Nonempty finite subsemigroup of group is subgroup
From Groupprops
(Redirected from Subsemigroup of finite group is subgroup)
This article gives the statement, and possibly proof, of a basic fact in group theory.
View a complete list of basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this
Statement
Verbal statement
Any nonempty multiplicatively closed finite subset (or equivalently, nonempty finite subsemigroup) of a group is a subgroup.
Symbolic statement
Let be a group and be a nonempty finite subset such that . Then, is a subgroup of .
Proof
Lemma
Statement of lemma: For any :
- All the positive powers of are in
- There exists a positive integer , dependent on , such that .
Proof: is closed under multiplication, so we get that the positive power of are all in . This proves (1).
Since is finite, the sequence must have some repeated element. Thus, there are positive integers such that . Multiplying both sides by , we get . Set , and we get . Since , is a positive integer.
The proof
We prove that satisfies the three conditions for being a subgroup, i.e., it is closed under all the group operations:
- Binary operation: Closure under the binary operation is already given to us.
- Identity element : Since is nonempty, there exists some element . Set in the lemma. Applying part (2) of the lemma, we get that is a positive power of , so by part (1) of the lemma, .
- Inverses : Set in the lemma. We make two cases:
- Case : In this case forcing .
- Case : In this case is a positive power of , hence by Part (1) of the lemma, .
Related results
External links
Mathlinks discussion forum page for Subgroup of a finite group