Tour:Nonempty finite subsemigroup of group is subgroup: Difference between revisions

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{{guided tour|beginners|Introduction two|Sufficiency of subgroup criterion|Finite group}}
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{{quotation|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).<br>''WHAT YOU NEED TO DO''': Understand the statement and proof below.}}
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{{quotation|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).<br>'''WHAT YOU NEED TO DO''': Understand the statement and proof below.}}


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{{#lst:Subsemigroup of finite group is subgroup|main}}
{{guided tour-bottom|beginners|Introduction two|Sufficiency of subgroup criterion|Finite group}}
 
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Latest revision as of 07:58, 4 April 2013

This article adapts material from the main article: subsemigroup of finite group is subgroup

This page is part of the Groupprops guided tour for beginners (Jump to beginning of tour)
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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 G be a group and H be a nonempty finite subset such that a,bHabH. Then, H is a subgroup of G.

Proof

Lemma

Statement of lemma: For any xH:

  1. All the positive powers of x are in H
  2. There exists a positive integer n(x), dependent on x, such that xn(x)=e.

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

Since H is finite, the sequence x,x2,x3, must have some repeated element. Thus, there are positive integers k>l such that xk=xl. Multiplying both sides by xl, we get xkl=e. Set n(x)=kl, and we get xn(x)=e. Since k>l, n(x) is a positive integer.

The proof

We prove that H 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 eH: Since H is nonempty, there exists some element uH. Set x=u in the lemma. Applying part (2) of the lemma, we get that e is a positive power of u, so by part (1) of the lemma, eH.
  • Inverses gHg1H: Set x=g in the lemma. We make two cases:
    • Case n(g)=1: In this case g=e forcing g1=e=gH.
    • Case n(g)>1: In this case g1=gn(g)1 is a positive power of g, hence by Part (1) of the lemma, g1H.

Related results


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