Nonempty finite subsemigroup of group is subgroup
From Groupprops
This article gives the statement, and possibly proof, of a basic fact in group theory.
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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,
.
- Case
Related results
External links
Mathlinks discussion forum page for Subgroup of a finite group