After defining positive, zero, and negative powers, we now prove the basic laws of powers. These laws are familiar from school algebra, but in group theory they must be justified from the definition of powers and associativity. The most important point is that all powers in the same formula must be powers of the same group element. In that case, powers combine exactly as expected. In this lesson, students will prove the product law, the power-of-a-power law, the inverse-power law, and the identity-power law for positive exponents. :::theorem[Laws of Powers] Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. Then: (i) $a^m\circ a^n=a^{m+n}\quad \forall\, m,n\in\mathbb{N}$. (ii) $(a^m)^n=a^{mn}\quad \forall\, m,n\in\mathbb{N}$. (iii) $(a^m)^{-1}=a^{-m}\quad \forall\, m\in\mathbb{N}$. (iv) $e^m=e\quad \forall\, m\in\mathbb{N}$. ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a\in G$. To prove the stated laws of powers. [1] To prove that $a^m\circ a^n=a^{m+n}\quad \forall\, m,n\in\mathbb{N}$. Let $m,n\in\mathbb{N}$. By the definition of positive power, $$\begin{align*} a^m\circ a^n &=\underbrace{a\circ a\circ\cdots\circ a}_{m\text{ times}}\circ\underbrace{a\circ a\circ\cdots\circ a}_{n\text{ times}}\\ &=\underbrace{a\circ a\circ\cdots\circ a}_{m+n\text{ times}}\\ &=a^{m+n}. \end{align*}$$ Therefore $a^m\circ a^n=a^{m+n}$. [2] To prove that $(a^m)^n=a^{mn}\quad \forall\, m,n\in\mathbb{N}$. Let $m,n\in\mathbb{N}$. Then $$\begin{align*} (a^m)^n &=\underbrace{a^m\circ a^m\circ\cdots\circ a^m}_{n\text{ times}}\\ &=\underbrace{a\circ a\circ\cdots\circ a}_{mn\text{ times}}\\ &=a^{mn}. \end{align*}$$ Therefore $(a^m)^n=a^{mn}$. [3] To prove that $(a^m)^{-1}=a^{-m}\quad \forall\, m\in\mathbb{N}$. Let $m\in\mathbb{N}$. Since $a^{-m}=(a^{-1})^m$, we verify that $a^{-m}$ is the inverse of $a^m$. Now $$\begin{align*} a^m\circ a^{-m} &=a^m\circ(a^{-1})^m\\ &=\underbrace{a\circ\cdots\circ a}_{m\text{ times}}\circ\underbrace{a^{-1}\circ\cdots\circ a^{-1}}_{m\text{ times}}\\ &=e. \end{align*}$$ Also, $$\begin{align*} a^{-m}\circ a^m &=(a^{-1})^m\circ a^m\\ &=\underbrace{a^{-1}\circ\cdots\circ a^{-1}}_{m\text{ times}}\circ\underbrace{a\circ\cdots\circ a}_{m\text{ times}}\\ &=e. \end{align*}$$ Therefore $a^{-m}$ is the inverse of $a^m$. Thus $$\begin{equation*} (a^m)^{-1}=a^{-m}. \end{equation*}$$ [4] To prove that $e^m=e\quad \forall\, m\in\mathbb{N}$. Let $m\in\mathbb{N}$. Then $$\begin{align*} e^m &=\underbrace{e\circ e\circ\cdots\circ e}_{m\text{ times}}\\ &=e. \end{align*}$$ Hence, all stated laws of powers hold. $\square$ ::: The proof of the inverse-power law deserves attention. The expression $$\begin{equation*} a^m\circ a^{-m}=e \end{equation*}$$ is not a commutative cancellation argument. It works because the factors are arranged as $m$ copies of $a$ followed by $m$ copies of $a^{-1}$, and associativity allows successive cancellation from the middle outward. For example, $$\begin{equation*} a^3\circ a^{-3}=a\circ a\circ a\circ a^{-1}\circ a^{-1}\circ a^{-1}=e. \end{equation*}$$ :::example Let $(\mathbb{R}-\{0\},\cdot)$ be the multiplicative group of nonzero real numbers and let $a=2$. Then $$\begin{equation*} 2^3\cdot2^4=2^{3+4}=2^7. \end{equation*}$$ Also, $$\begin{equation*} (2^3)^4=2^{12}. \end{equation*}$$ These are numerical instances of the general power laws in a group. ::: :::example Let $(G,\circ)$ be a group and let $a\in G$. Then $$\begin{equation*} (a^5)^{-1}=a^{-5}. \end{equation*}$$ Indeed, $$\begin{equation*} a^{-5}=a^{-1}\circ a^{-1}\circ a^{-1}\circ a^{-1}\circ a^{-1}. \end{equation*}$$ This element is the inverse of $a^5$. ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. Simplify $$\begin{equation*} a^3\circ a^5. \end{equation*}$$ ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. To simplify $a^3\circ a^5$. Using the product law for powers, $$\begin{align*} a^3\circ a^5 &=a^{3+5}\\ &=a^8. \end{align*}$$ Hence, $$\begin{equation*} a^3\circ a^5=a^8. \end{equation*}$$ $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. Simplify $$\begin{equation*} (a^4)^3\circ(a^2)^{-1}. \end{equation*}$$ ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. To simplify $$\begin{equation*} (a^4)^3\circ(a^2)^{-1}. \end{equation*}$$ Using the laws of powers, $$\begin{align*} (a^4)^3\circ(a^2)^{-1} &=a^{12}\circ a^{-2}\\ &=a^{10}. \end{align*}$$ Hence, $$\begin{equation*} (a^4)^3\circ(a^2)^{-1}=a^{10}. \end{equation*}$$ $\square$ ::: :::remark The formula $(a\circ b)^n=a^n\circ b^n$ is not valid in every group. It is valid when $a\circ b=b\circ a$, but it can fail in a non-commutative group. The power laws in this lesson involve powers of one fixed element $a$. A common mistake is to extend these laws to products of different elements without checking commutativity. ::: The power laws combine exponents while keeping the same base. We observe two familiar identities: $a^m a^n=a^{m+n}$ and $(a^m)^n=a^{mn}$. Change $m$ and $n$ and compare the exponent produced by addition with the exponent produced by multiplication. This helps distinguish the product law from the power-of-a-power law. :::scientific-preview[Exponent Law Comparison] @libraries: plotly ::: Use the calculator to check exponent laws for ordinary nonzero numbers. Enter a base $a$ and positive integers $m,n$. The calculator compares $a^m a^n$ with $a^{m+n}$ and $(a^m)^n$ with $a^{mn}$. The numerical example supports the abstract theorem, but the theorem itself holds in every group for powers of one fixed element. :::calculator[Power Law Test] ::: :::exercise [1] State the product law for powers of one group element. [2] State the power-of-a-power law. [3] Prove that $(a^m)^{-1}=a^{-m}$ for $m\in\mathbb{N}$. [4] Simplify $a^6\circ a^7$. [5] Simplify $(a^3)^5\circ(a^4)^{-1}$. ::: :::answer [1] $a^m\circ a^n=a^{m+n}$. [2] $(a^m)^n=a^{mn}$. [3] Since $a^m\circ a^{-m}=a^{-m}\circ a^m=e$, $a^{-m}$ is the inverse of $a^m$. [4] $a^{13}$. [5] $(a^3)^5\circ(a^4)^{-1}=a^{15}\circ a^{-4}=a^{11}$. ::: :::faq Q: Do the power laws require commutativity of the group? A: No. The laws in this lesson involve powers of a single element, so commutativity of the group is not needed. Q: Is $(a\circ b)^n=a^n\circ b^n$ always true? A: No. It is generally false unless $a$ and $b$ commute. Q: Why is associativity needed? A: Associativity makes repeated products and regrouping of powers meaningful. Q: What is $(a^m)^{-1}$? A: It is $a^{-m}$. ::: :::call-to-action[Integral Power Laws] subtitle: Continue to integral powers and extend the power laws to negative and zero exponents. button: Next Lesson | /abstract-algebra/introduction-to-groups/powers-of-elements-in-groups/integral-power-laws button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/powers-of-elements-in-groups/positive-and-negative-powers :::