View source for Nonempty finite subsemigroup of group is subgroup

{{basic fact}}
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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 <math>G</math> be a [[group]] and <math>H</math> be a nonempty finite subset such that <math>a,b \in H \implies ab \in H</math>. Then, <math>H</math> is a subgroup of <math>G</math>.

==Proof==

<center>{{#widget:YouTube|id=1F2Qo9bfMfc}}</center>

===Lemma===

'''Statement of lemma''': For any <math>x \in H</math>:

# All the positive powers of <math>x</math> are in <math>H</math>
# There exists a positive integer <math>n(x)</math>, dependent on <math>x</math>, such that <math>x^{n(x)} = e</math>.

'''Proof''': <math>H</math> is closed under multiplication, so we get that the positive power of <math>x</math> are all in <math>H</math>. This proves (1).

Since <math>H</math> is finite, the sequence <math>x,x^2,x^3, \ldots</math> must have some repeated element. Thus, there are positive integers <math>k > l</math> such that <math>x^k = x^l</math>. Multiplying both sides by <math>x^{-l}</math>, we get <math>x^{k-l} = e</math>. Set <math>n(x) = k - l</math>, and we get <math>x^{n(x)} = e</math>. Since <math>k > l</math>, <math>n(x)</math> is a positive integer.

===The proof===

We prove that <math>H</math> 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''' <math>e \in H</math>: Since <math>H</math> is nonempty, there exists some element <math>u \in H</math>. Set <math>x = u</math> in the lemma. Applying part (2) of the lemma, we get that <math>e</math> is a positive power of <math>u</math>, so by part (1) of the lemma, <math>e \in H</math>.
* '''Inverses''' <math>g \in H \implies g^{-1} \in H</math>: Set <math>x = g</math> in the lemma. We make two cases:
** Case <math>n(g) = 1</math>: In this case <math>g = e</math> forcing <math>g^{-1} = e = g \in H</math>.
** Case <math>n(g) > 1</math>: In this case <math>g^{-1} = g^{n(g) - 1}</math> is a positive power of <math>g</math>, hence by Part (1) of the lemma, <math>g^{-1} \in H</math>.

==Related results==

* [[Sufficiency of subgroup condition]]
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==External links==

===Links to related riders===

{{mathlinks|134285|Subgroup of a finite group}}