After proving the laws of positive powers, we now extend power notation to all integer exponents. This extension is one of the reasons groups are so useful: every element has an inverse, so negative exponents are meaningful. Once zero and negative exponents are included, the familiar exponent laws continue to hold for a fixed group element. In this lesson, students will learn how to combine positive, zero, and negative powers and why powers of the same element behave like an integer-indexed system inside the group. :::definition[Integral Power] Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. For $n\in\mathbb{Z}$, the $integral power$ $a^n$ is defined as follows: (i) If $n>0$, then $a^n=\underbrace{a\circ a\circ\cdots\circ a}_{n\text{ times}}$. (ii) If $n=0$, then $a^0=e$. (iii) If $n<0$, then $a^n=(a^{-1})^{-n}$. ::: This definition packages all powers into one notation. Positive exponents mean repeated products of $a$. The exponent $0$ gives the identity element. Negative exponents mean repeated products of $a^{-1}$. For example, $$\begin{equation*} a^{-4}=(a^{-1})^4. \end{equation*}$$ The symbol $-n$ in the negative case is positive because $n<0$. :::lemma Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. If $n\in\mathbb{N}$, then $$\begin{equation*} a^n\circ a^{-n}=a^{-n}\circ a^n=e. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$, $a\in G$, and $n\in\mathbb{N}$. To prove that $a^n\circ a^{-n}=a^{-n}\circ a^n=e$. By the definition of negative power, $$\begin{equation*} a^{-n}=(a^{-1})^n. \end{equation*}$$ Now $a^{-n}$ is the inverse of $a^n$, because $$\begin{equation*} (a^n)^{-1}=a^{-n}. \end{equation*}$$ Therefore $$\begin{equation*} a^n\circ a^{-n}=e \end{equation*}$$ and $$\begin{equation*} a^{-n}\circ a^n=e. \end{equation*}$$ Hence, $$\begin{equation*} a^n\circ a^{-n}=a^{-n}\circ a^n=e. \end{equation*}$$ $\square$ ::: :::theorem[Integral Power Law] Let $(G,\circ)$ be a group and let $a\in G$. If $m,n\in\mathbb{Z}$, then $$\begin{equation*} a^m\circ a^n=a^{m+n}. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $a\in G$, and $m,n\in\mathbb{Z}$. To prove that $$\begin{equation*} a^m\circ a^n=a^{m+n}. \end{equation*}$$ If $m=0$, then $$\begin{align*} a^m\circ a^n &=a^0\circ a^n\\ &=e\circ a^n\\ &=a^n\\ &=a^{0+n}\\ &=a^{m+n}. \end{align*}$$ If $n=0$, then $$\begin{align*} a^m\circ a^n &=a^m\circ a^0\\ &=a^m\circ e\\ &=a^m\\ &=a^{m+0}\\ &=a^{m+n}. \end{align*}$$ If $m>0$ and $n>0$, then the result follows from the positive power law. If $m<0$ and $n<0$, let $m=-r$ and $n=-s$, where $r,s\in\mathbb{N}$. Then $$\begin{align*} a^m\circ a^n &=a^{-r}\circ a^{-s}\\ &=(a^{-1})^r\circ(a^{-1})^s\\ &=(a^{-1})^{r+s}\\ &=a^{-(r+s)}\\ &=a^{m+n}. \end{align*}$$ It remains to consider the case where one exponent is positive and the other is negative. Let $m>0$ and $n=-s$, where $s\in\mathbb{N}$. If $m\geq s$, then $$\begin{align*} a^m\circ a^{-s} &=a^{m-s}\circ a^s\circ a^{-s}\\ &=a^{m-s}\circ e\\ &=a^{m-s}\\ &=a^{m+n}. \end{align*}$$ If $m<s$, then $$\begin{align*} a^m\circ a^{-s} &=a^m\circ a^{-m}\circ a^{-(s-m)}\\ &=e\circ a^{-(s-m)}\\ &=a^{-(s-m)}\\ &=a^{m-s}\\ &=a^{m+n}. \end{align*}$$ The case $m<0$ and $n>0$ is similar, using the same cancellation of equal positive and negative powers of $a$. Hence, $$\begin{equation*} a^m\circ a^n=a^{m+n}\quad \forall\, m,n\in\mathbb{Z}. \end{equation*}$$ $\square$ ::: The theorem says that the powers of a fixed group element behave like integer addition in their exponents. This does not mean that the whole group is commutative. It only says that powers of the same element combine predictably. For example, $a^3\circ a^{-5}=a^{-2}$. The cancellation occurs because $a^3$ cancels three of the inverse factors appearing inside $a^{-5}$. :::corollary Let $(G,\circ)$ be a group and let $a\in G$. If $m,n\in\mathbb{Z}$, then $$\begin{equation*} (a^m)^n=a^{mn}. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $a\in G$, and $m,n\in\mathbb{Z}$. To prove that $(a^m)^n=a^{mn}$. If $n>0$, then by repeated use of the integral power law, $$\begin{align*} (a^m)^n &=\underbrace{a^m\circ a^m\circ\cdots\circ a^m}_{n\text{ times}}\\ &=a^{m+m+\cdots+m}\\ &=a^{mn}. \end{align*}$$ If $n=0$, then $$\begin{equation*} (a^m)^0=e=a^0=a^{m\cdot0}. \end{equation*}$$ If $n<0$, then $n=-r$ for some $r\in\mathbb{N}$. Thus $$\begin{align*} (a^m)^n &=(a^m)^{-r}\\ &=\big((a^m)^{-1}\big)^r\\ &=(a^{-m})^r\\ &=a^{-mr}\\ &=a^{mn}. \end{align*}$$ Hence, $$\begin{equation*} (a^m)^n=a^{mn}. \end{equation*}$$ $\square$ ::: :::example Let $(G,\circ)$ be a group and let $a\in G$. Then $$\begin{align*} a^7\circ a^{-3} &=a^{7-3}\\ &=a^4. \end{align*}$$ Also, $$\begin{align*} a^{-8}\circ a^5 &=a^{-8+5}\\ &=a^{-3}. \end{align*}$$ These calculations use only powers of the same element. ::: :::solved-problem Let $(G,\circ)$ be a group and let $a\in G$. Simplify $$\begin{equation*} a^{-4}\circ a^9\circ a^{-2}. \end{equation*}$$ ::: :::solution Let $(G,\circ)$ be a group and let $a\in G$. To simplify $$\begin{equation*} a^{-4}\circ a^9\circ a^{-2}. \end{equation*}$$ Using the integral power law, $$\begin{align*} a^{-4}\circ a^9\circ a^{-2} &=a^{-4+9-2}\\ &=a^3. \end{align*}$$ Hence, $$\begin{equation*} a^{-4}\circ a^9\circ a^{-2}=a^3. \end{equation*}$$ $\square$ ::: :::remark The integral power law applies only when the base is the same. It is correct to combine $a^{-4}\circ a^9$ because both are powers of $a$. It is not correct to combine $a^m\circ b^n$ into a single power unless additional information is given about the relation between $a$ and $b$. ::: The integral power law reduces a product of powers of the same element to addition on the integer number line. We observe a running total of exponents as three powers are multiplied. Change the three exponents and notice that positive and negative steps can cancel. This visualizes why $a^{-4}\circ a^9\circ a^{-2}$ simplifies to $a^3$. :::scientific-preview[Running Exponent Sum] @libraries: plotly ::: Use the calculator to simplify powers of a single symbol by adding integer exponents. Enter three integer exponents. The calculator returns the combined exponent for $a^r\circ a^s\circ a^t$. This models the integral power law in any group. :::calculator[Integral Exponent Simplifier] ::: :::exercise [1] Define $a^n$ for $n\in\mathbb{Z}$. [2] Simplify $a^5\circ a^{-2}$. [3] Simplify $a^{-6}\circ a^4$. [4] Simplify $(a^{-3})^4$. [5] Explain why $a^m\circ a^n=a^{m+n}$ does not require commutativity of $G$. ::: :::answer [1] If $n>0$, $a^n$ is the product of $n$ copies of $a$; if $n=0$, $a^0=e$; if $n<0$, $a^n=(a^{-1})^{-n}$. [2] $a^3$. [3] $a^{-2}$. [4] $a^{-12}$. [5] The formula uses powers of one fixed element, so only associativity and inverse cancellation are needed. ::: :::faq Q: Do exponent laws hold for negative exponents in a group? A: Yes, for powers of a fixed element. Q: What is $a^{-n}$ when $n>0$? A: It is $(a^{-1})^n$. Q: What is $a^m\circ a^n$ for integers $m,n$? A: It is $a^{m+n}$. Q: Can we combine powers of different elements? A: Not without additional information. ::: :::call-to-action[Conjugation and Powers] subtitle: Continue to conjugation power formulas, where powers interact with expressions of the form $aba^{-1}$. button: Next Lesson | /abstract-algebra/introduction-to-groups/powers-of-elements-in-groups/conjugation-and-powers button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/powers-of-elements-in-groups/laws-of-powers-in-groups :::