subsemigroup of a cyclic semigroup

subsemigroup of a cyclic semigroup

It is a well-known fact that the

subgroup

of a

cyclic group

is cyclic. Is this true for

semigroups

? The answer is clearly no. For example, take the

cyclic semigroup

of

positive integers

under

addition

, and the subsemigroup generated by, say, 7 and 17. If it were cyclic, generated by some positive integer

n

, then

n

must divide both 7 and 17, which implies that

n

=

1

. But there are no positive integers

p

and

q

such that

1

=

7

⁢

p

+

17

⁢

q

, and the result follows. However, the following does hold:

Proposition 1

.

Every subsemigroup of a cyclic semigroup is

finitely generated

.

Proof.

Let

S

be a cyclic semigroup. The result is obvious if

S

is finite. So assume that

S

is

infinite

. Since every infinite cyclic semigroup is

isomorphic

to the semigroup of positive integers under addition, we may as well assume that

S

=

{

1

,

2

,

…

}

. Let

T

be a subsemigroup of

S

. Since

S

is

well-ordered

, so is

T

. Take the least

element

p

1

of

T

. If

⟨

p

1

⟩

=

T

, then we are done. Otherwise,

T

-

⟨

p

1

⟩

is non-empty, and thus has a least element

p

2

. So by the

Euclidean algorithm

,

p

2

=

q

1

⁢

p

1

+

r

1

, where

0

<

r

1

<

p

1

. If

⟨

p

1

,

p

2

⟩

=

T

, then we are done. Otherwise, continue this process. Along the way, we collect the

quotients

{

q

1

,

q

2

,

…

}

, as well as the

remainders

{

r

1

,

r

2

,

…

}

. We make the following observations:

1.

Since

p

1

<

p

2

<

⋯

, the quotients are non-decreasing:

q

1

≤

q

2

≤

⋯

. For if

q

j

+

1

<

q

j

, then

p

j

+

2

=

q

j

+

1

⁢

p

1

+

r

j

+

1

<

q

j

+

1

⁢

p

1

+

p

1

=

(

1

+

q

j

+

1

)

⁢

p

1

≤

q

j

⁢

p

1

<

q

j

⁢

p

1

+

r

j

=

p

j

+

1

,

which is a

contradiction

.

2.

Elements of

{

r

1

,

r

2

,

…

}

are pairwise distinct, for if

r

i

=

r

j

for

i

<

j

, then

p

j

+

1

=

q

j

⁢

p

1

+

r

j

=

q

j

⁢

p

1

+

r

i

=

(

q

i

⁢

p

1

+

r

i

)

+

(

q

j

-

q

i

)

⁢

p

1

=

p

i

+

1

+

(

q

j

-

q

i

)

⁢

p

1

.

Since

q

i

≤

q

j

by the first observation,

p

j

+

1

lies in the subsemigroup generated by

p

i

+

1

and

p

1

, contradicting the construction of

p

j

+

1

.

Since the remainders are

integers

between

0

and

p

1

, the set

{

r

1

,

r

2

,

…

}

of remainders can not be infinite. Suppose the set has

k

elements. Then we must have

⟨

p

1

,

p

2

,

…

,

p

k

+

1

⟩

=

T

. Otherwise, we may continue the process and find

p

k

+

2

,

q

k

+

1

, and

r

k

+

1

. This means that

r

k

+

1

is among one of

r

1

,

…

,

r

k

. But by the second observation, this means that

p

k

+

2

∈

⟨

p

1

,

p

2

,

…

,

p

k

+

1

⟩

after all.
∎

In fact, the above proof shows that there is an

algorithm

for finding a

finite set

of

generators

for the subsemigroup

T

.

The following result is immediate:

Corollary 1

.

Any submonoid of a cyclic monoid is finitely generated.

As an application to

formal language

theory, we have the following:

Corollary 2

.

The

Kleene star

of any

language

over a singleton

alphabet

is

regular

(

http://planetmath.org/RegularLanguage

).

Proof.

The Kleene star of a language

L

over

{

a

}

is just a submonoid of the cyclic monoid

⟨

a

⟩

=

{

a

}

*

, and hence is generated by some finite set

F

by the

proposition

above. Since every finite set is regular, so is

F

*

=

L

*

.
∎

References

1

A. Salomaa,

Formal Languages

, Academic Press, New York (1973).

Title

subsemigroup of a cyclic semigroup

Canonical name

SubsemigroupOfACyclicSemigroup

Date of creation

2013-03-22 19:01:48

Last modified on

2013-03-22 19:01:48

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

6

Author

CWoo (3771)

Entry type

Result

Classification

msc 20M99

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