After proving the formula for the order of a power, we now study the special case of prime order. Prime order elements are especially simple because a positive integer is either divisible by the prime or relatively prime to it. This gives a sharp dichotomy for powers: a power is either the identity, or it has the same prime order as the original element. This result is important later when studying cyclic groups of prime order and generators. In this lesson, students will prove the main prime-order consequences and solve standard examples. :::theorem Let $(G,\circ)$ be a group with identity element $e$, let $a\in G$, and let $p$ be a prime number. If $o(a)=p$, then $$\begin{equation*} o(a^k)=p \quad \forall\, k\in\mathbb{N},\; 1\leq k<p. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$, $a\in G$, and $p$ is a prime number. Given that $$\begin{equation*} o(a)=p. \end{equation*}$$ To prove that $$\begin{equation*} o(a^k)=p \quad \forall\, k\in\mathbb{N},\; 1\leq k<p. \end{equation*}$$ Let $k\in\mathbb{N}$ such that $$\begin{equation*} 1\leq k<p. \end{equation*}$$ Since $p$ is prime and $1\leq k<p$, we get $$\begin{equation*} \operatorname{gcd}(k,p)=1. \end{equation*}$$ Using the order of a power formula, $$\begin{align*} o(a^k) &=\frac{o(a)}{\operatorname{gcd}(k,o(a))}\\ &=\frac{p}{\operatorname{gcd}(k,p)}\\ &=\frac{p}{1}\\ &=p. \end{align*}$$ Hence, $$\begin{equation*} o(a^k)=p \quad \forall\, k\in\mathbb{N},\; 1\leq k<p. \end{equation*}$$ $\square$ ::: The theorem says that if $a$ has prime order $p$, then all nontrivial powers $a,a^2,\ldots,a^{p-1}$ also have order $p$. The identity appears at $a^p$. This is one of the reasons prime order groups are structurally simple. There is no smaller positive divisor of $p$ available to create a smaller order for a non-identity power. :::theorem Let $(G,\circ)$ be a group with identity element $e$, let $a\in G$, and let $p$ be a prime number. If $o(a)=p$, then for every $k\in\mathbb{N}$, either $$\begin{equation*} a^k=e \end{equation*}$$ or $$\begin{equation*} o(a^k)=p. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$, $a\in G$, and $p$ is a prime number. Given that $$\begin{equation*} o(a)=p. \end{equation*}$$ To prove that for every $k\in\mathbb{N}$, either $a^k=e$ or $o(a^k)=p$. Let $k\in\mathbb{N}$. Since $p$ is prime, either $$\begin{equation*} p\mid k \end{equation*}$$ or $$\begin{equation*} \operatorname{gcd}(k,p)=1. \end{equation*}$$ [1] Let $p\mid k$. Then there exists $q\in\mathbb{N}$ such that $$\begin{equation*} k=pq. \end{equation*}$$ Now $$\begin{align*} a^k &=a^{pq}\\ &=(a^p)^q\\ &=e^q\\ &=e. \end{align*}$$ Therefore $a^k=e$. [2] Let $\operatorname{gcd}(k,p)=1$. Using the order of a power formula, $$\begin{align*} o(a^k) &=\frac{p}{\operatorname{gcd}(k,p)}\\ &=\frac{p}{1}\\ &=p. \end{align*}$$ Therefore $o(a^k)=p$. Hence, for every $k\in\mathbb{N}$, either $a^k=e$ or $o(a^k)=p$. $\square$ ::: :::example Let $o(a)=5$. Then $$\begin{equation*} o(a)=o(a^2)=o(a^3)=o(a^4)=5. \end{equation*}$$ Also, $$\begin{equation*} a^5=e,\quad a^{10}=e,\quad a^{15}=e. \end{equation*}$$ Thus every positive power of $a$ is either the identity or has order $5$. ::: :::example In the group $\{1,\omega,\omega^2\}$ under complex multiplication, suppose $\omega^3=1$ and $\omega\neq1$. Then $$\begin{equation*} o(\omega)=3. \end{equation*}$$ Since $3$ is prime, the non-identity powers $\omega$ and $\omega^2$ both have order $3$. Indeed, $$\begin{equation*} (\omega^2)^3=\omega^6=(\omega^3)^2=1 \end{equation*}$$ and $\omega^2\neq1$. ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. If $o(a)=7$, find $o(a^3)$. ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. Given that $$\begin{equation*} o(a)=7. \end{equation*}$$ To find $o(a^3)$. Since $7$ is prime and $1\leq3<7$, the prime-order theorem gives $$\begin{equation*} o(a^3)=7. \end{equation*}$$ Hence, $o(a^3)=7$. $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. If $o(a)=7$, determine whether $a^{21}=e$. ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. Given that $$\begin{equation*} o(a)=7. \end{equation*}$$ To determine whether $a^{21}=e$. By the divisibility criterion, $$\begin{equation*} a^{21}=e \iff 7\mid21. \end{equation*}$$ Since $$\begin{equation*} 21=7\cdot3, \end{equation*}$$ we get $$\begin{equation*} 7\mid21. \end{equation*}$$ Therefore $$\begin{equation*} a^{21}=e. \end{equation*}$$ Hence, $a^{21}=e$. $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. If $o(a)=p$, where $p$ is prime, prove that $a^k=e$ if and only if $p\mid k$. ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. Given that $o(a)=p$, where $p$ is prime. To prove that $$\begin{equation*} a^k=e \iff p\mid k. \end{equation*}$$ This follows directly from the divisibility criterion for powers equal to the identity. Since $o(a)=p$, for every $k\in\mathbb{N}$, $$\begin{equation*} a^k=e \iff p\mid k. \end{equation*}$$ Hence, $$\begin{equation*} a^k=e \iff p\mid k. \end{equation*}$$ $\square$ ::: :::remark Prime order does not mean every power is different. The powers repeat after $p$ steps: $$\begin{equation*} a^{p+1}=a,\quad a^{p+2}=a^2. \end{equation*}$$ Prime order means that before reaching $a^p=e$, no positive smaller exponent gives the identity. Therefore $a,a^2,\ldots,a^{p-1}$ are non-identity elements, and each has order $p$. ::: Prime order creates a two-outcome pattern for powers: identity at multiples of $p$, and full order $p$ everywhere else. We observe this by plotting the order of $a^k$ for several exponents $k$. Change the prime $p$ and notice that every non-multiple of $p$ stays at height $p$. This visual pattern comes from the fact that a prime number has no positive divisors except $1$ and itself. :::scientific-preview[Prime Power Outcome Explorer] @libraries: plotly ::: Use the calculator for the prime-order case. Enter a prime $p$ and a positive exponent $k$. The calculator decides whether $a^k=e$ or $o(a^k)=p$ under the assumption $o(a)=p$. Test exponents just below, equal to, and just above multiples of $p$. :::calculator[Prime Order Test] ::: :::exercise [1] If $o(a)=11$, find $o(a^4)$. [2] If $o(a)=11$, determine whether $a^{33}=e$. [3] If $o(a)=11$, determine whether $a^{35}=e$. [4] Prove that if $o(a)=p$ and $1\leq k<p$, then $a^k\neq e$. [5] If $o(a)=p$ and $p\nmid k$, find $o(a^k)$. ::: :::answer [1] $o(a^4)=11$. [2] Yes, because $11\mid33$. [3] No, because $11\nmid35$. [4] If $a^k=e$, then $p\mid k$, which is impossible for $1\leq k<p$. [5] $o(a^k)=p$. ::: :::faq Q: If $o(a)=p$ is prime, what are the possible orders of $a^k$? A: The power $a^k$ is either the identity, or it has order $p$. Q: When is $a^k=e$ in the prime-order case? A: Exactly when $p\mid k$. Q: What is the order of $a^k$ when $1\leq k<p$? A: It is $p$. Q: Why does prime order simplify the formula for $o(a^k)$? A: Because $\operatorname{gcd}(k,p)$ is either $1$ or $p$. ::: :::call-to-action[Products and Conjugates] subtitle: Continue to products and conjugates, where order interacts with commuting elements and conjugation. button: Next Lesson | /abstract-algebra/introduction-to-groups/order-of-an-element/orders-of-products-and-conjugates button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/order-of-an-element/order-of-powers :::