Abelian Group

After studying cyclic groups, we have already seen an important source of commutativity: every cyclic group is abelian. We now study abelian groups directly. An abelian group is a group in which the order of multiplication does not matter. This condition is simple to state, but it strongly affects proofs, inverse formulas, power formulas, and examples. In this lesson, students will learn the definition of an abelian group, compare abelian and non-abelian examples, and understand why commutativity must be checked as a separate property.

Let

(G,\circ)

be a group. Then

(G,\circ)

is called an

\textbf{abelian group}

\textbf{commutative group}

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

Thus an abelian group is a group whose operation is commutative.

Let

(G,\circ)

be a group. Then

(G,\circ)

is called an

\textbf{abelian group}

\textbf{commutative group}

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

Thus an abelian group is a group whose operation is commutative.

The definition says that every pair of elements must commute. It is not enough to check a few selected pairs. If even one pair

a,b\in G

satisfies

a\circ b\neq b\circ a

, then the group is not abelian. In additive notation, the operation is usually written as

+

, and commutativity appears as

a+b=b+a

. In multiplicative notation, the condition appears as

ab=ba

The group

(\mathbb{Z},+)

is abelian. Let

m,n\in\mathbb{Z}

. Then ordinary addition gives

\begin{equation*} m+n=n+m. \end{equation*}

Therefore

\begin{equation*} m+n=n+m \quad \forall\, m,n\in\mathbb{Z}. \end{equation*}

Hence,

(\mathbb{Z},+)

is an abelian group.

The group

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

is abelian. Let

a,b\in\mathbb{Q}-\{0\}

. Since ordinary multiplication of rational numbers is commutative,

\begin{equation*} ab=ba. \end{equation*}

Therefore

\begin{equation*} a\cdot b=b\cdot a \quad \forall\, a,b\in\mathbb{Q}-\{0\}. \end{equation*}

Hence,

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

is an abelian group.

The symmetric group

\mathbb{S}_3

is not abelian. Let

\begin{equation*} \sigma=(123),\quad \tau=(12). \end{equation*}

Using right-to-left composition, we get

\begin{equation*} \sigma\circ\tau=(13) \end{equation*}

and

\begin{equation*} \tau\circ\sigma=(23). \end{equation*}

Since

\begin{equation*} (13)\neq(23), \end{equation*}

we have

\begin{equation*} \sigma\circ\tau\neq\tau\circ\sigma. \end{equation*}

Hence,

\mathbb{S}_3

is not abelian.

Let

(G,\circ)

be a cyclic group. Then

G

is abelian.

Given that

(G,\circ)

is a cyclic group. To prove that

G

is abelian. Since

G

is cyclic, there exists

a\in G

such that

\begin{equation*} G=\langle a\rangle. \end{equation*}

Let

x,y\in G

. Then there exist

m,n\in\mathbb{Z}

such that

\begin{equation*} x=a^m,\quad y=a^n. \end{equation*}

Now

\begin{align*} x\circ y &=a^m\circ a^n\\ &=a^{m+n}\\ &=a^{n+m}\\ &=a^n\circ a^m\\ &=y\circ x. \end{align*}

Therefore

\begin{equation*} x\circ y=y\circ x \quad \forall\, x,y\in G. \end{equation*}

Hence,

G

is abelian.

\square

This theorem shows that cyclic groups are a special family of abelian groups. The converse is false. The Klein four group is abelian, but it is not cyclic because no element has order

4

. Thus commutativity does not automatically mean that one element generates the entire group. A common student error is to confuse “abelian” with “cyclic.” Every cyclic group is abelian, but abelian groups may require more than one generator.

Let

(G,\circ)

be a group with exactly two elements. Prove that

G

is abelian.

Let

(G,\circ)

be a group with identity element

e

. Given that

G

has exactly two elements. To prove that

G

is abelian. Since

e\in G

, there exists

a\in G

such that

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

and

a\neq e

. Now

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

Since

a\in G

and

G

is closed under

\circ

, we get

\begin{equation*} a\circ a\in G. \end{equation*}

If possible let

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

Then

\begin{align*} a\circ a=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, since

a\neq e

. Hence,

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

Thus all products in

G

satisfy

\begin{equation*} x\circ y=y\circ x \quad \forall\, x,y\in G. \end{equation*}

Hence,

G

is abelian.

\square

A group of two elements is necessarily cyclic and abelian. But this is a special small case. Larger groups may be abelian or non-abelian, cyclic or non-cyclic. The correct method is always to check the operation, not only the size of the set.

Use the calculator below to compare commutativity in two small groups. The cyclic group of order

4

is abelian, while the displayed sample from

\mathbb{S}_3

is non-abelian. Select a structure and a pair of elements. The calculator reports whether the selected pair commutes.

[1] Define an abelian group. [2] Give two examples of abelian groups. [3] Give one example of a non-abelian group. [4] Prove that every cyclic group is abelian. [5] Prove that every group with exactly two elements is abelian.

[1] A group

(G,\circ)

is abelian if

a\circ b=b\circ a

for all

a,b\in G

. [2] Examples are

(\mathbb{Z},+)

and

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

. [3] The group

\mathbb{S}_3

is non-abelian. [4] If

G=\langle a\rangle

, then

x=a^m

and

y=a^n

, so

x\circ y=a^{m+n}=a^{n+m}=y\circ x

. [5] If

G=\{e,a\}

, then

a\circ a=e

; the multiplication table is symmetric, so

G

is abelian.

It is a group in which every pair of elements commutes.

Yes. Every cyclic group is abelian.

No. The Klein four group is abelian but not cyclic.

No. Some permutations in

\mathbb{S}_3

do not commute.

BMLabs Mathematics

BMLabs Mathematics

Abelian Group

Abelian Group

ABELIAN GROUP

COMMUTATIVITY TEST

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Coordinates and Locus

Invariants under Orthogonal Transformation