Semigroups

After proving the elementary properties of groups, we now step slightly backward and study semigroups. A semigroup has only associativity as its structural law, so it is weaker than a group. This is useful because many group theorems can be understood by asking how much extra information must be added to a semigroup in order to recover a group. In this section, the central question is: when does a semigroup become a group? In this lesson, students will learn the definition of semigroup, compare it with the definition of group, and prepare for the criteria involving equations, cancellation laws, and idempotent elements.

Let

S

be a non-empty set and let

\circ

be a binary operation on

S

. Then

(S,\circ)

is called a

\textbf{semigroup}

\circ

is associative on

S

, that is,

\begin{equation*} (a\circ b)\circ c=a\circ(b\circ c) \quad \forall\, a,b,c\in S. \end{equation*}

Thus a semigroup is a non-empty set with an associative binary operation.

Let

S

be a non-empty set and let

\circ

be a binary operation on

S

. Then

(S,\circ)

is called a

\textbf{semigroup}

\circ

is associative on

S

, that is,

\begin{equation*} (a\circ b)\circ c=a\circ(b\circ c) \quad \forall\, a,b,c\in S. \end{equation*}

Thus a semigroup is a non-empty set with an associative binary operation.

The difference between a semigroup and a group is important. A group requires closure, associativity, an identity element, and inverse elements. A semigroup requires only a binary operation and associativity. Since a binary operation already includes closure, the additional visible condition in a semigroup is associativity. Therefore every group is a semigroup, but not every semigroup is a group. The missing parts are the identity element and inverse elements.

Let

(G,\circ)

be a group. Then

(G,\circ)

is a semigroup.

Given that

(G,\circ)

is a group. To prove that

(G,\circ)

is a semigroup. Since

(G,\circ)

is a group,

\circ

is a binary operation on

G

. Also, by the associativity axiom of a group,

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

Therefore

\circ

is associative on

G

. Hence,

(G,\circ)

is a semigroup.

\square

The converse of this theorem is false. A semigroup need not have an identity element, and even if it has an identity element, it need not have inverses. For example,

(\mathbb{N},+)

is a semigroup because addition is associative and the sum of two positive integers is positive. But if

\mathbb{N}=\{1,2,3,\ldots\}

, then

(\mathbb{N},+)

has no additive identity inside the set. Hence it is not a group.

Let

\mathbb{N}=\{1,2,3,\ldots\}

. Then

(\mathbb{N},+)

is a semigroup. Let

a,b,c\in\mathbb{N}

. Since

a+b\in\mathbb{N}

, addition is a binary operation on

\mathbb{N}

. Also,

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

Therefore addition is associative on

\mathbb{N}

. Hence,

(\mathbb{N},+)

is a semigroup.

Let

S=\mathbb{Z}

and let

\cdot

be ordinary multiplication. Then

(\mathbb{Z},\cdot)

is a semigroup. Let

a,b,c\in\mathbb{Z}

. Since

ab\in\mathbb{Z}

, multiplication is a binary operation on

\mathbb{Z}

. Also,

\begin{equation*} (ab)c=a(bc). \end{equation*}

Therefore multiplication is associative on

\mathbb{Z}

. Hence,

(\mathbb{Z},\cdot)

is a semigroup.

Let

S

be any non-empty set, and let

\mathcal{F}(S)

be the set of all functions from

S

S

. Then

\mathcal{F}(S)

is a semigroup under composition of functions. If

f,g\in\mathcal{F}(S)

, then

f\circ g\in\mathcal{F}(S)

. Also, for all

f,g,h\in\mathcal{F}(S)

\begin{equation*} (f\circ g)\circ h=f\circ(g\circ h). \end{equation*}

Thus composition is associative. Hence,

\mathcal{F}(S)

is a semigroup under composition.

Let

S=\{0,x\}

and define

\circ

S

\begin{equation*} 0\circ0=0,\quad 0\circ x=0,\quad x\circ0=0,\quad x\circ x=0. \end{equation*}

Prove that

(S,\circ)

is a semigroup.

Let

S=\{0,x\}

and let

\circ

be defined by

\begin{equation*} 0\circ0=0,\quad 0\circ x=0,\quad x\circ0=0,\quad x\circ x=0. \end{equation*}

To prove that

(S,\circ)

is a semigroup. [1] To prove closure. For all

a,b\in S

, the operation gives

\begin{equation*} a\circ b=0. \end{equation*}

Since

0\in S

, we get

a\circ b\in S

for all

a,b\in S

. [2] To prove associativity. Let

a,b,c\in S

. Since

a\circ b=0

, we get

\begin{align*} (a\circ b)\circ c&=0\circ c\\ &=0. \end{align*}

Also, since

b\circ c=0

, we get

\begin{align*} a\circ(b\circ c)&=a\circ0\\ &=0. \end{align*}

Therefore

\begin{equation*} (a\circ b)\circ c=a\circ(b\circ c) \quad \forall\, a,b,c\in S. \end{equation*}

Hence,

(S,\circ)

is a semigroup.

\square

This example is a useful warning. The operation is associative, so the structure is a semigroup. But it is not a group because there is no identity element. Indeed,

0\circ x=0\neq x

x\circ0=0\neq x

, and

x\circ x=0\neq x

. Thus neither

0

nor

x

can act as an identity element. A semigroup may be very far from being a group.

Use the calculator below to check associativity for a small operation table on the set

\{0,1\}

. Enter the four products

0\circ0

0\circ1

1\circ0

, and

1\circ1

. The calculator tests all triples from

\{0,1\}

. This illustrates that a semigroup condition is entirely about associativity once closure on the set is built into the table.

[1] Define a semigroup. [2] Prove that every group is a semigroup. [3] Prove that

(\mathbb{N},+)

is a semigroup when

\mathbb{N}=\{1,2,3,\ldots\}

. [4] Determine whether

(\mathbb{Z},-)

is a semigroup. [5] Give an example of a semigroup that is not a group.

[1] A semigroup is a non-empty set with an associative binary operation. [2] Every group has an associative binary operation, so every group is a semigroup. [3] Addition is closed and associative on

\mathbb{N}

. [4] No. Subtraction is not associative because

(5-3)-1=1

, while

5-(3-1)=3

. [5]

(\mathbb{N},+)

is a semigroup but not a group when

\mathbb{N}=\{1,2,3,\ldots\}

Yes. A group has an associative binary operation, so it is a semigroup.

No. A semigroup may fail to have an identity element or inverse elements.

Associativity of the binary operation.

Semigroups help us understand which extra conditions force a weaker algebraic structure to become a group.

BMLabs Mathematics

BMLabs Mathematics

Semigroups

Semigroups

SEMIGROUP

SEMIGROUP TEST

GROUP CRITERION BY EQUATIONS

Coordinates and Locus

Invariants under Orthogonal Transformation