After defining finite and infinite order, we now compare the order of an element with the order of its inverse. Since inverse elements undo the original element, it is natural to expect that they return to the identity after the same number of repetitions. This is correct. If $a$ has finite order $n$, then $a^{-1}$ also has order $n$. If $a$ has infinite order, then $a^{-1}$ also has infinite order. In this lesson, students will prove the equality $o(a)=o(a^{-1})$ and use it in examples. :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. Then $$\begin{equation*} o(a)=o(a^{-1}). \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a\in G$. To prove that $$\begin{equation*} o(a)=o(a^{-1}). \end{equation*}$$ First suppose that $o(a)=n$, where $n\in\mathbb{N}$. Then $$\begin{equation*} a^n=e \end{equation*}$$ and $n$ is the least positive integer with this property. Taking inverses on both sides, $$\begin{align*} a^n=e &\implies (a^n)^{-1}=e^{-1}\\ &\implies (a^{-1})^n=e. \end{align*}$$ Therefore $a^{-1}$ has finite order, and $$\begin{equation*} o(a^{-1})\leq n. \end{equation*}$$ Let $o(a^{-1})=m$. Then $$\begin{equation*} (a^{-1})^m=e. \end{equation*}$$ Taking inverses on both sides, $$\begin{align*} (a^{-1})^m=e &\implies \big((a^{-1})^m\big)^{-1}=e^{-1}\\ &\implies a^m=e. \end{align*}$$ Since $o(a)=n$, the least positive exponent giving $e$ is $n$. Hence, $$\begin{equation*} n\leq m. \end{equation*}$$ Also $m=o(a^{-1})\leq n$. Therefore $$\begin{equation*} m=n. \end{equation*}$$ Thus $o(a^{-1})=o(a)$ when $o(a)$ is finite. Now suppose that $o(a)$ is infinite. If possible let $o(a^{-1})=k$ for some $k\in\mathbb{N}$. Then $$\begin{equation*} (a^{-1})^k=e. \end{equation*}$$ Taking inverses gives $$\begin{equation*} a^k=e. \end{equation*}$$ This contradicts that $a$ has infinite order. Therefore $a^{-1}$ has infinite order. Hence, $$\begin{equation*} o(a)=o(a^{-1}). \end{equation*}$$ $\square$ ::: The proof has two parts because order may be finite or infinite. In the finite case, the argument shows two inequalities and then concludes equality. In the infinite case, the proof uses contradiction. A common mistake is to prove only that $(a^{-1})^n=e$ and then immediately claim the order is $n$. That is incomplete, because one must also prove that no smaller positive exponent works. :::example In the group $S=\{1,i,-1,-i\}$ under multiplication, the inverse of $i$ is $-i$. Since $$\begin{equation*} i^4=1 \end{equation*}$$ and no smaller positive power of $i$ is $1$, we have $o(i)=4$. Also, $$\begin{equation*} (-i)^1=-i,\quad (-i)^2=-1,\quad (-i)^3=i,\quad (-i)^4=1. \end{equation*}$$ Thus $$\begin{equation*} o(-i)=4. \end{equation*}$$ Hence, $$\begin{equation*} o(i)=o(i^{-1}). \end{equation*}$$ ::: :::example In $(\mathbb{Z},+)$, the inverse of $5$ is $-5$. Since $5$ has infinite order under addition, the theorem says that $-5$ also has infinite order. Indeed, there is no positive integer $n$ such that $$\begin{equation*} n(-5)=0. \end{equation*}$$ Hence, $o(5)=o(-5)=\infty$ in $(\mathbb{Z},+)$. ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. If $$\begin{equation*} o(a)=6, \end{equation*}$$ find $o(a^{-1})$. ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. Given that $$\begin{equation*} o(a)=6. \end{equation*}$$ To find $o(a^{-1})$. By the theorem, $$\begin{equation*} o(a^{-1})=o(a). \end{equation*}$$ Therefore $$\begin{equation*} o(a^{-1})=6. \end{equation*}$$ Hence, $o(a^{-1})=6$. $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. If $$\begin{equation*} o(a^{-1})=9, \end{equation*}$$ find $o(a)$. ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. Given that $$\begin{equation*} o(a^{-1})=9. \end{equation*}$$ To find $o(a)$. Since $$\begin{equation*} o(a)=o(a^{-1}), \end{equation*}$$ we get $$\begin{equation*} o(a)=9. \end{equation*}$$ Hence, $o(a)=9$. $\square$ ::: :::remark The equality $o(a)=o(a^{-1})$ is often used silently in later problems. For example, if an element has order $2$, then $a^{-1}=a$ because $a^2=e$. If an element has order $3$, then $a^{-1}=a^2$, and both $a$ and $a^2$ have order $3$. The theorem gives a reliable way to compare an element and its inverse without recomputing everything. ::: The additive group modulo $n$ lets us see why an element and its inverse have the same order. We compare the residues $ka$ and $k(-a)$ modulo $n$ for the same values of $k$. The two paths may move in opposite directions around the modular cycle, but they return to the identity at the same first step. Try changing $a$ and notice that the two plotted sequences always hit $0$ together. :::scientific-preview[Inverse Residue Comparison] @libraries: plotly ::: Use the calculator to compare the order of an element and the order of its inverse in the additive group modulo $n$. In additive notation, the inverse of $a$ is $-a$. The calculator computes both orders and displays them side by side. This numerical check reinforces the theorem $o(a)=o(a^{-1})$. :::calculator[Inverse Order Test] ::: :::exercise [1] State the theorem relating $o(a)$ and $o(a^{-1})$. [2] If $o(a)=10$, find $o(a^{-1})$. [3] If $o(a^{-1})=15$, find $o(a)$. [4] If $a$ has infinite order, prove that $a^{-1}$ has infinite order. [5] In $\{1,i,-1,-i\}$, find $o((-i)^{-1})$. ::: :::answer [1] For every element $a$ of a group, $o(a)=o(a^{-1})$. [2] $o(a^{-1})=10$. [3] $o(a)=15$. [4] If $(a^{-1})^k=e$ for some $k\in\mathbb{N}$, then taking inverses gives $a^k=e$, contradicting infinite order. [5] Since $(-i)^{-1}=i$ and $o(i)=4$, we get $o((-i)^{-1})=4$. ::: :::faq Q: Do an element and its inverse always have the same order? A: Yes. In any group, $o(a)=o(a^{-1})$. Q: What if $a$ has infinite order? A: Then $a^{-1}$ also has infinite order. Q: Is it enough to prove $(a^{-1})^n=e$? A: No. To prove the order is $n$, one must also prove that no smaller positive exponent gives the identity. Q: What is special about order $2$? A: If $o(a)=2$, then $a^{-1}=a$. ::: :::call-to-action[Powers Equal to the Identity] subtitle: Continue to the divisibility criterion for powers equal to the identity. button: Next Lesson | /abstract-algebra/introduction-to-groups/order-of-an-element/powers-equal-to-identity button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/order-of-an-element/finite-and-infinite-order :::