After studying powers of elements in a group, the next natural question is when those powers return to the identity element. This leads to the order of an element. The order of an element measures the first positive power that gives the identity, when such a power exists. This idea is central in abstract algebra because it connects powers, cyclic subgroups, roots of identity, and later results such as Lagrange’s theorem. In this lesson, students will learn finite order, infinite order, and how to compute the order of elements in familiar groups. :::definition[Order] Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. If there exists a least positive integer $n$ such that $$\begin{equation*} a^n=e, \end{equation*}$$ then $a$ is said to have $finite order$. The least positive integer $n$ is called the $order$ of $a$ and is denoted by $$\begin{equation*} o(a)=n. \end{equation*}$$ ::: :::definition[Infinite Order] Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. If there does not exist any positive integer $n$ such that $$\begin{equation*} a^n=e, \end{equation*}$$ then $a$ is said to have $infinite order$. In many texts this is denoted by $$\begin{equation*} o(a)=\infty. \end{equation*}$$ Some older notes also write $o(a)=0$, but $o(a)=\infty$ is the clearer modern notation. ::: The word least in the definition of finite order is essential. If $a^6=e$, then it does not automatically mean that $o(a)=6$. A smaller positive power may already be the identity. For example, if $a^2=e$, then $a^6=(a^2)^3=e$, but the order is $2$, not $6$. Therefore the order is not just any exponent that gives the identity; it is the first positive exponent that gives the identity. :::example In the group $(\mathbb{Z},+)$, the identity element is $0$. The additive notation for powers is repeated addition. For $a=3$, we have $$\begin{equation*} n\cdot3=3n. \end{equation*}$$ There is no positive integer $n$ such that $3n=0$. Hence, $3$ has infinite order in $(\mathbb{Z},+)$. In fact, every nonzero integer has infinite order under addition, while $0$ has order $1$. ::: :::example In the multiplicative group $\{1,-1\}$, the identity element is $1$. We have $$\begin{equation*} 1^1=1 \end{equation*}$$ and $$\begin{equation*} (-1)^1=-1,\quad (-1)^2=1. \end{equation*}$$ Therefore $$\begin{equation*} o(1)=1,\quad o(-1)=2. \end{equation*}$$ ::: :::example Let $S=\{1,i,-1,-i\}$ under complex multiplication. The identity element is $1$. We compute: $$\begin{equation*} i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1. \end{equation*}$$ Since $4$ is the least positive integer for which $i^4=1$, we get $$\begin{equation*} o(i)=4. \end{equation*}$$ Also, $$\begin{equation*} o(-1)=2,\quad o(-i)=4,\quad o(1)=1. \end{equation*}$$ ::: :::theorem Let $(G,\circ)$ be a group with identity element $e$. Then $$\begin{equation*} o(e)=1. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$. To prove that $o(e)=1$. Since $e$ is the identity element, $$\begin{equation*} e^1=e. \end{equation*}$$ The positive integer $1$ is the least positive integer. Therefore the least positive integer $n$ such that $e^n=e$ is $1$. Hence, $$\begin{equation*} o(e)=1. \end{equation*}$$ $\square$ ::: :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. If $$\begin{equation*} o(a)=1, \end{equation*}$$ then $$\begin{equation*} a=e. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a\in G$. Given that $$\begin{equation*} o(a)=1. \end{equation*}$$ To prove that $a=e$. By the definition of order, $$\begin{equation*} a^1=e. \end{equation*}$$ Since $a^1=a$, we get $$\begin{equation*} a=e. \end{equation*}$$ Hence, $$\begin{equation*} a=e. \end{equation*}$$ $\square$ ::: :::solved-problem Determine the order of $-1$ in $(\mathbb{R}-\{0\},\cdot)$. ::: :::solution Let $(\mathbb{R}-\{0\},\cdot)$ be the multiplicative group of nonzero real numbers. To determine the order of $-1$. The identity element is $1$. Now $$\begin{equation*} (-1)^1=-1\neq1 \end{equation*}$$ and $$\begin{equation*} (-1)^2=1. \end{equation*}$$ Therefore the least positive integer $n$ such that $(-1)^n=1$ is $2$. Hence, $$\begin{equation*} o(-1)=2. \end{equation*}$$ $\square$ ::: :::solved-problem Determine the order of $2$ in $(\mathbb{Q}-\{0\},\cdot)$. ::: :::solution Let $(\mathbb{Q}-\{0\},\cdot)$ be the multiplicative group of nonzero rational numbers. To determine the order of $2$. The identity element is $1$. For every $n\in\mathbb{N}$, $$\begin{equation*} 2^n\neq1. \end{equation*}$$ Thus there is no positive integer $n$ such that $2^n=1$. Therefore $2$ has infinite order in $(\mathbb{Q}-\{0\},\cdot)$. Hence, $$\begin{equation*} o(2)=\infty. \end{equation*}$$ $\square$ ::: :::remark The order of an element always depends on the group operation. In an additive group, the identity is $0$, and order means the least positive integer $n$ such that $na=0$. In a multiplicative group, the identity is usually $1$, and order means the least positive integer $n$ such that $a^n=1$. Confusing these two notations is a common early mistake. ::: The additive group modulo $n$ gives a concrete model for finite order. We observe the running residues $a,2a,3a,\ldots$ modulo $n$ until the residue becomes $0$, the identity. Change the modulus and element to see how the first return to $0$ becomes the order. Notice that the same element can have different orders when the modulus changes, so order is always relative to the group operation and the group itself. :::scientific-preview[Additive Order Cycle Explorer] @libraries: plotly ::: Use the calculator to compute the order of an element under addition modulo $n$. Enter a modulus $n$ and an element $a$. The calculator searches for the least positive integer $k$ such that $ka\equiv0\pmod n$. This is the additive version of the order of an element. :::calculator[Order Modulo n] ::: :::exercise [1] Define the order of an element in a group. [2] What is the order of the identity element? [3] Find the order of $-1$ in $(\mathbb{Q}-\{0\},\cdot)$. [4] Find the order of $i$ in $\{1,i,-1,-i\}$ under multiplication. [5] Determine whether $5$ has finite or infinite order in $(\mathbb{Z},+)$. ::: :::answer [1] The order of $a$ is the least positive integer $n$ such that $a^n=e$, if such an integer exists. [2] The order of the identity element is $1$. [3] The order of $-1$ is $2$. [4] The order of $i$ is $4$. [5] The element $5$ has infinite order in $(\mathbb{Z},+)$. ::: :::faq Q: What does finite order mean? A: It means that some positive power of the element equals the identity, and the least such positive exponent is the order. Q: What does infinite order mean? A: It means that no positive power of the element equals the identity. Q: Can the identity element have order greater than $1$? A: No. The identity element has order $1$. Q: Does order depend on the operation? A: Yes. The order of an element is always computed relative to the group operation. ::: :::call-to-action[Order of the Inverse] subtitle: Continue to the order of the inverse element and prove that an element and its inverse have the same order. button: Next Lesson | /abstract-algebra/introduction-to-groups/order-of-an-element/order-of-inverse-element button-ghost: Previous Section | /abstract-algebra/introduction-to-groups/powers-of-elements-in-groups/power-equations-in-groups :::