After defining abelian groups, we now study a useful test for commutativity. In every group, the inverse of a product reverses order: $$\begin{equation*} (a\circ b)^{-1}=b^{-1}\circ a^{-1}. \end{equation*}$$ In an abelian group, reversing the order has no effect, so the formula becomes $$\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}. \end{equation*}$$ This lesson proves that the second formula is not only a consequence of commutativity, but also a criterion for an abelian group. :::theorem Let $(G,\circ)$ be a group with identity element $e$. Then $G$ is abelian if and only if $$\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$. To prove that $G$ is abelian if and only if $$\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}$$ [1] To prove that if $G$ is abelian, then $(a\circ b)^{-1}=a^{-1}\circ b^{-1}$ for all $a,b\in G$. Let $G$ be abelian and let $a,b\in G$. In every group, $$\begin{equation*} (a\circ b)^{-1}=b^{-1}\circ a^{-1}. \end{equation*}$$ Since $G$ is abelian, $b^{-1}\circ a^{-1}=a^{-1}\circ b^{-1}$. Therefore $$\begin{align*} (a\circ b)^{-1} &=b^{-1}\circ a^{-1}\\ &=a^{-1}\circ b^{-1} \quad [\because\, G \text{ is abelian}]. \end{align*}$$ Thus $$\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}$$ [2] To prove that if $(a\circ b)^{-1}=a^{-1}\circ b^{-1}$ for all $a,b\in G$, then $G$ is abelian. Let $a,b\in G$. Given that $$\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}. \end{equation*}$$ But the general inverse law gives $$\begin{equation*} (a\circ b)^{-1}=b^{-1}\circ a^{-1}. \end{equation*}$$ Therefore $$\begin{equation*} a^{-1}\circ b^{-1}=b^{-1}\circ a^{-1}. \end{equation*}$$ Taking inverses on both sides, we get $$\begin{align*} (a^{-1}\circ b^{-1})^{-1} &=(b^{-1}\circ a^{-1})^{-1}\\ b\circ a&=a\circ b. \end{align*}$$ Therefore $$\begin{equation*} a\circ b=b\circ a \quad \forall\, a,b\in G. \end{equation*}$$ Hence, $G$ is abelian. $\square$ ::: The theorem is a useful reminder that the order of factors matters in a non-abelian group. The formula $(a\circ b)^{-1}=b^{-1}\circ a^{-1}$ always holds. The formula $(a\circ b)^{-1}=a^{-1}\circ b^{-1}$ is special; it holds for every pair exactly when the group is abelian. Students often memorize the wrong order for the inverse of a product. This theorem explains why the wrong-looking order becomes correct only in the commutative case. :::example In the abelian group $(\mathbb{R}-\{0\},\cdot)$, let $a=2$ and $b=5$. Then $$\begin{equation*} (ab)^{-1}=10^{-1}=\frac{1}{10} \end{equation*}$$ and $$\begin{equation*} a^{-1}b^{-1}=\frac{1}{2}\cdot\frac{1}{5}=\frac{1}{10}. \end{equation*}$$ Since real multiplication is commutative, the order of inverse factors does not matter. ::: :::example In a matrix group under multiplication, the formula $$\begin{equation*} (AB)^{-1}=B^{-1}A^{-1} \end{equation*}$$ holds for invertible matrices $A$ and $B$. In general, one cannot replace the right-hand side by $A^{-1}B^{-1}$ unless $A$ and $B$ commute. Thus the inverse product criterion detects the difference between commutative and non-commutative multiplication. ::: :::solved-problem Let $(G,\circ)$ be a group. Suppose that $$\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}$$ Prove directly that $a\circ b=b\circ a$ for all $a,b\in G$. ::: :::solution Let $(G,\circ)$ be a group and let $a,b\in G$. Given that $$\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}$$ To prove that $$\begin{equation*} a\circ b=b\circ a. \end{equation*}$$ In every group, $$\begin{equation*} (a\circ b)^{-1}=b^{-1}\circ a^{-1}. \end{equation*}$$ Therefore $$\begin{equation*} a^{-1}\circ b^{-1}=b^{-1}\circ a^{-1}. \end{equation*}$$ Taking inverses on both sides, $$\begin{align*} (a^{-1}\circ b^{-1})^{-1} &=(b^{-1}\circ a^{-1})^{-1}\\ b\circ a&=a\circ b. \end{align*}$$ Thus $$\begin{equation*} a\circ b=b\circ a. \end{equation*}$$ Hence, $G$ is abelian. $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group. If $G$ is abelian and $a,b,c\in G$, prove that $$\begin{equation*} (a\circ b\circ c)^{-1}=a^{-1}\circ b^{-1}\circ c^{-1}. \end{equation*}$$ ::: :::solution Let $(G,\circ)$ be an abelian group and let $a,b,c\in G$. To prove that $$\begin{equation*} (a\circ b\circ c)^{-1}=a^{-1}\circ b^{-1}\circ c^{-1}. \end{equation*}$$ In every group, the inverse of a product reverses order. Therefore $$\begin{align*} (a\circ b\circ c)^{-1} &=c^{-1}\circ b^{-1}\circ a^{-1}. \end{align*}$$ Since $G$ is abelian, the inverse factors commute. Hence $$\begin{align*} c^{-1}\circ b^{-1}\circ a^{-1} &=a^{-1}\circ b^{-1}\circ c^{-1}. \end{align*}$$ Therefore $$\begin{equation*} (a\circ b\circ c)^{-1}=a^{-1}\circ b^{-1}\circ c^{-1}. \end{equation*}$$ $\square$ ::: :::remark The theorem must be read with the phrase “for all $a,b\in G$.” It is possible that one special pair satisfies $(a\circ b)^{-1}=a^{-1}\circ b^{-1}$ even in a non-abelian group. The criterion says that the formula must hold for every pair of elements. ::: We observe that the inverse of a product always reverses the order of the factors. Choose the arbitrary-group setting to keep the guaranteed identity $(a\circ b)^{-1}=b^{-1}\circ a^{-1}$. Choose the abelian setting to see why the reversed product can be rearranged as $a^{-1}\circ b^{-1}$. The criterion matters because the unreversed formula is valid for every pair exactly when the group is abelian. :::calculator[Inverse Product Test] ::: Notice that the order-reversing identity is universal, while the unreversed identity is a commutativity test. The phrase “for every pair” is essential because one accidental commuting pair does not make the whole group abelian. :::exercise [1] State the inverse product criterion for abelian groups. [2] In an arbitrary group, what is $(a\circ b)^{-1}$? [3] If $G$ is abelian, simplify $(a\circ b\circ c)^{-1}$. [4] If $(a\circ b)^{-1}=a^{-1}\circ b^{-1}$ for all $a,b\in G$, prove that $a^{-1}\circ b^{-1}=b^{-1}\circ a^{-1}$. [5] Explain why the condition must hold for all pairs. ::: :::answer [1] A group $G$ is abelian if and only if $(a\circ b)^{-1}=a^{-1}\circ b^{-1}$ for all $a,b\in G$. [2] $(a\circ b)^{-1}=b^{-1}\circ a^{-1}$. [3] $(a\circ b\circ c)^{-1}=a^{-1}\circ b^{-1}\circ c^{-1}$. [4] Since $(a\circ b)^{-1}$ is also $b^{-1}\circ a^{-1}$, equality with $a^{-1}\circ b^{-1}$ gives the result. [5] A special pair may commute accidentally; abelian means every pair commutes. ::: :::faq Q: What is the inverse of a product in any group? A: It is the product of the inverses in reverse order. Q: When is $(a\circ b)^{-1}=a^{-1}\circ b^{-1}$ valid for all $a,b$? A: Exactly when the group is abelian. Q: Does the theorem require finite groups? A: No. It holds for all groups. Q: Why does taking inverses help prove commutativity? A: It turns equality of inverse products into equality of the original products in reversed order. ::: :::call-to-action[Groups of Exponent Two] subtitle: Continue to groups in which every element is its own inverse. button: Next Lesson | published/abstract-algebra/introduction-to-groups/abelian-groups/groups-of-exponent-two button-ghost: Previous Lesson | published/abstract-algebra/introduction-to-groups/abelian-groups/abelian-group :::