Laws of Powers

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.

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}

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}

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*}

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.

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

Let

(G,\circ)

be a group and let

a\in G

. Simplify

\begin{equation*} a^3\circ a^5. \end{equation*}

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

Let

(G,\circ)

be a group and let

a\in G

. Simplify

\begin{equation*} (a^4)^3\circ(a^2)^{-1}. \end{equation*}

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

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.

Use the calculator below 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.

[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}

[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}

No. The laws in this lesson involve powers of a single element, so commutativity of the group is not needed.

No. It is generally false unless

a

and

b

commute.

Associativity makes repeated products and regrouping of powers meaningful.

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