After proving the divisibility criterion for powers equal to the identity, we now determine the order of a power $a^m$. If $a$ has finite order $n$, then the order of $a^m$ is controlled by the greatest common divisor of $m$ and $n$. This result is one of the most frequently used formulas in cyclic groups and finite group computations. It explains when a power of an element has the same order as the element and when the order becomes smaller. In this lesson, students will prove the formula $o(a^m)=\dfrac{n}{\operatorname{gcd}(m,n)}$ and apply it in examples. :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. If $o(a)=n$, then for every positive integer $m$, $$\begin{equation*} o(a^m)=\frac{n}{\operatorname{gcd}(m,n)}. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a\in G$. Given that $$\begin{equation*} o(a)=n. \end{equation*}$$ Let $m\in\mathbb{N}$. To prove that $$\begin{equation*} o(a^m)=\frac{n}{\operatorname{gcd}(m,n)}. \end{equation*}$$ Let $$\begin{equation*} d=\operatorname{gcd}(m,n). \end{equation*}$$ Then there exist relatively prime positive integers $r$ and $s$ such that $$\begin{equation*} m=dr,\quad n=ds,\quad \operatorname{gcd}(r,s)=1. \end{equation*}$$ We first prove that $(a^m)^s=e$. Now $$\begin{align*} (a^m)^s &=a^{ms}\\ &=a^{drs}\\ &=a^{rn}\\ &=(a^n)^r\\ &=e. \end{align*}$$ Therefore $a^m$ has finite order and $$\begin{equation*} o(a^m)\mid s. \end{equation*}$$ Let $$\begin{equation*} o(a^m)=t. \end{equation*}$$ Then $$\begin{equation*} (a^m)^t=e. \end{equation*}$$ Thus $$\begin{align*} (a^m)^t=e &\implies a^{mt}=e. \end{align*}$$ Since $o(a)=n$, the divisibility criterion gives $$\begin{equation*} n\mid mt. \end{equation*}$$ Using $m=dr$ and $n=ds$, we get $$\begin{align*} ds\mid drt &\implies s\mid rt. \end{align*}$$ Since $\operatorname{gcd}(r,s)=1$, Euclid's lemma gives $$\begin{equation*} s\mid t. \end{equation*}$$ We already have $t\mid s$. Therefore $$\begin{equation*} t=s. \end{equation*}$$ Since $$\begin{equation*} s=\frac{n}{d}=\frac{n}{\operatorname{gcd}(m,n)}, \end{equation*}$$ we get $$\begin{equation*} o(a^m)=\frac{n}{\operatorname{gcd}(m,n)}. \end{equation*}$$ Hence, $$\begin{equation*} o(a^m)=\frac{n}{\operatorname{gcd}(m,n)}. \end{equation*}$$ $\square$ ::: The formula shows two important cases. If $\operatorname{gcd}(m,n)=1$, then $o(a^m)=n$, so $a^m$ has the same order as $a$. If $m$ divides $n$, then $o(a^m)=\dfrac{n}{m}$. For example, if $o(a)=12$, then $o(a^5)=12$ because $\operatorname{gcd}(5,12)=1$, while $o(a^4)=3$ because $\operatorname{gcd}(4,12)=4$. :::corollary Let $(G,\circ)$ be a group and let $a\in G$. If $o(a)=n$ and $\operatorname{gcd}(m,n)=1$, then $$\begin{equation*} o(a^m)=n. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $a\in G$, $o(a)=n$, and $\operatorname{gcd}(m,n)=1$. To prove that $$\begin{equation*} o(a^m)=n. \end{equation*}$$ Using the order of a power formula, $$\begin{align*} o(a^m) &=\frac{n}{\operatorname{gcd}(m,n)}\\ &=\frac{n}{1}\\ &=n. \end{align*}$$ Hence, $$\begin{equation*} o(a^m)=n. \end{equation*}$$ $\square$ ::: :::corollary Let $(G,\circ)$ be a group and let $a\in G$. If $o(a)$ is infinite, then $o(a^m)$ is infinite for every positive integer $m$. ::: :::proof Given that $(G,\circ)$ is a group, $a\in G$, and $o(a)$ is infinite. Let $m\in\mathbb{N}$. To prove that $o(a^m)$ is infinite. If possible let $o(a^m)$ be finite. Then there exists $k\in\mathbb{N}$ such that $$\begin{equation*} (a^m)^k=e. \end{equation*}$$ Therefore $$\begin{align*} (a^m)^k=e &\implies a^{mk}=e. \end{align*}$$ Since $mk\in\mathbb{N}$, this contradicts that $a$ has infinite order. A contradiction. Hence, $o(a^m)$ is infinite for every positive integer $m$. $\square$ ::: :::example Let $o(a)=12$. Then $$\begin{align*} o(a^4) &=\frac{12}{\operatorname{gcd}(4,12)}\\ &=\frac{12}{4}\\ &=3. \end{align*}$$ Thus $a^4$ has order $3$. ::: :::example Let $o(a)=12$. Then $$\begin{align*} o(a^5) &=\frac{12}{\operatorname{gcd}(5,12)}\\ &=\frac{12}{1}\\ &=12. \end{align*}$$ Thus $a^5$ has the same order as $a$. ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. If $o(a)=18$, find $o(a^6)$. ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. Given that $$\begin{equation*} o(a)=18. \end{equation*}$$ To find $o(a^6)$. Using the order of a power formula, $$\begin{align*} o(a^6) &=\frac{18}{\operatorname{gcd}(6,18)}\\ &=\frac{18}{6}\\ &=3. \end{align*}$$ Hence, $$\begin{equation*} o(a^6)=3. \end{equation*}$$ $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. If $o(a)=20$, determine all positive integers $m$ with $1\leq m<20$ such that $o(a^m)=20$. ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. Given that $$\begin{equation*} o(a)=20. \end{equation*}$$ To determine all positive integers $m$ with $1\leq m<20$ such that $o(a^m)=20$. By the formula, $$\begin{equation*} o(a^m)=\frac{20}{\operatorname{gcd}(m,20)}. \end{equation*}$$ Thus $o(a^m)=20$ if and only if $$\begin{equation*} \operatorname{gcd}(m,20)=1. \end{equation*}$$ The positive integers $m$ with $1\leq m<20$ and $\operatorname{gcd}(m,20)=1$ are $$\begin{equation*} 1,3,7,9,11,13,17,19. \end{equation*}$$ Hence, the required values are $$\begin{equation*} 1,3,7,9,11,13,17,19. \end{equation*}$$ $\square$ ::: The formula $o(a^m)=n/\operatorname{gcd}(m,n)$ creates a repeating pattern as $m$ varies. We observe the order of each power $a^m$ for a fixed original order $n$. Change $n$ and notice that powers relatively prime to $n$ keep the full order, while powers sharing a larger common divisor have smaller order. This connects a group-theoretic question to the arithmetic structure of divisors of $n$. :::scientific-preview[Power Order Pattern Explorer] @libraries: plotly ::: Use the calculator to compute the order of $a^m$ from $o(a)=n$. Enter $n$ and $m$. The calculator applies the formula $o(a^m)=n/\operatorname{gcd}(m,n)$. Try values of $m$ that are relatively prime to $n$ and compare them with values that divide $n$. :::calculator[Order of a Power] ::: :::exercise [1] State the formula for $o(a^m)$ when $o(a)=n$. [2] If $o(a)=15$, find $o(a^5)$. [3] If $o(a)=15$, find $o(a^4)$. [4] If $o(a)=24$, find all $m$ with $1\leq m<24$ such that $o(a^m)=24$. [5] Prove that if $a$ has infinite order, then $a^m$ has infinite order for every $m\in\mathbb{N}$. ::: :::answer [1] $o(a^m)=\dfrac{n}{\operatorname{gcd}(m,n)}$. [2] $o(a^5)=15/\operatorname{gcd}(5,15)=3$. [3] $o(a^4)=15/\operatorname{gcd}(4,15)=15$. [4] The values are $1,5,7,11,13,17,19,23$. [5] If $(a^m)^k=e$, then $a^{mk}=e$, contradicting that $a$ has infinite order. ::: :::faq Q: What is the order of $a^m$ when $o(a)=n$? A: It is $n/\operatorname{gcd}(m,n)$. Q: When does $a^m$ have the same order as $a$? A: When $\operatorname{gcd}(m,n)=1$. Q: Can a power have smaller order than the original element? A: Yes. For example, if $o(a)=12$, then $o(a^4)=3$. Q: What happens if $a$ has infinite order? A: Every positive power $a^m$ also has infinite order. ::: :::call-to-action[Prime Order Elements] subtitle: Continue to elements of prime order and see how the order formula simplifies in the prime case. button: Next Lesson | /abstract-algebra/introduction-to-groups/order-of-an-element/prime-order-elements button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/order-of-an-element/powers-equal-to-identity :::