Inverse Product Criterion

After defining abelian groups, we now study a useful test for commutativity. In every group, the inverse of a product reverses order:

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

In an abelian group, reversing the order has no effect, so the formula becomes

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

This lesson proves that the second formula is not only a consequence of commutativity, but also a criterion for an abelian group.

Let

(G,\circ)

be a group with identity element

e

. Then

G

is abelian if and only if

\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}

After defining abelian groups, we now study a useful test for commutativity. In every group, the inverse of a product reverses order:

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

In an abelian group, reversing the order has no effect, so the formula becomes

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

This lesson proves that the second formula is not only a consequence of commutativity, but also a criterion for an abelian group.

Let

(G,\circ)

be a group with identity element

e

. Then

G

is abelian if and only if

\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}

Given that

(G,\circ)

is a group with identity element

e

. To prove that

G

is abelian if and only if

\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}

[1] To prove that if

G

is abelian, then

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

for all

a,b\in G

. Let

G

be abelian and let

a,b\in G

. In every group,

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

Since

G

is abelian,

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

. Therefore

\begin{align*} (a\circ b)^{-1} &=b^{-1}\circ a^{-1}\\ &=a^{-1}\circ b^{-1} \quad [\because\, G \text{ is abelian}]. \end{align*}

Thus

\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}

[2] To prove that if

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

for all

a,b\in G

, then

G

is abelian. Let

a,b\in G

. Given that

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

But the general inverse law gives

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

Therefore

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

Taking inverses on both sides, we get

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

Therefore

\begin{equation*} a\circ b=b\circ a \quad \forall\, a,b\in G. \end{equation*}

Hence,

G

is abelian.

\square

The theorem is a useful reminder that the order of factors matters in a non-abelian group. The formula

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

always holds. The formula

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

is special; it holds for every pair exactly when the group is abelian. Students often memorize the wrong order for the inverse of a product. This theorem explains why the wrong-looking order becomes correct only in the commutative case.

In the abelian group

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

, let

a=2

and

b=5

. Then

\begin{equation*} (ab)^{-1}=10^{-1}=\frac{1}{10} \end{equation*}

and

\begin{equation*} a^{-1}b^{-1}=\frac{1}{2}\cdot\frac{1}{5}=\frac{1}{10}. \end{equation*}

Since real multiplication is commutative, the order of inverse factors does not matter.

In a matrix group under multiplication, the formula

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

holds for invertible matrices

A

and

B

. In general, one cannot replace the right-hand side by

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

unless

A

and

B

commute. Thus the inverse product criterion detects the difference between commutative and non-commutative multiplication.

Let

(G,\circ)

be a group. Suppose that

\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}

Prove directly that

a\circ b=b\circ a

for all

a,b\in G

Let

(G,\circ)

be a group and let

a,b\in G

. Given that

\begin{equation*} (a\circ b)^{-1}=a^{-1}\circ b^{-1}\quad \forall\, a,b\in G. \end{equation*}

To prove that

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

In every group,

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

Therefore

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

Taking inverses on both sides,

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

Thus

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

Hence,

G

is abelian.

\square

Let

(G,\circ)

be a group. If

G

is abelian and

a,b,c\in G

, prove that

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

Let

(G,\circ)

be an abelian group and let

a,b,c\in G

. To prove that

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

In every group, the inverse of a product reverses order. Therefore

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

Since

G

is abelian, the inverse factors commute. Hence

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

Therefore

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

\square

The theorem must be read with the phrase “for all

a,b\in G

.” It is possible that one special pair satisfies

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

even in a non-abelian group. The criterion says that the formula must hold for every pair of elements.

Use the calculator below to compare the two inverse-product orders symbolically. Choose whether the group is abelian. The calculator explains whether the order-reversing formula can be rewritten without reversing the factors.

[1] State the inverse product criterion for abelian groups. [2] In an arbitrary group, what is

(a\circ b)^{-1}

? [3] If

G

is abelian, simplify

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

. [4] If

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

for all

a,b\in G

, prove that

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

. [5] Explain why the condition must hold for all pairs.

[1] A group

G

is abelian if and only if

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

for all

a,b\in G

. [2]

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

. [3]

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

. [4] Since

(a\circ b)^{-1}

is also

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

, equality with

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

gives the result. [5] A special pair may commute accidentally; abelian means every pair commutes.

It is the product of the inverses in reverse order.

Exactly when the group is abelian.

No. It holds for all groups.

It turns equality of inverse products into equality of the original products in reversed order.

BMLabs Mathematics

BMLabs Mathematics

Inverse Product Criterion

Inverse Product Criterion

INVERSE PRODUCT TEST

GROUPS OF EXPONENT TWO

Coordinates and Locus

Invariants under Orthogonal Transformation