After proving that an element and its inverse have the same order, we now study exactly which powers of an element are equal to the identity. If $o(a)=n$, then $a^n=e$, but many other powers may also be equal to $e$. The precise rule is divisibility: $a^p=e$ exactly when $n$ divides $p$. This theorem is one of the most important tools for working with orders because it turns a group equation into an arithmetic divisibility statement. In this lesson, students will prove this criterion and use it to decide when powers equal the identity. :::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 $p$, $$\begin{equation*} a^p=e \iff n\mid p. \end{equation*}$$ ::: :::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 $p\in\mathbb{N}$. To prove that $$\begin{equation*} a^p=e \iff n\mid p. \end{equation*}$$ [1] To prove that $a^p=e \implies n\mid p$. By the division algorithm, there exist integers $q$ and $r$ such that $$\begin{equation*} p=nq+r,\quad 0\leq r<n. \end{equation*}$$ Now $$\begin{align*} a^p &=a^{nq+r}\\ &=(a^n)^q\circ a^r\\ &=e^q\circ a^r\\ &=a^r. \end{align*}$$ If $a^p=e$, then $$\begin{equation*} a^r=e. \end{equation*}$$ Since $0\leq r<n$ and $n$ is the least positive integer such that $a^n=e$, we must have $$\begin{equation*} r=0. \end{equation*}$$ Therefore $$\begin{equation*} p=nq. \end{equation*}$$ Thus $$\begin{equation*} n\mid p. \end{equation*}$$ [2] To prove that $n\mid p \implies a^p=e$. Let $$\begin{equation*} n\mid p. \end{equation*}$$ Then there exists $q\in\mathbb{N}$ such that $$\begin{equation*} p=nq. \end{equation*}$$ Now $$\begin{align*} a^p &=a^{nq}\\ &=(a^n)^q\\ &=e^q\\ &=e. \end{align*}$$ Therefore $$\begin{equation*} n\mid p \implies a^p=e. \end{equation*}$$ Hence, $$\begin{equation*} a^p=e \iff n\mid p. \end{equation*}$$ $\square$ ::: The first half of the proof is often where mistakes occur. One cannot simply say that $a^p=e$ and $o(a)=n$ imply $n\mid p$ without justification. The division algorithm supplies the missing step. If $p=nq+r$ with $0\leq r<n$, then $a^p=a^r$. Since $a^p=e$, we get $a^r=e$. The minimality of $n$ forces $r=0$, and this is exactly the divisibility condition. :::corollary Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. If $o(a)=n$, then $$\begin{equation*} a^r\neq e \quad \forall\, r\in\mathbb{N},\; 1\leq r<n. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$, $a\in G$, and $o(a)=n$. To prove that $$\begin{equation*} a^r\neq e \quad \forall\, r\in\mathbb{N},\; 1\leq r<n. \end{equation*}$$ If possible let there exist $r\in\mathbb{N}$ such that $1\leq r<n$ and $$\begin{equation*} a^r=e. \end{equation*}$$ By the previous theorem, $$\begin{equation*} n\mid r. \end{equation*}$$ But $1\leq r<n$, so $n\nmid r$. A contradiction. Hence, $$\begin{equation*} a^r\neq e \quad \forall\, r\in\mathbb{N},\; 1\leq r<n. \end{equation*}$$ $\square$ ::: :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. If $o(a)=n$, then $$\begin{equation*} a,a^2,\ldots,a^n \end{equation*}$$ are all distinct. ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a\in G$. Given that $$\begin{equation*} o(a)=n. \end{equation*}$$ To prove that $a,a^2,\ldots,a^n$ are all distinct. If possible let $$\begin{equation*} a^r=a^s \end{equation*}$$ for some $r,s\in\mathbb{N}$ such that $$\begin{equation*} 1\leq r<s\leq n. \end{equation*}$$ Then $$\begin{align*} a^r=a^s &\implies a^s\circ a^{-r}=e\\ &\implies a^{s-r}=e. \end{align*}$$ Since $1\leq r<s\leq n$, we get $$\begin{equation*} 0<s-r<n. \end{equation*}$$ Thus $a^{s-r}=e$ for a positive integer $s-r<n$, which contradicts $o(a)=n$. A contradiction. Hence, $a,a^2,\ldots,a^n$ are all distinct. $\square$ ::: :::example Let $o(a)=6$. Then $a^p=e$ exactly when $6\mid p$. Thus $$\begin{equation*} a^{12}=e,\quad a^{18}=e,\quad a^{24}=e. \end{equation*}$$ But $$\begin{equation*} a^7\neq e,\quad a^{10}\neq e,\quad a^{17}\neq e. \end{equation*}$$ ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. If $o(a)=8$, determine whether $a^{36}=e$. ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. Given that $$\begin{equation*} o(a)=8. \end{equation*}$$ To determine whether $a^{36}=e$. By the divisibility criterion, $$\begin{equation*} a^{36}=e \iff 8\mid36. \end{equation*}$$ But $$\begin{equation*} 8\nmid36. \end{equation*}$$ Therefore $$\begin{equation*} a^{36}\neq e. \end{equation*}$$ Hence, $a^{36}\neq e$. $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. If $o(a)=8$, determine whether $a^{40}=e$. ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. Given that $$\begin{equation*} o(a)=8. \end{equation*}$$ To determine whether $a^{40}=e$. By the divisibility criterion, $$\begin{equation*} a^{40}=e \iff 8\mid40. \end{equation*}$$ Since $$\begin{equation*} 40=8\cdot5, \end{equation*}$$ we get $$\begin{equation*} 8\mid40. \end{equation*}$$ Therefore $$\begin{equation*} a^{40}=e. \end{equation*}$$ Hence, $a^{40}=e$. $\square$ ::: The divisibility criterion turns the group question $a^p=e$ into the arithmetic question $n\mid p$. We observe the exponent line and mark exactly the multiples of the order $n$. Move the order slider and watch the identity exponents spread farther apart or closer together. This makes the phrase first positive power concrete: the first marked exponent is the order, and all later marked exponents are its multiples. :::scientific-preview[Identity Exponent Pattern] @libraries: plotly ::: Use the calculator to apply the divisibility criterion. Enter the order $n$ of an element and an exponent $p$. The calculator checks whether $a^p=e$ must hold when $o(a)=n$. Try values just below and just above multiples of $n$ to see why divisibility, not size, decides the result. :::calculator[Identity Power Test] ::: :::exercise [1] State the divisibility criterion for $a^p=e$ when $o(a)=n$. [2] If $o(a)=5$, determine whether $a^{20}=e$. [3] If $o(a)=5$, determine whether $a^{22}=e$. [4] If $o(a)=9$, list three positive exponents $p$ for which $a^p=e$. [5] Prove that if $o(a)=n$, then $a,a^2,\ldots,a^n$ are all distinct. ::: :::answer [1] If $o(a)=n$, then $a^p=e$ if and only if $n\mid p$. [2] Yes, because $5\mid20$. [3] No, because $5\nmid22$. [4] Examples are $9,18,27$. [5] If $a^r=a^s$ with $1\leq r<s\leq n$, then $a^{s-r}=e$ with $0<s-r<n$, contradicting $o(a)=n$. ::: :::faq Q: If $a^p=e$, must $p=o(a)$? A: No. The exponent $p$ may be any positive multiple of $o(a)$. Q: If $o(a)=n$, when is $a^p=e$? A: Exactly when $n$ divides $p$. Q: Why are $a,a^2,\ldots,a^n$ distinct? A: If two of them were equal, a smaller positive power would equal the identity. Q: Does the theorem apply when $a$ has infinite order? A: No. The theorem assumes $o(a)=n$ is finite. ::: :::call-to-action[Order of Powers] subtitle: Continue to the formula for the order of a power of an element. button: Next Lesson | /abstract-algebra/introduction-to-groups/order-of-an-element/order-of-powers button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/order-of-an-element/order-of-inverse-element :::