After studying prime-order cyclic groups, we now examine a special feature of cyclic groups of even order. If a cyclic group has order $2m$, then the element halfway around the cycle, namely $a^m$, has order $2$. More strongly, it is the only element of order $2$. This result is important because elements of order $2$ appear in symmetry, permutation groups, and later subgroup arguments. In this lesson, students will prove the uniqueness of the element of order $2$ in an even cyclic group. :::theorem Let $(G,\circ)$ be a cyclic group with identity element $e$. If $$\begin{equation*} |G|=2m \end{equation*}$$ for some $m\in\mathbb{N}$, then $G$ has exactly one element of order $2$. ::: :::proof Given that $(G,\circ)$ is a cyclic group with identity element $e$ and $|G|=2m$ for some $m\in\mathbb{N}$. To prove that $G$ has exactly one element of order $2$. Since $G$ is cyclic, there exists $a\in G$ such that $$\begin{equation*} G=\langle a\rangle. \end{equation*}$$ Since $|G|=2m$, we get $$\begin{equation*} o(a)=2m. \end{equation*}$$ Now $$\begin{align*} (a^m)^2 &=a^{2m}\\ &=e. \end{align*}$$ Also $$\begin{equation*} a^m\neq e, \end{equation*}$$ because $m<2m$ and $o(a)=2m$. Therefore $$\begin{equation*} o(a^m)=2. \end{equation*}$$ Thus $G$ has at least one element of order $2$. Now let $x\in G$ such that $$\begin{equation*} o(x)=2. \end{equation*}$$ Since $G=\langle a\rangle$, there exists $r\in\mathbb{Z}$ such that $$\begin{equation*} x=a^r. \end{equation*}$$ Since $o(x)=2$, we have $$\begin{equation*} x^2=e. \end{equation*}$$ Therefore $$\begin{align*} x^2=e &\implies (a^r)^2=e\\ &\implies a^{2r}=e. \end{align*}$$ Since $o(a)=2m$, the divisibility criterion gives $$\begin{equation*} 2m\mid2r. \end{equation*}$$ Therefore $$\begin{equation*} m\mid r. \end{equation*}$$ Thus there exists $q\in\mathbb{Z}$ such that $$\begin{equation*} r=mq. \end{equation*}$$ If $q$ is even, then $q=2t$ for some $t\in\mathbb{Z}$, and $$\begin{align*} x &=a^r\\ &=a^{mq}\\ &=a^{2mt}\\ &=e, \end{align*}$$ which contradicts $o(x)=2$. Therefore $q$ is odd. Then $q=2t+1$ for some $t\in\mathbb{Z}$, and $$\begin{align*} x &=a^{mq}\\ &=a^{m(2t+1)}\\ &=a^{2mt+m}\\ &=(a^{2m})^t\circ a^m\\ &=a^m. \end{align*}$$ Therefore every element of order $2$ is equal to $a^m$. Hence, $G$ has exactly one element of order $2$. $\square$ ::: The proof has two parts. First, it constructs the element $a^m$ and shows that its square is the identity. Second, it proves uniqueness by writing any element of order $2$ as a power $a^r$ and forcing $r$ to be congruent to $m$ modulo $2m$. The cyclic assumption is essential because it lets every element be written as a power of one generator. In a cyclic group of even order $n=2m$, the unique element of order $2$ is the halfway power $a^m$. Choose an even order and the preview highlights the element halfway around the cycle. Squaring that element moves another halfway step, reaching $a^{2m}=e$. :::scientific-preview[Halfway Element Explorer] ::: :::example Let $G=\langle a\rangle$ and $|G|=8$. Then $$\begin{equation*} m=4. \end{equation*}$$ The unique element of order $2$ is $$\begin{equation*} a^4. \end{equation*}$$ Indeed, $$\begin{equation*} (a^4)^2=a^8=e \end{equation*}$$ and $a^4\neq e$. ::: :::example In the additive group $(\mathbb{Z}_{10},+_{10})$, the unique element of order $2$ is $\overline{5}$. Indeed, $$\begin{equation*} 2\overline{5}=\overline{10}=\overline{0} \end{equation*}$$ and $\overline{5}\neq\overline{0}$. No other element has additive order $2$. ::: :::solved-problem Find the unique element of order $2$ in a cyclic group $G=\langle a\rangle$ of order $18$. ::: :::solution Let $G=\langle a\rangle$ be a cyclic group of order $18$. To find the unique element of order $2$. Since $$\begin{equation*} 18=2\cdot9, \end{equation*}$$ we have $m=9$. By the theorem, the unique element of order $2$ is $$\begin{equation*} a^9. \end{equation*}$$ Indeed, $$\begin{align*} (a^9)^2 &=a^{18}\\ &=e. \end{align*}$$ Since $o(a)=18$, we have $a^9\neq e$. Hence, the unique element of order $2$ is $a^9$. $\square$ ::: :::solved-problem Let $(G,\circ)$ be a cyclic group with only one generator. Prove that either $|G|=1$ or $|G|=2$. ::: :::solution Let $(G,\circ)$ be a cyclic group with only one generator. To prove that either $|G|=1$ or $|G|=2$. Since $G$ is cyclic, there exists $a\in G$ such that $$\begin{equation*} G=\langle a\rangle. \end{equation*}$$ By the inverse generator theorem, $$\begin{equation*} G=\langle a^{-1}\rangle. \end{equation*}$$ Since $G$ has only one generator, we get $$\begin{equation*} a=a^{-1}. \end{equation*}$$ Therefore $$\begin{align*} a=a^{-1} &\implies a^2=e. \end{align*}$$ Thus $$\begin{equation*} o(a)=1 \end{equation*}$$ or $$\begin{equation*} o(a)=2. \end{equation*}$$ If $o(a)=1$, then $a=e$, so $$\begin{equation*} G=\langle a\rangle=\{e\}. \end{equation*}$$ Therefore $|G|=1$. If $o(a)=2$, then $$\begin{equation*} G=\langle a\rangle=\{e,a\}. \end{equation*}$$ Therefore $|G|=2$. Hence, either $|G|=1$ or $|G|=2$. $\square$ ::: :::remark An even cyclic group has exactly one element of order $2$, but a non-cyclic group may behave differently. The Klein four group has three elements of order $2$. This gives another way to see that $V_4$ is not cyclic. ::: Use the calculator below to find the unique element of order $2$ in a cyclic group of even order $n$. Enter an even integer $n$. The calculator returns the exponent $n/2$, so the element is $a^{n/2}$. :::calculator[Unique Order Two Element Calculator] ::: :::exercise [1] State the theorem about elements of order $2$ in an even cyclic group. [2] Find the unique element of order $2$ in a cyclic group $\langle a\rangle$ of order $14$. [3] Find the unique element of order $2$ in $(\mathbb{Z}_{12},+_{12})$. [4] Explain why $V_4$ is not cyclic using elements of order $2$. [5] If a cyclic group has only one generator, prove that its order is $1$ or $2$. ::: :::answer [1] If $G=\langle a\rangle$ and $|G|=2m$, then $a^m$ is the unique element of order $2$. [2] The element is $a^7$. [3] The element is $\overline{6}$. [4] The group $V_4$ has three elements of order $2$, while an even cyclic group has exactly one. [5] If $a$ is the only generator, then $a^{-1}$ is also a generator, so $a=a^{-1}$. Hence $a^2=e$, and $|G|=1$ or $|G|=2$. ::: :::faq Q: Which element has order $2$ in a cyclic group of order $2m$? A: The unique element of order $2$ is $a^m$. Q: Why is the element unique? A: Any element is a power $a^r$, and the condition $(a^r)^2=e$ forces $r$ to be congruent to $m$ modulo $2m$ for a non-identity element. Q: Does every even group have exactly one element of order $2$? A: No. This theorem is for cyclic groups. Q: Why is the Klein four group not cyclic? A: It has three elements of order $2$, but an even cyclic group has exactly one. ::: :::call-to-action[Subgroups] subtitle: Continue to subgroups, where cyclic groups become a source of important subgroup examples. button: Next Section | /abstract-algebra/introduction-to-groups/subgroups/subgroup-definition button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/cyclic-groups/prime-order-cyclic-groups :::