After defining cyclic groups and generators, the next structural result is that cyclic groups are always abelian. This theorem is powerful because it turns the existence of a single generator into a commutativity law for the whole group. The proof is short but conceptually important: if every element is a power of the same element, then any two elements commute because integer addition in the exponents is commutative. In this lesson, students will prove that every cyclic group is abelian and learn why the converse is false. :::theorem Let $(G,\circ)$ be a group. If $G$ is cyclic, then $G$ is abelian. ::: :::proof Given that $(G,\circ)$ is a group and $G$ is cyclic. To prove that $G$ is abelian. Since $G$ is cyclic, there exists $a\in G$ such that $$\begin{equation*} G=\langle a\rangle. \end{equation*}$$ Let $x,y\in G$. Since $G=\langle a\rangle$, there exist $m,n\in\mathbb{Z}$ such that $$\begin{equation*} x=a^m,\quad y=a^n. \end{equation*}$$ Now $$\begin{align*} x\circ y &=a^m\circ a^n\\ &=a^{m+n}\\ &=a^{n+m}\\ &=a^n\circ a^m\\ &=y\circ x. \end{align*}$$ Therefore $$\begin{equation*} x\circ y=y\circ x \quad \forall\, x,y\in G. \end{equation*}$$ Hence, $G$ is abelian. $\square$ ::: The proof that cyclic groups are abelian is really a statement about exponents. If $x=a^m$ and $y=a^n$, then multiplying in either order gives the same final exponent because $m+n=n+m$. Choose two exponents and compare both paths. The preview shows that both products land at the same power of the generator. :::scientific-preview[Power Commutativity Explorer] ::: The proof uses only the power law $a^m\circ a^n=a^{m+n}$ and the commutativity of integer addition. It does not assume that $G$ was already abelian. This is an important logical point. The commutativity of the group operation is derived from the fact that both elements are powers of the same generator. :::example The group $(\mathbb{Z},+)$ is cyclic because $\mathbb{Z}=\langle 1\rangle$. Therefore it is abelian. Indeed, $$\begin{equation*} m+n=n+m \quad \forall\, m,n\in\mathbb{Z}. \end{equation*}$$ This example is familiar, but the theorem shows that the same conclusion holds for every cyclic group, not only for integer addition. ::: :::example The additive group $(\mathbb{Z}_n,+_n)$ is cyclic because $$\begin{equation*} \mathbb{Z}_n=\langle \overline{1}\rangle. \end{equation*}$$ Therefore $(\mathbb{Z}_n,+_n)$ is abelian. In concrete terms, $$\begin{equation*} \overline{a}+_n\overline{b}=\overline{b}+_n\overline{a} \quad \forall\, \overline{a},\overline{b}\in\mathbb{Z}_n. \end{equation*}$$ ::: :::remark The converse of the theorem is false. Every cyclic group is abelian, but not every abelian group is cyclic. The Klein four group $V_4$ is abelian, but it is not cyclic because no element has order $4$. This distinction is important: commutativity alone does not guarantee that one element generates the whole group. ::: :::solved-problem Let $(G,\circ)$ be a cyclic group generated by $a$. If $x=a^4$ and $y=a^{-7}$, prove that $x\circ y=y\circ x$. ::: :::solution Let $(G,\circ)$ be a cyclic group generated by $a$, and let $$\begin{equation*} x=a^4,\quad y=a^{-7}. \end{equation*}$$ To prove that $x\circ y=y\circ x$. Using the integral power law, $$\begin{align*} x\circ y &=a^4\circ a^{-7}\\ &=a^{-3}. \end{align*}$$ Also, $$\begin{align*} y\circ x &=a^{-7}\circ a^4\\ &=a^{-3}. \end{align*}$$ Therefore $$\begin{equation*} x\circ y=y\circ x. \end{equation*}$$ Hence, $x\circ y=y\circ x$. $\square$ ::: :::solved-problem Show that the Klein four group $V_4$ is abelian but not cyclic. ::: :::solution Let $$\begin{equation*} V_4=\{e,a,b,c\} \end{equation*}$$ be the Klein four group. To show that $V_4$ is abelian but not cyclic. The operation table of $V_4$ is symmetric, so $$\begin{equation*} x\circ y=y\circ x \quad \forall\, x,y\in V_4. \end{equation*}$$ Therefore $V_4$ is abelian. Now $$\begin{equation*} o(e)=1,\quad o(a)=2,\quad o(b)=2,\quad o(c)=2. \end{equation*}$$ If $V_4$ were cyclic, then it would have an element of order $4$. But no element of $V_4$ has order $4$. Therefore $V_4$ is not cyclic. Hence, $V_4$ is abelian but not cyclic. $\square$ ::: Use the calculator below to multiply powers of a single generator symbolically. Enter two integer exponents $m$ and $n$. The calculator compares $a^m a^n$ with $a^n a^m$. This models the proof that powers of one generator commute. :::calculator[Cyclic Commutativity Calculator] ::: :::exercise [1] Prove that every cyclic group is abelian. [2] Give an example of a cyclic abelian group. [3] Give an example of an abelian group that is not cyclic. [4] If $G=\langle a\rangle$, write arbitrary elements $x,y\in G$ in terms of powers of $a$. [5] If $x=a^{-3}$ and $y=a^{10}$, simplify $x\circ y$ and $y\circ x$. ::: :::answer [1] If $x=a^m$ and $y=a^n$, then $x\circ y=a^{m+n}=a^{n+m}=y\circ x$. [2] $(\mathbb{Z},+)$ is cyclic and abelian. [3] The Klein four group $V_4$ is abelian but not cyclic. [4] There exist $m,n\in\mathbb{Z}$ such that $x=a^m$ and $y=a^n$. [5] Both products equal $a^7$. ::: :::faq Q: Is every cyclic group abelian? A: Yes. Every cyclic group is abelian. Q: Is every abelian group cyclic? A: No. The Klein four group is abelian but not cyclic. Q: Why do elements commute in a cyclic group? A: They are powers of the same element, and exponents add in either order. Q: Does the proof require the group to be finite? A: No. The theorem holds for both finite and infinite cyclic groups. ::: :::call-to-action[Finite and Infinite Cyclic Groups] subtitle: Continue to the relation between cyclic groups and the order of a generator. button: Next Lesson | /abstract-algebra/introduction-to-groups/cyclic-groups/finite-and-infinite-cyclic-groups button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/cyclic-groups/cyclic-group :::