After studying semigroup conditions for groups, we return to groups and develop the language of powers. Powers are repeated products of a single group element. This notation is essential because many later concepts, such as order of an element, cyclic group, generator, and roots of identity, are expressed using powers. In ordinary arithmetic, powers are familiar, but in an abstract group they must be defined through the group operation. In this lesson, students will learn positive powers, zero power, and negative powers of an element in a group. :::definition[Positive Power] Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. For $n\in\mathbb{N}$, the $positive power$ $a^n$ is defined by $$\begin{equation*} a^n=\underbrace{a\circ a\circ\cdots\circ a}_{n\text{ times}}. \end{equation*}$$ Also, $$\begin{equation*} a^0=e. \end{equation*}$$ ::: The notation $a^n$ means repeated use of the group operation. If the operation is written multiplicatively, then $a^3=a\circ a\circ a$. If the operation is addition, the same idea is usually written as $na$ rather than $a^n$. For example, in $(\mathbb{Z},+)$, the expression $3a$ means $a+a+a$. In this section, we use multiplicative-style notation because it is the standard notation for abstract groups. :::definition[Negative Power] Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. For $n\in\mathbb{N}$, the $negative power$ $a^{-n}$ is defined by $$\begin{equation*} a^{-n}=(a^{-1})^n. \end{equation*}$$ Thus $a^{-n}$ is the $n$-th positive power of the inverse element $a^{-1}$. ::: Negative powers are possible only because every element of a group has an inverse. The expression $a^{-3}$ means $$\begin{equation*} a^{-3}=a^{-1}\circ a^{-1}\circ a^{-1}. \end{equation*}$$ Students often read $a^{-n}$ as a fraction. That interpretation is safe only in numerical multiplicative groups. In a general group, $a^{-n}$ means repeated product of $a^{-1}$, not a quotient in the ordinary sense. :::example Let $(\mathbb{R}-\{0\},\cdot)$ be the multiplicative group of nonzero real numbers. If $a=2$, then $$\begin{equation*} a^3=2^3=8 \end{equation*}$$ and $$\begin{equation*} a^{-3}=2^{-3}=\left(\frac{1}{2}\right)^3=\frac{1}{8}. \end{equation*}$$ Here the abstract definition agrees with ordinary numerical powers. ::: :::example Let $(\mathbb{Z},+)$ be the additive group of integers. If the operation is addition, repeated products are written as repeated sums. Thus the multiplicative notation $a^3$ corresponds to $$\begin{equation*} a+a+a=3a. \end{equation*}$$ The inverse element of $a$ is $-a$, so the negative power notation corresponds to $$\begin{equation*} a^{-3}=(-a)+(-a)+(-a)=-3a \end{equation*}$$ when translated into additive notation. ::: :::theorem Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. Then $$\begin{equation*} a^1=a \end{equation*}$$ and $$\begin{equation*} a^0=e. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group with identity element $e$ and $a\in G$. To prove that $a^1=a$ and $a^0=e$. By the definition of positive power, $a^1$ is the product consisting of one factor $a$. Therefore $$\begin{equation*} a^1=a. \end{equation*}$$ By the definition of zero power in a group, $$\begin{equation*} a^0=e. \end{equation*}$$ Hence, $a^1=a$ and $a^0=e$. $\square$ ::: :::solved-problem Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. Write $a^4$, $a^{-2}$, and $a^0$ using the group operation. ::: :::solution Let $(G,\circ)$ be a group with identity element $e$ and let $a\in G$. To write $a^4$, $a^{-2}$, and $a^0$ using the group operation. By the definition of positive power, $$\begin{equation*} a^4=a\circ a\circ a\circ a. \end{equation*}$$ By the definition of negative power, $$\begin{equation*} a^{-2}=a^{-1}\circ a^{-1}. \end{equation*}$$ By the definition of zero power, $$\begin{equation*} a^0=e. \end{equation*}$$ Hence, $$\begin{equation*} a^4=a\circ a\circ a\circ a,\quad a^{-2}=a^{-1}\circ a^{-1},\quad a^0=e. \end{equation*}$$ $\square$ ::: :::remark The expression $a^0=e$ is not a numerical convention copied into groups. It is the correct identity-compatible definition. If we want the law $a^m\circ a^n=a^{m+n}$ to remain valid when one exponent is $0$, then $a^0$ must act as the identity element. For example, $a^m\circ a^0$ should equal $a^m$, and this happens exactly when $a^0=e$. ::: Integral exponents split into three cases: positive powers, the zero power, and negative powers. We observe these cases on an exponent number line centered at $0$. Move the exponent and notice how the interpretation changes from repeated products of $a^{-1}$ to the identity element and then to repeated products of $a$. This helps prevent the common mistake of treating negative powers as ordinary fractions in every group. :::scientific-preview[Exponent Type Number Line] @libraries: plotly ::: Use the calculator to view powers of a number in the multiplicative group of nonzero rational numbers. Enter a nonzero integer base and an integer exponent. The calculator displays the interpretation of positive, zero, and negative powers. This is a numerical model for the abstract group definition. :::calculator[Power Notation] ::: :::exercise [1] Define $a^n$ for $n\in\mathbb{N}$ in a group. [2] Define $a^0$ in a group. [3] Define $a^{-n}$ for $n\in\mathbb{N}$ in a group. [4] Write $a^{-4}$ as a product of inverse elements. [5] Translate $a^5$ into additive notation. ::: :::answer [1] $a^n=\underbrace{a\circ a\circ\cdots\circ a}_{n\text{ times}}$. [2] $a^0=e$, where $e$ is the identity element. [3] $a^{-n}=(a^{-1})^n$. [4] $a^{-4}=a^{-1}\circ a^{-1}\circ a^{-1}\circ a^{-1}$. [5] In additive notation, $a^5$ corresponds to $5a=a+a+a+a+a$. ::: :::faq Q: Why is $a^0$ defined as $e$? A: It makes the power notation compatible with the identity element and exponent laws. Q: What does $a^{-n}$ mean in a group? A: It means $(a^{-1})^n$, the product of $n$ copies of the inverse of $a$. Q: Does negative power require the group property? A: Yes. Negative powers require inverse elements. Q: Is power notation the same in additive groups? A: The idea is the same, but additive groups usually use $na$ instead of $a^n$. ::: :::call-to-action[Laws of Powers] subtitle: Continue to the exponent laws for powers of a fixed group element. button: Next Lesson | /abstract-algebra/introduction-to-groups/powers-of-elements-in-groups/laws-of-powers-in-groups button-ghost: Previous Section | /abstract-algebra/introduction-to-groups/semigroup-conditions-for-groups/unique-idempotent-semigroups :::