After studying orders of powers and prime-order elements, we now examine two important ways order interacts with other group operations. First, if two elements commute and have relatively prime finite orders, then the product has order equal to the product of the two orders. Second, conjugate elements have the same order. These results are used throughout finite group theory, cyclic groups, conjugacy, and symmetry computations. In this lesson, students will prove both results and learn why commutativity is required in the product theorem but not in the conjugation theorem. :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. If $o(a)=m$, $o(b)=n$, $\operatorname{gcd}(m,n)=1$, and $$\begin{equation*} a\circ b=b\circ a, \end{equation*}$$ then $$\begin{equation*} o(a\circ b)=mn. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a,b\in G$. Given that $$\begin{equation*} o(a)=m,\quad o(b)=n,\quad \operatorname{gcd}(m,n)=1, \end{equation*}$$ and $$\begin{equation*} a\circ b=b\circ a. \end{equation*}$$ To prove that $$\begin{equation*} o(a\circ b)=mn. \end{equation*}$$ Since $a$ and $b$ commute, we have $$\begin{align*} (a\circ b)^{mn} &=a^{mn}\circ b^{mn}\\ &=(a^m)^n\circ(b^n)^m\\ &=e^n\circ e^m\\ &=e. \end{align*}$$ Therefore $o(a\circ b)$ exists and $$\begin{equation*} o(a\circ b)\mid mn. \end{equation*}$$ Let $$\begin{equation*} o(a\circ b)=r. \end{equation*}$$ Then $$\begin{equation*} (a\circ b)^r=e. \end{equation*}$$ Since $a$ and $b$ commute, $$\begin{align*} (a\circ b)^r=e &\implies a^r\circ b^r=e\\ &\implies a^r=b^{-r}. \end{align*}$$ Let $$\begin{equation*} c=a^r=b^{-r}. \end{equation*}$$ Now $$\begin{align*} c^m &=(a^r)^m\\ &=(a^m)^r\\ &=e. \end{align*}$$ Also, $$\begin{align*} c^n &=(b^{-r})^n\\ &=(b^n)^{-r}\\ &=e. \end{align*}$$ Since $\operatorname{gcd}(m,n)=1$, there exist integers $u,v$ such that $$\begin{equation*} um+vn=1. \end{equation*}$$ Using $c^m=e$ and $c^n=e$, we get $$\begin{align*} c &=c^{um+vn}\\ &=(c^m)^u\circ(c^n)^v\\ &=e. \end{align*}$$ Therefore $$\begin{equation*} a^r=e \end{equation*}$$ and $$\begin{equation*} b^r=e. \end{equation*}$$ Since $o(a)=m$, we get $$\begin{equation*} m\mid r. \end{equation*}$$ Since $o(b)=n$, we get $$\begin{equation*} n\mid r. \end{equation*}$$ Since $\operatorname{gcd}(m,n)=1$, it follows that $$\begin{equation*} mn\mid r. \end{equation*}$$ But $r=o(a\circ b)$ and $r\mid mn$. Therefore $$\begin{equation*} r=mn. \end{equation*}$$ Hence, $$\begin{equation*} o(a\circ b)=mn. \end{equation*}$$ $\square$ ::: The commutativity condition $a\circ b=b\circ a$ is essential. It allows the formula $$\begin{equation*} (a\circ b)^r=a^r\circ b^r. \end{equation*}$$ Without commutativity, this formula can fail. The coprime condition is also essential because it lets us conclude from $m\mid r$ and $n\mid r$ that $mn\mid r$. If $m$ and $n$ are not relatively prime, the order of the product may be smaller than $mn$. :::theorem Let $(G,\circ)$ be a group with identity element $e$, let $a\in G$, and let $x\in G$. Then $$\begin{equation*} o(a)=o(x\circ a\circ x^{-1}). \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$, $a\in G$, and $x\in G$. To prove that $$\begin{equation*} o(a)=o(x\circ a\circ x^{-1}). \end{equation*}$$ First suppose that $o(a)=m$, where $m\in\mathbb{N}$. Using the conjugation power formula, $$\begin{align*} (x\circ a\circ x^{-1})^m &=x\circ a^m\circ x^{-1}\\ &=x\circ e\circ x^{-1}\\ &=e. \end{align*}$$ Therefore $$\begin{equation*} o(x\circ a\circ x^{-1})\leq m. \end{equation*}$$ Let $$\begin{equation*} o(x\circ a\circ x^{-1})=r. \end{equation*}$$ Then $$\begin{equation*} (x\circ a\circ x^{-1})^r=e. \end{equation*}$$ Using the conjugation power formula again, $$\begin{align*} (x\circ a\circ x^{-1})^r=e &\implies x\circ a^r\circ x^{-1}=e. \end{align*}$$ Multiplying on the left by $x^{-1}$ and on the right by $x$, we get $$\begin{align*} x\circ a^r\circ x^{-1}=e &\implies a^r=x^{-1}\circ e\circ x\\ &\implies a^r=e. \end{align*}$$ Since $o(a)=m$, we get $$\begin{equation*} m\leq r. \end{equation*}$$ Thus $r=m$, and hence $$\begin{equation*} o(x\circ a\circ x^{-1})=o(a). \end{equation*}$$ If $a$ has infinite order, then $x\circ a\circ x^{-1}$ cannot have finite order. If possible let $o(x\circ a\circ x^{-1})=r$. Then the same calculation gives $a^r=e$, contradicting that $a$ has infinite order. Therefore both orders are infinite. Hence, $$\begin{equation*} o(a)=o(x\circ a\circ x^{-1}). \end{equation*}$$ $\square$ ::: :::example Let $o(a)=3$ and $o(b)=4$. Suppose $a\circ b=b\circ a$. Since $$\begin{equation*} \operatorname{gcd}(3,4)=1, \end{equation*}$$ the product theorem gives $$\begin{equation*} o(a\circ b)=12. \end{equation*}$$ ::: :::example Let $A$ be an invertible matrix and let $B$ be a matrix of finite order under multiplication. Then $ABA^{-1}$ has the same order as $B$. This is the matrix version of the conjugation theorem. It says that similarity by an invertible matrix preserves the order of a matrix element in the general linear group. ::: :::solved-problem Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. If $o(a)=2$, $o(b)=3$, and $a\circ b=b\circ a$, find $o(a\circ b)$. ::: :::solution Let $(G,\circ)$ be a group with identity element $e$ and let $a,b\in G$. Given that $$\begin{equation*} o(a)=2,\quad o(b)=3, \end{equation*}$$ and $$\begin{equation*} a\circ b=b\circ a. \end{equation*}$$ To find $o(a\circ b)$. Since $$\begin{equation*} \operatorname{gcd}(2,3)=1, \end{equation*}$$ the product theorem gives $$\begin{align*} o(a\circ b) &=2\cdot3\\ &=6. \end{align*}$$ Hence, $$\begin{equation*} o(a\circ b)=6. \end{equation*}$$ $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group and let $a,x\in G$. If $o(a)=8$, find $$\begin{equation*} o(x\circ a\circ x^{-1}). \end{equation*}$$ ::: :::solution Let $(G,\circ)$ be a group and let $a,x\in G$. Given that $$\begin{equation*} o(a)=8. \end{equation*}$$ To find $$\begin{equation*} o(x\circ a\circ x^{-1}). \end{equation*}$$ By the conjugation theorem, $$\begin{equation*} o(x\circ a\circ x^{-1})=o(a). \end{equation*}$$ Therefore $$\begin{equation*} o(x\circ a\circ x^{-1})=8. \end{equation*}$$ Hence, $$\begin{equation*} o(x\circ a\circ x^{-1})=8. \end{equation*}$$ $\square$ ::: :::remark The product theorem and conjugation theorem have different hypotheses. For products, commutativity and coprime orders are needed. For conjugates, no commutativity assumption is needed. Conjugation preserves order because the powers of $x\circ a\circ x^{-1}$ remain conjugates of the powers of $a$. ::: The product theorem predicts $o(a\circ b)=mn$ only when the elements commute and the orders $m$ and $n$ are relatively prime. We observe the predicted product order and compare it with the least common multiple of $m$ and $n$. When $m$ and $n$ are coprime, these two numbers agree; when they are not coprime, the theorem does not apply. This helps separate the arithmetic hypothesis from the group-theoretic commutativity hypothesis. :::scientific-preview[Product Order Hypothesis Explorer] @libraries: plotly ::: Use the calculator to compute the order of a product when the two orders are coprime and the elements commute. Enter $m=o(a)$ and $n=o(b)$. The calculator checks whether the coprime hypothesis holds and returns the predicted product order. Remember that the calculator cannot verify commutativity; that remains a group-theoretic condition. :::calculator[Product Order Test] ::: :::exercise [1] State the theorem for the order of a product of commuting elements with coprime orders. [2] If $o(a)=5$, $o(b)=6$, and $a\circ b=b\circ a$, find $o(a\circ b)$. [3] If $o(a)=4$, $o(b)=6$, and $a\circ b=b\circ a$, can the product theorem be applied directly? [4] Prove that conjugate elements have the same order. [5] If $o(a)=12$, find $o(x\circ a\circ x^{-1})$. ::: :::answer [1] If $o(a)=m$, $o(b)=n$, $\operatorname{gcd}(m,n)=1$, and $a\circ b=b\circ a$, then $o(a\circ b)=mn$. [2] $o(a\circ b)=30$. [3] No, because $\operatorname{gcd}(4,6)=2$. [4] Use $(x\circ a\circ x^{-1})^r=x\circ a^r\circ x^{-1}$ and compare the least positive exponents giving the identity. [5] $o(x\circ a\circ x^{-1})=12$. ::: :::faq Q: When is $o(a\circ b)=o(a)o(b)$? A: This holds when $a$ and $b$ commute and their orders are relatively prime. Q: Why must $a$ and $b$ commute in the product theorem? A: The proof uses $(a\circ b)^r=a^r\circ b^r$, which requires commutativity between $a$ and $b$. Q: Do conjugate elements always have the same order? A: Yes. Conjugation preserves order in every group. Q: Does the conjugation theorem require an abelian group? A: No. It holds in arbitrary groups. ::: :::call-to-action[Standard Examples of Groups] subtitle: Continue to standard examples of groups, where these order results appear in concrete number and matrix examples. button: Next Section | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/groups-from-number-systems button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/order-of-an-element/prime-order-elements :::