positive multiple gives $\overline{0}$. ::: :::solved-problem Find the inverse of $\overline{7}$ in $(\mathbb{Z}_{12},+_{12})$. ::: :::solution Let $\overline{7}\in\mathbb{Z}_{12}$. To find the inverse of $\overline{7}$ under addition modulo $12$. We need $\overline{x}\in\mathbb{Z}_{12}$ such that $$\begin{equation*} \overline{7}+_{12}\overline{x}=\overline{0}. \end{equation*}$$ This means $$\begin{equation*} \overline{7+x}=\overline{0}. \end{equation*}$$ Taking $x=5$, we get $$\begin{equation*} \overline{7}+_{12}\overline{5}=\overline{12}=\overline{0}. \end{equation*}$$ Therefore the inverse of $\overline{7}$ is $\overline{5}$. Hence, $$\begin{equation*} -\overline{7}=\overline{5} \end{equation*}$$ in $\mathbb{Z}_{12}$. $\square$ ::: :::solved-problem Find the order of $\overline{4}$ in $(\mathbb{Z}_{10},+_{10})$. ::: :::solution Let $\overline{4}\in\mathbb{Z}_{10}$. To find the order of $\overline{4}$ under addition modulo $10$. We need the least positive integer $k$ such that $$\begin{equation*} k\overline{4}=\overline{0}. \end{equation*}$$ This means $$\begin{equation*} \overline{4k}=\overline{0}, \end{equation*}$$ or equivalently, $$\begin{equation*} 10\mid4k. \end{equation*}$$ The least positive such $k$ is $$\begin{equation*} k=\frac{10}{\operatorname{gcd}(4,10)}=\frac{10}{2}=5. \end{equation*}$$ Hence, $$\begin{equation*} o(\overline{4})=5. \end{equation*}$$ $\square$ ::: :::remark The additive group $(\mathbb{Z}_n,+_n)$ exists for every $n\geq2$. Multiplication modulo $n$ on all of $\mathbb{Z}_n$ is different: it is not usually a group because $\overline{0}$ has no multiplicative inverse. This distinction between addition modulo $n$ and multiplication modulo $n$ is a frequent source of errors. ::: Addition in $\mathbb{Z}_n$ moves around a finite cycle of residues. We observe the running sums $a,2a,3a,\ldots$ modulo $n$ until the identity residue $0$ appears. Change $n$ and $a$ to see how the additive inverse and the order of $\overline{a}$ are controlled by modular arithmetic. The first return to $0$ gives the order of $\overline{a}$. :::scientific-preview[Modulo Addition Cycle Explorer] @libraries: plotly ::: Use the calculator to compute addition and inverses in $\mathbb{Z}_n$. Enter $n$, $a$, and $b$. The calculator gives $\overline{a}+_n\overline{b}$, the additive inverse of $\overline{a}$, and the order of $\overline{a}$. This supports the group proof by making identity and inverse behavior explicit. :::calculator[Addition Modulo n] ::: :::exercise [1] Define $\mathbb{Z}_n$. [2] Prove that $\overline{0}$ is the identity element of $(\mathbb{Z}_n,+_n)$. [3] Find the inverse of $\overline{9}$ in $\mathbb{Z}_{14}$. [4] Find the order of $\overline{6}$ in $\mathbb{Z}_{15}$. [5] Write the addition table of $\mathbb{Z}_2$. ::: :::answer [1] $\mathbb{Z}_n=\{\overline{0},\overline{1},\ldots,\overline{n-1}\}$. [2] For every $\overline{a}\in\mathbb{Z}_n$, $\overline{a}+_n\overline{0}=\overline{0}+_n\overline{a}=\overline{a}$. [3] The inverse is $\overline{5}$ because $\overline{9}+_{14}\overline{5}=\overline{0}$. [4] $o(\overline{6})=15/\operatorname{gcd}(6,15)=5$. [5] The table has $\overline{0}+_2\overline{0}=\overline{0}$, $\overline{0}+_2\overline{1}=\overline{1}$, $\overline{1}+_2\overline{0}=\overline{1}$, and $\overline{1}+_2\overline{1}=\overline{0}$. ::: :::faq Q: Is $(\mathbb{Z}_n,+_n)$ always a group? A: Yes. It is a finite group for every integer $n\geq2$. Q: What is the identity element in $\mathbb{Z}_n$ under addition? A: The identity element is $\overline{0}$. Q: What is the inverse of $\overline{a}$? A: The inverse is $\overline{-a}$. Q: How many elements does $\mathbb{Z}_n$ have? A: It has $n$ elements. ::: :::call-to-action[Permutation Groups] subtitle: Continue to permutation groups, where the operation is composition of functions. button: Next Lesson | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/permutation-groups button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/roots-of-unity-groups :::