Tour:Nonempty finite subsemigroup of group is subgroup
This article adapts material from the main article: subsemigroup of finite group is subgroup
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In a finite group, any nonempty subset that is closed under the operation of multiplication is, in fact, a subgroup. This isn't true for infinite groups (think for a moment about positive integers inside all integers).
WHAT YOU NEED TO DO: Understand the statement and proof below.
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
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