Tour:Nonempty finite subsemigroup of group is subgroup

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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 G be a group and H be a nonempty finite subset such that a,b \in H \implies ab \in H. Then, H is a subgroup of G.

Proof

Lemma

Statement of lemma: For any x \in H:

  1. All the positive powers of x are in H
  2. There exists a positive integer n(x), dependent on x, such that x^{n(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,x^2,x^3, \ldots must have some repeated element. Thus, there are positive integers k > l such that x^k = x^l. Multiplying both sides by x^{-l}, we get x^{k-l} = e. Set n(x) = k - l, and we get x^{n(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 e \in H: Since H is nonempty, there exists some element u \in H. 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, e \in H.
  • Inverses g \in H \implies g^{-1} \in H: Set x = g in the lemma. We make two cases:
    • Case n(g) = 1: In this case g = e forcing g^{-1} = e = g \in H.
    • Case n(g) > 1: In this case g^{-1} = g^{n(g) - 1} is a positive power of g, hence by Part (1) of the lemma, g^{-1} \in H.

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