Properties of Inverses

After studying equations in groups, we now collect the main properties of inverses. These inverse laws are used constantly in group theory. They tell us that the identity is its own inverse, the inverse of an inverse returns the original element, and the inverse of a product reverses the order of the factors. The last rule is especially important in non-commutative groups, where reversing order cannot be ignored. In this lesson, students will prove the standard inverse properties and apply them to products of two or more group elements.

Let

(G,\circ)

be a group with identity element

e

. Then: (i)

e^{-1}=e

. (ii)

(a^{-1})^{-1}=a \quad \forall\, a\in G

. (iii)

(a\circ b)^{-1}=b^{-1}\circ a^{-1} \quad \forall\, a,b\in G

Let

(G,\circ)

be a group with identity element

e

. Then: (i)

e^{-1}=e

. (ii)

(a^{-1})^{-1}=a \quad \forall\, a\in G

. (iii)

(a\circ b)^{-1}=b^{-1}\circ a^{-1} \quad \forall\, a,b\in G

Given that

(G,\circ)

is a group with identity element

e

. To prove the stated properties of inverses. [1] To prove that

e^{-1}=e

. Since

e

is the identity element,

\begin{equation*} e\circ e=e. \end{equation*}

Thus

e

is an inverse element of

e

. Since the inverse of an element is unique,

\begin{equation*} e^{-1}=e. \end{equation*}

Therefore

e^{-1}=e

. [2] To prove that

(a^{-1})^{-1}=a \quad \forall\, a\in G

. Let

a\in G

. Then

\begin{equation*} a\circ a^{-1}=e \end{equation*}

and

\begin{equation*} a^{-1}\circ a=e. \end{equation*}

Therefore

a

is an inverse element of

a^{-1}

. Since the inverse of an element is unique,

\begin{equation*} (a^{-1})^{-1}=a. \end{equation*}

[3] To prove that

(a\circ b)^{-1}=b^{-1}\circ a^{-1} \quad \forall\, a,b\in G

. Let

a,b\in G

. Now

\begin{align*} (a\circ b)\circ(b^{-1}\circ a^{-1}) &=a\circ\big(b\circ(b^{-1}\circ a^{-1})\big) \quad [\because\, \circ \text{ is associative}]\\ &=a\circ\big((b\circ b^{-1})\circ a^{-1}\big) \quad [\because\, \circ \text{ is associative}]\\ &=a\circ(e\circ a^{-1}) \quad [\because\, b\circ b^{-1}=e]\\ &=a\circ a^{-1} \quad [\because\, e \text{ is the identity element}]\\ &=e. \end{align*}

Also

\begin{align*} (b^{-1}\circ a^{-1})\circ(a\circ b) &=b^{-1}\circ\big(a^{-1}\circ(a\circ b)\big) \quad [\because\, \circ \text{ is associative}]\\ &=b^{-1}\circ\big((a^{-1}\circ a)\circ b\big) \quad [\because\, \circ \text{ is associative}]\\ &=b^{-1}\circ(e\circ b) \quad [\because\, a^{-1}\circ a=e]\\ &=b^{-1}\circ b \quad [\because\, e \text{ is the identity element}]\\ &=e. \end{align*}

Therefore

b^{-1}\circ a^{-1}

is an inverse element of

a\circ b

. Since the inverse of an element is unique,

\begin{equation*} (a\circ b)^{-1}=b^{-1}\circ a^{-1}. \end{equation*}

Hence, all stated properties of inverses hold in

G

\square

The formula for the inverse of a product is often the first place where students feel the difference between commutative and non-commutative algebra. The inverse of

a\circ b

is not generally

a^{-1}\circ b^{-1}

. The correct order is reversed:

\begin{equation*} (a\circ b)^{-1}=b^{-1}\circ a^{-1}. \end{equation*}

This is similar to undoing two actions. If

a

is performed first and then

b

, the undoing must begin with

b^{-1}

and then proceed to

a^{-1}

Let

(G,\circ)

be a group and let

a,b,c\in G

. Then

\begin{equation*} (a\circ b\circ c)^{-1}=c^{-1}\circ b^{-1}\circ a^{-1}. \end{equation*}

Given that

(G,\circ)

is a group and

a,b,c\in G

. To prove that

\begin{equation*} (a\circ b\circ c)^{-1}=c^{-1}\circ b^{-1}\circ a^{-1}. \end{equation*}

Using associativity, write

\begin{equation*} a\circ b\circ c=(a\circ b)\circ c. \end{equation*}

Now

\begin{align*} (a\circ b\circ c)^{-1} &=\big((a\circ b)\circ c\big)^{-1}\\ &=c^{-1}\circ(a\circ b)^{-1} \quad [\because\, (u\circ v)^{-1}=v^{-1}\circ u^{-1}]\\ &=c^{-1}\circ(b^{-1}\circ a^{-1}) \quad [\because\, (a\circ b)^{-1}=b^{-1}\circ a^{-1}]\\ &=c^{-1}\circ b^{-1}\circ a^{-1} \quad [\because\, \circ \text{ is associative}]. \end{align*}

Hence,

\begin{equation*} (a\circ b\circ c)^{-1}=c^{-1}\circ b^{-1}\circ a^{-1}. \end{equation*}

\square

In the multiplicative group

(\mathbb{Q}-\{0\},\cdot)

\begin{equation*} (2\cdot3)^{-1}=6^{-1}=\frac{1}{6}. \end{equation*}

Also,

\begin{equation*} 3^{-1}\cdot2^{-1}=\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}. \end{equation*}

Thus

\begin{equation*} (2\cdot3)^{-1}=3^{-1}\cdot2^{-1}. \end{equation*}

Since rational multiplication is commutative, reversing the order is not visible here, but the rule still reverses the order.

Let

A

and

B

be invertible matrices of the same size. Under matrix multiplication,

\begin{equation*} (AB)^{-1}=B^{-1}A^{-1}. \end{equation*}

Indeed,

\begin{align*} (AB)(B^{-1}A^{-1}) &=A(BB^{-1})A^{-1}\\ &=AIA^{-1}\\ &=AA^{-1}\\ &=I. \end{align*}

Also,

\begin{align*} (B^{-1}A^{-1})(AB) &=B^{-1}(A^{-1}A)B\\ &=B^{-1}IB\\ &=B^{-1}B\\ &=I. \end{align*}

Hence,

B^{-1}A^{-1}

is the inverse of

AB

Let

(G,\circ)

be a group and let

a,b\in G

. Prove that

\begin{equation*} (a^{-1}\circ b^{-1})^{-1}=b\circ a. \end{equation*}

Let

(G,\circ)

be a group and let

a,b\in G

. To prove that

\begin{equation*} (a^{-1}\circ b^{-1})^{-1}=b\circ a. \end{equation*}

Using the inverse of a product rule,

\begin{align*} (a^{-1}\circ b^{-1})^{-1} &=(b^{-1})^{-1}\circ(a^{-1})^{-1}\\ &=b\circ a. \end{align*}

Hence,

\begin{equation*} (a^{-1}\circ b^{-1})^{-1}=b\circ a. \end{equation*}

\square

Let

(G,\circ)

be a group and let

a_1,a_2,\ldots,a_p\in G

, where

p\in\mathbb{N}

. Prove that

\begin{equation*} (a_1\circ a_2\circ\cdots\circ a_p)^{-1} =a_p^{-1}\circ a_{p-1}^{-1}\circ\cdots\circ a_1^{-1}. \end{equation*}

Let

(G,\circ)

be a group and let

a_1,a_2,\ldots,a_p\in G

, where

p\in\mathbb{N}

. To prove that

\begin{equation*} (a_1\circ a_2\circ\cdots\circ a_p)^{-1} =a_p^{-1}\circ a_{p-1}^{-1}\circ\cdots\circ a_1^{-1}. \end{equation*}

We prove the result by mathematical induction on

p

. For

p=1

\begin{equation*} (a_1)^{-1}=a_1^{-1}. \end{equation*}

Therefore the result is true for

p=1

. Assume that the result is true for

p=k

, that is,

\begin{equation*} (a_1\circ a_2\circ\cdots\circ a_k)^{-1} =a_k^{-1}\circ a_{k-1}^{-1}\circ\cdots\circ a_1^{-1}. \end{equation*}

To prove that the result is true for

p=k+1

. Now

\begin{align*} (a_1\circ a_2\circ\cdots\circ a_k\circ a_{k+1})^{-1} &=\big((a_1\circ a_2\circ\cdots\circ a_k)\circ a_{k+1}\big)^{-1}\\ &=a_{k+1}^{-1}\circ(a_1\circ a_2\circ\cdots\circ a_k)^{-1}\\ &=a_{k+1}^{-1}\circ a_k^{-1}\circ a_{k-1}^{-1}\circ\cdots\circ a_1^{-1}. \end{align*}

Therefore the result is true for

p=k+1

. Hence, by mathematical induction,

\begin{equation*} (a_1\circ a_2\circ\cdots\circ a_p)^{-1} =a_p^{-1}\circ a_{p-1}^{-1}\circ\cdots\circ a_1^{-1}. \end{equation*}

\square

Use the calculator below to verify the inverse of a product for nonzero rational numbers. Enter two nonzero rational numbers as fractions. The calculator computes the inverse of the product and compares it with the product of the inverses in reversed order. In rational multiplication the order does not affect the value, but the displayed formula keeps the group-theoretic order.

[1] Prove that

e^{-1}=e

. [2] Prove that

(a^{-1})^{-1}=a

. [3] Prove that

(a\circ b)^{-1}=b^{-1}\circ a^{-1}

. [4] Simplify

(a^{-1}\circ b)^{-1}

. [5] Find the inverse of

A_1\circ A_2\circ A_3\circ A_4

in a group.

[1] Since

e\circ e=e

e

is its own inverse. [2] Since

a\circ a^{-1}=a^{-1}\circ a=e

a

is the inverse of

a^{-1}

. [3] The inverse is verified by multiplying

a\circ b

with

b^{-1}\circ a^{-1}

from both sides and getting

e

. [4]

(a^{-1}\circ b)^{-1}=b^{-1}\circ a

. [5]

(A_1\circ A_2\circ A_3\circ A_4)^{-1}=A_4^{-1}\circ A_3^{-1}\circ A_2^{-1}\circ A_1^{-1}

The inverse of the identity element is the identity element itself.

The inverse of

a^{-1}

a

To undo

a\circ b

, one must undo

b

first and then undo

a

Not in general. The correct formula is

(a\circ b)^{-1}=b^{-1}\circ a^{-1}

BMLabs Mathematics

BMLabs Mathematics

Properties of Inverses

Properties of Inverses

PROPERTIES OF INVERSES

PRODUCT INVERSE TEST

SEMIGROUP CONDITIONS FOR GROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation