Groups of Exponent Two

After proving the inverse product criterion, we now study a common sufficient condition for a group to be abelian. Suppose every element satisfies

a^2=e

. Then every element is its own inverse. This strong condition forces commutativity. Groups with this property appear naturally in examples such as the Klein four group and vector spaces over the field with two elements. In this lesson, students will prove that a group in which every element has square equal to the identity must be abelian.

Let

(G,\circ)

be a group with identity element

e

. We say that

G

satisfies the

\textbf{exponent two condition}

\begin{equation*} a^2=e \quad \forall\, a\in G. \end{equation*}

Equivalently, every element of

G

is its own inverse.

After proving the inverse product criterion, we now study a common sufficient condition for a group to be abelian. Suppose every element satisfies

a^2=e

Let

(G,\circ)

be a group with identity element

e

. We say that

G

satisfies the

\textbf{exponent two condition}

\begin{equation*} a^2=e \quad \forall\, a\in G. \end{equation*}

Equivalently, every element of

G

is its own inverse.

a^2=e

, then

a\circ a=e

, so

a

is an inverse of itself. By uniqueness of inverse,

a^{-1}=a

. Thus the exponent two condition says that inversion does not change any element. This is a strong restriction. It does not say that every element has order exactly

2

, because the identity has order

1

. It says that every non-identity element has order

2

Let

(G,\circ)

be a group with identity element

e

. If

\begin{equation*} a^2=e \quad \forall\, a\in G, \end{equation*}

then

G

is abelian.

Given that

(G,\circ)

is a group with identity element

e

. Given that

\begin{equation*} a^2=e \quad \forall\, a\in G. \end{equation*}

To prove that

G

is abelian. Let

a,b\in G

. Since

a\circ b\in G

, the hypothesis gives

\begin{equation*} (a\circ b)^2=e. \end{equation*}

Therefore

\begin{align*} (a\circ b)^2=e &\implies a\circ b=(a\circ b)^{-1} \quad [\because\, a\circ b \text{ is its own inverse}]\\ &\implies a\circ b=b^{-1}\circ a^{-1} \quad [\because\, \text{inverse of a product}]\\ &\implies a\circ b=b\circ a \quad [\because\, a^{-1}=a,\; b^{-1}=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 proof uses the hypothesis twice: once for the product

a\circ b

, and once for the elements

a

and

b

separately. This is why the condition must hold for every element of the group. It is not enough that each of

a

and

b

has square equal to

e

; we also need

(a\circ b)^2=e

The Klein four group

\begin{equation*} V_4=\{e,a,b,c\} \end{equation*}

satisfies

\begin{equation*} e^2=e,\quad a^2=e,\quad b^2=e,\quad c^2=e. \end{equation*}

Therefore

V_4

satisfies the exponent two condition. By the theorem,

V_4

is abelian.

The group

(\mathbb{Z}_2,+_2)

satisfies the additive form of the exponent two condition:

\begin{equation*} 2\overline{x}=\overline{0}\quad \forall\, \overline{x}\in\mathbb{Z}_2. \end{equation*}

Thus every element is its own additive inverse. The group is abelian.

Let

(G,\circ)

be a group with identity element

e

. Suppose that

\begin{equation*} a^2=e \quad \forall\, a\in G. \end{equation*}

Prove that

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

Let

(G,\circ)

be a group with identity element

e

. Given that

\begin{equation*} a^2=e \quad \forall\, a\in G. \end{equation*}

To prove that

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

Let

a\in G

. Then

\begin{equation*} a^2=e. \end{equation*}

Thus

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

Also

a\circ a=e

gives both left and right inverse equations for

a

with itself. Therefore

a

is an inverse of

a

. Since the inverse of

a

is unique,

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

Hence,

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

\square

Let

(G,\circ)

be a group with exactly two elements. Prove that every element of

G

satisfies

x^2=e

Let

(G,\circ)

be a group with identity element

e

and exactly two elements. To prove that every element of

G

satisfies

x^2=e

. Write

\begin{equation*} G=\{e,a\}, \end{equation*}

where

a\neq e

. For

x=e

\begin{equation*} e^2=e. \end{equation*}

Now consider

a

. Since

G

is closed under

\circ

\begin{equation*} a^2\in G. \end{equation*}

If possible let

\begin{equation*} a^2=a. \end{equation*}

Then

\begin{align*} a^2=a &\implies a\circ a=a\circ e \quad [\because\, a\circ e=a]\\ &\implies a=e \quad [\because\, \text{left cancellation law}]. \end{align*}

A contradiction. Hence,

\begin{equation*} a^2=e. \end{equation*}

Therefore every element of

G

satisfies

x^2=e

\square

Let

(G,\circ)

be a group. If

|G|=2

, then

G

is abelian.

Given that

(G,\circ)

is a group and

|G|=2

. To prove that

G

is abelian. By the previous solved result, every element

x\in G

satisfies

\begin{equation*} x^2=e. \end{equation*}

By the exponent two theorem,

G

is abelian. Hence, every group of order

2

is abelian.

\square

The exponent two theorem gives a fast way to prove commutativity in many small groups. However, it is only a sufficient condition. There are many abelian groups that do not satisfy

a^2=e

for all elements. For example,

(\mathbb{Z}_3,+_3)

is abelian, but

\overline{1}+\overline{1}\neq\overline{0}

Use the calculator below to test the exponent two condition in

\mathbb{Z}_n

under addition. In additive notation, the condition is

2x=0

for every element. Try

n=2

n=4

, and

n=6

to see when the full condition holds.

[1] State the exponent two condition. [2] Prove that if

a^2=e

for every

a\in G

, then

a^{-1}=a

for every

a\in G

. [3] Prove that a group satisfying the exponent two condition is abelian. [4] Give an example of an abelian group that does not satisfy the exponent two condition. [5] Prove that every group of order

2

is abelian.

[1] The condition is

a^2=e

for every

a\in G

. [2] Since

a\circ a=e

, the element

a

is its own inverse. [3] For

a,b\in G

(a\circ b)^2=e

, so

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

. [4] The group

(\mathbb{Z}_3,+_3)

is abelian but does not satisfy

2x=0

for every element. [5] If

G=\{e,a\}

, then

a^2=e

, so the group satisfies the exponent two condition and is abelian.

It means every element is its own inverse.

Yes. Every group satisfying it is abelian.

No. Many abelian groups have elements of order greater than

2

No. The identity has order

1

BMLabs Mathematics

BMLabs Mathematics

Groups of Exponent Two

Groups of Exponent Two

EXPONENT TWO CONDITION

EXPONENT TWO TEST

POWER CRITERIA FOR COMMUTATIVITY

Coordinates and Locus

Invariants under Orthogonal Transformation