After studying cyclic groups, we have already seen an important source of commutativity: every cyclic group is abelian. We now study abelian groups directly. An abelian group is a group in which the order of multiplication does not matter. This condition is simple to state, but it strongly affects proofs, inverse formulas, power formulas, and examples. In this lesson, students will learn the definition of an abelian group, compare abelian and non-abelian examples, and understand why commutativity must be checked as a separate property. :::definition[Abelian Group] Let $(G,\circ)$ be a group. Then $(G,\circ)$ is called an $abelian group$ or $commutative group$ if $$\begin{equation*} a\circ b=b\circ a \quad \forall\, a,b\in G. \end{equation*}$$ Thus an abelian group is a group whose operation is commutative. ::: The definition says that every pair of elements must commute. It is not enough to check a few selected pairs. If even one pair $a,b\in G$ satisfies $a\circ b\neq b\circ a$, then the group is not abelian. In additive notation, the operation is usually written as $+$, and commutativity appears as $a+b=b+a$. In multiplicative notation, the condition appears as $ab=ba$. :::example The group $(\mathbb{Z},+)$ is abelian. Let $m,n\in\mathbb{Z}$. Then ordinary addition gives $$\begin{equation*} m+n=n+m. \end{equation*}$$ Therefore $$\begin{equation*} m+n=n+m \quad \forall\, m,n\in\mathbb{Z}. \end{equation*}$$ Hence, $(\mathbb{Z},+)$ is an abelian group. ::: :::example The group $(\mathbb{Q}-\{0\},\cdot)$ is abelian. Let $a,b\in\mathbb{Q}-\{0\}$. Since ordinary multiplication of rational numbers is commutative, $$\begin{equation*} ab=ba. \end{equation*}$$ Therefore $$\begin{equation*} a\cdot b=b\cdot a \quad \forall\, a,b\in\mathbb{Q}-\{0\}. \end{equation*}$$ Hence, $(\mathbb{Q}-\{0\},\cdot)$ is an abelian group. ::: :::example The symmetric group $\mathbb{S}_3$ is not abelian. Let $$\begin{equation*} \sigma=(123),\quad \tau=(12). \end{equation*}$$ Using right-to-left composition, we get $$\begin{equation*} \sigma\circ\tau=(13) \end{equation*}$$ and $$\begin{equation*} \tau\circ\sigma=(23). \end{equation*}$$ Since $$\begin{equation*} (13)\neq(23), \end{equation*}$$ we have $$\begin{equation*} \sigma\circ\tau\neq\tau\circ\sigma. \end{equation*}$$ Hence, $\mathbb{S}_3$ is not abelian. ::: :::theorem Let $(G,\circ)$ be a cyclic group. Then $G$ is abelian. ::: :::proof Given that $(G,\circ)$ is a cyclic group. 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 $x,y\in G$. Then 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. $\square$ ::: This theorem shows that cyclic groups are a special family of abelian groups. The converse is false. The Klein four group is abelian, but it is not cyclic because no element has order $4$. Thus commutativity does not automatically mean that one element generates the entire group. A common student error is to confuse “abelian” with “cyclic.” Every cyclic group is abelian, but abelian groups may require more than one generator. :::solved-problem Let $(G,\circ)$ be a group with exactly two elements. Prove that $G$ is abelian. ::: :::solution Let $(G,\circ)$ be a group with identity element $e$. Given that $G$ has exactly two elements. To prove that $G$ is abelian. Since $e\in G$, there exists $a\in G$ such that $$\begin{equation*} G=\{e,a\} \end{equation*}$$ and $a\neq e$. Now $$\begin{equation*} e\circ e=e,\quad e\circ a=a,\quad a\circ e=a. \end{equation*}$$ Since $a\in G$ and $G$ is closed under $\circ$, we get $$\begin{equation*} a\circ a\in G. \end{equation*}$$ If possible let $$\begin{equation*} a\circ a=a. \end{equation*}$$ Then $$\begin{align*} a\circ a=a &\implies a\circ a=a\circ e \quad [\because\, a\circ e=a]\\ &\implies a=e \quad [\because\, \text{left cancellation law}]. \end{align*}$$ A contradiction, since $a\neq e$. Hence, $$\begin{equation*} a\circ a=e. \end{equation*}$$ Thus all products in $G$ satisfy $$\begin{equation*} x\circ y=y\circ x \quad \forall\, x,y\in G. \end{equation*}$$ Hence, $G$ is abelian. $\square$ ::: :::remark A group of two elements is necessarily cyclic and abelian. But this is a special small case. Larger groups may be abelian or non-abelian, cyclic or non-cyclic. The correct method is always to check the operation, not only the size of the set. ::: We observe commutativity by comparing $x\circ y$ with $y\circ x$ for chosen elements. Select the cyclic group $\mathbb{Z}_4$ to see that every pair gives the same result in both orders. Select $\mathbb{S}_3$ to test permutation composition, where the order of composition can change the answer. Try choosing $(123)$ and $(12)$ in different orders and observe that one non-commuting pair is enough to prove that a group is not abelian. :::calculator[Commutativity Test] ::: Notice that the cyclic example always gives equal products because both elements come from powers of one generator. In contrast, the permutation example shows that a single failure of $x\circ y=y\circ x$ disproves abelianity for the whole group. :::exercise [1] Define an abelian group. [2] Give two examples of abelian groups. [3] Give one example of a non-abelian group. [4] Prove that every cyclic group is abelian. [5] Prove that every group with exactly two elements is abelian. ::: :::answer [1] A group $(G,\circ)$ is abelian if $a\circ b=b\circ a$ for all $a,b\in G$. [2] Examples are $(\mathbb{Z},+)$ and $(\mathbb{Q}-\{0\},\cdot)$. [3] The group $\mathbb{S}_3$ is non-abelian. [4] If $G=\langle a\rangle$, then $x=a^m$ and $y=a^n$, so $x\circ y=a^{m+n}=a^{n+m}=y\circ x$. [5] If $G=\{e,a\}$, then $a\circ a=e$; the multiplication table is symmetric, so $G$ is abelian. ::: :::faq Q: What is an abelian group? A: It is a group in which every pair of elements commutes. 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: Is $\mathbb{S}_3$ abelian? A: No. Some permutations in $\mathbb{S}_3$ do not commute. ::: :::call-to-action[Inverse Product Criterion] subtitle: Continue to a useful test for abelian groups using the inverse of a product. button: Next Lesson | published/abstract-algebra/introduction-to-groups/abelian-groups/inverse-product-criterion button-ghost: Previous Section | published/abstract-algebra/introduction-to-groups/cyclic-groups/even-cyclic-groups :::