Lagrange's theorem restricts the orders of subgroups. Since every element $a$ of a group generates a cyclic subgroup $\langle a\rangle$, it also restricts the order of every element. This is one of the most useful consequences of Lagrange's theorem. Instead of studying $a$ alone, we study the subgroup generated by $a$. The size of this subgroup is exactly the order of the element, so it must divide the order of the whole group. :::definition[Order of an Element] Let $(G,\circ)$ be a group with identity element $e$, and let $a\in G$. If there exists a least positive integer $n$ such that $$\begin{equation*} a^n=e, \end{equation*}$$ then $a$ is said to have $finite order$ and the order of $a$ is $$\begin{equation*} o(a)=n. \end{equation*}$$ If no such positive integer exists, then $a$ is said to have infinite order. ::: The order of an element is the length of the cycle created by repeatedly applying that element. In $\mathbb{Z}_n$ under addition, this means repeatedly adding the chosen element until the cycle returns to $0$. Change $n$ and the element $a$ to see the generated subgroup $\langle a\rangle$. Notice that the cycle length always divides $n$ in this finite cyclic example. :::scientific-preview[Element Order Cycle Explorer] ::: :::example[Possible Element Orders in a Group of Order Twelve] Let $(G,\circ)$ be a finite group with $|G|=12$. If $a\in G$, then $$\begin{equation*} o(a)\mid 12. \end{equation*}$$ Therefore the possible orders of elements are among $$\begin{equation*} 1,2,3,4,6,12. \end{equation*}$$ An element of order $5$ cannot occur in such a group. ::: :::example[Order of an Element in Integers Modulo Eight] Let $G=\mathbb{Z}_8$ under addition modulo $8$. The element $2$ generates $$\begin{equation*} \langle 2\rangle=\{0,2,4,6\}. \end{equation*}$$ Thus $$\begin{equation*} o(2)=4. \end{equation*}$$ Since $|G|=8$, we have $$\begin{equation*} 4\mid 8. \end{equation*}$$ This agrees with Lagrange's theorem. ::: :::solved-problem[Ruling Out an Element Order] Let $(G,\circ)$ be a finite group with $|G|=30$. Prove that no element of $G$ has order $7$. ::: :::solution[Solution] Let $(G,\circ)$ be a finite group with $|G|=30$. If $a\in G$, then by Lagrange's theorem applied to $\langle a\rangle$, $$\begin{equation*} o(a)\mid 30. \end{equation*}$$ But $$\begin{equation*} 7\nmid 30. \end{equation*}$$ Therefore no element of $G$ can have order $7$. ::: :::solved-problem[Finding Possible Element Orders] Let $(G,\circ)$ be a finite group with $|G|=20$. List all possible element orders allowed by Lagrange's theorem. ::: :::solution[Solution] Let $(G,\circ)$ be a finite group with $|G|=20$. By Lagrange's theorem applied to cyclic subgroups, the order of every element divides $20$. The positive divisors of $20$ are $$\begin{equation*} 1,2,4,5,10,20. \end{equation*}$$ Therefore the possible element orders allowed by Lagrange's theorem are $$\begin{equation*} 1,2,4,5,10,20. \end{equation*}$$ ::: :::exercise[Exercises] 1. Let $|G|=21$. List all possible element orders allowed by Lagrange's theorem. 2. Let $|G|=16$. Can $G$ contain an element of order $6$? Justify your answer. 3. Let $a$ be an element of a finite group $G$ and suppose $o(a)=9$. What can be concluded about $|G|$? ::: :::answer[Answers] 1. The possible element orders are $$\begin{equation*} 1,3,7,21. \end{equation*}$$ 2. No. Since $$\begin{equation*} 6\nmid 16, \end{equation*}$$ there is no element of order $6$ in $G$. 3. Since $o(a)\mid |G|$, we get $$\begin{equation*} 9\mid |G|. \end{equation*}$$ ::: :::faq Q: Why does Lagrange's theorem apply to an element? A: It applies to the cyclic subgroup generated by the element. Q: Is $o(a)$ always equal to $|G|$? A: No. The order of $a$ divides $|G|$, but it may be smaller. Q: If $d$ divides $|G|$, must there be an element of order $d$? A: No. Lagrange's theorem does not guarantee the existence of such an element. ::: :::call-to-action subtitle: Continue with the special case where the order of the whole group is prime. button: Next: Groups of Prime Order | published/abstract-algebra/introduction-to-groups/index-and-lagrange-theorem/groups-of-prime-order.md :::