Having learned how isomorphisms preserve structure, we can classify the simplest family of groups. The focus keyword classification of cyclic groups means that cyclic groups are determined, up to isomorphism, entirely by their order. A finite cyclic group of order $n$ is structurally the same as $\mathbb{Z}_n$, and an infinite cyclic group is structurally the same as $\mathbb{Z}$. This result is powerful because it converts an abstract generator into a familiar additive model. Students often think two cyclic groups differ because one is written multiplicatively and another additively. Classification shows that notation is secondary; the order of the generator controls the whole group. :::theorem[Finite Cyclic Groups] Every finite cyclic group of order $n$ is isomorphic to $(\mathbb{Z}_n,+_n)$. ::: :::proof Given that $G$ is a finite cyclic group of order $n$. To prove that $G\cong\mathbb{Z}_n$. Let $G=\langle a\rangle$ with $o(a)=n$. Define $f:\mathbb{Z}_n\to G$ by $$\begin{equation*} f([k])=a^k. \end{equation*}$$ First, $f$ is well-defined. If $[k]=[l]$, then $k-l=nt$ for some $t\in\mathbb{Z}$. Hence $$\begin{align*} a^k&=a^{l+nt}\\ &=a^l(a^n)^t\\ &=a^l. \end{align*}$$ Let $[k],[l]\in\mathbb{Z}_n$. Then $$\begin{align*} f([k]+_n[l])&=f([k+l])\\ &=a^{k+l}\\ &=a^ka^l\\ &=f([k])f([l]). \end{align*}$$ Thus $f$ is a homomorphism. Since $G=\langle a\rangle$, every element of $G$ is $a^k$ for some $k$, so $f$ is onto. If $f([k])=e$, then $a^k=e$. Since $o(a)=n$, $n\mid k$, so $[k]=[0]$. Hence $\ker f=\{[0]\}$, and $f$ is one-to-one. Therefore $f$ is an isomorphism. Hence $G\cong\mathbb{Z}_n$. $\square$ ::: :::theorem[Infinite Cyclic Groups] Every infinite cyclic group is isomorphic to $(\mathbb{Z},+)$. ::: :::proof Given that $G$ is an infinite cyclic group. To prove that $G\cong\mathbb{Z}$. Let $G=\langle a\rangle$, where $a$ has infinite order. Define $f:\mathbb{Z}\to G$ by $$\begin{equation*} f(m)=a^m. \end{equation*}$$ For $m,n\in\mathbb{Z}$, $$\begin{align*} f(m+n)&=a^{m+n}\\ &=a^ma^n\\ &=f(m)f(n). \end{align*}$$ Thus $f$ is a homomorphism. Since $G=\langle a\rangle$, every element of $G$ has the form $a^m$, so $f$ is onto. If $f(m)=f(n)$, then $a^m=a^n$, so $a^{m-n}=e$. Since $a$ has infinite order, $m-n=0$. Therefore $m=n$, and $f$ is one-to-one. Hence $f$ is an isomorphism. $\square$ ::: :::corollary Any two cyclic groups of the same order are isomorphic. ::: :::proof Given that $G$ and $H$ are cyclic groups of the same order. To prove that $G\cong H$. If both groups have finite order $n$, then $G\cong\mathbb{Z}_n$ and $H\cong\mathbb{Z}_n$ by the finite cyclic group theorem. Therefore $G\cong H$. If both groups are infinite, then $G\cong\mathbb{Z}$ and $H\cong\mathbb{Z}$ by the infinite cyclic group theorem. Therefore $G\cong H$. Hence any two cyclic groups of the same order are isomorphic. $\square$ ::: :::theorem[Groups of Prime Order] Every group of prime order $p$ is cyclic. Hence every group of order $p$ is isomorphic to $\mathbb{Z}_p$. ::: :::proof Given that $G$ is a group of prime order $p$. To prove that $G$ is cyclic and $G\cong\mathbb{Z}_p$. Choose $a\in G$ with $a\neq e$. By Lagrange's theorem, $o(a)\mid |G|=p$. Since $a\neq e$, $o(a)\neq 1$. Therefore $o(a)=p$. Thus $\langle a\rangle$ has $p$ elements, so $\langle a\rangle=G$. Hence $G$ is cyclic. By the classification of cyclic groups, $G\cong\mathbb{Z}_p$. $\square$ ::: :::example[Two Groups of Order Four] There are two familiar group structures of order $4$: the cyclic group $\mathbb{Z}_4$ and the Klein four-group $K_4$. The group $\mathbb{Z}_4$ has an element of order $4$, namely $[1]$. In $K_4$, every nonidentity element has order $2$. Since element orders are preserved by isomorphism, $$\begin{equation*} \mathbb{Z}_4\not\cong K_4. \end{equation*}$$ This example also shows that order alone does not classify all groups, but it does classify cyclic groups. ::: :::solved-problem Show that $S_3$ and $\mathbb{Z}_6$ are not isomorphic, even though both have order $6$. ::: :::solution Let $S_3$ be the symmetric group on three letters and let $\mathbb{Z}_6$ be the cyclic group of order $6$. The group $\mathbb{Z}_6$ is commutative. The group $S_3$ is not commutative. For example, in $S_3$, $$\begin{equation*} (12)(23)\neq(23)(12). \end{equation*}$$ Since commutativity is preserved by isomorphism, a commutative group cannot be isomorphic to a noncommutative group. Hence $S_3\not\cong\mathbb{Z}_6$. $\square$ ::: :::tip The classification of cyclic groups should be used only after confirming that the groups are cyclic. Two groups of the same finite order need not be isomorphic in general. But two cyclic groups of the same finite order are always isomorphic. ::: :::exercise 1. Prove that every cyclic group of order $8$ is isomorphic to $\mathbb{Z}_8$. 2. Prove that every infinite cyclic group has exactly two generators. 3. Explain why a group of order $5$ must be cyclic. ::: :::answer 1. This is the finite cyclic classification theorem with $n=8$. 2. If $G=\langle a\rangle$ is infinite cyclic, its generators are $a$ and $a^{-1}$, corresponding to the generators $1$ and $-1$ of $\mathbb{Z}$. 3. If $a\neq e$, then $o(a)$ divides $5$ and is not $1$, so $o(a)=5$. Hence $G=\langle a\rangle$. ::: :::faq[Frequently Asked Questions] Q: Are all groups of the same order isomorphic? A: No. The statement is true for cyclic groups of the same order, not for arbitrary groups. Q: Why is every infinite cyclic group like $\mathbb{Z}$? A: Each element is a unique integral power of a generator, just as each integer is a unique multiple of $1$. Q: Does $\mathbb{Z}_n$ always mean a cyclic group? A: Under addition modulo $n$, $\mathbb{Z}_n$ is cyclic and generated by $[1]$. ::: :::call-to-action[Continue to Cyclic Homomorphisms] subtitle: Use the image of a generator to list and understand homomorphisms between cyclic groups. button: Study Cyclic Homomorphisms | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/homomorphisms-between-cyclic-groups button-ghost: Review Isomorphism Tests | /abstract-algebra/homomorphisms-and-isomorphisms-of-groups/homomorphisms-of-groups/isomorphism-tests-and-examples :::