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 $G$ 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, $g^{-1}\in H$.

Related results

• Sufficiency of subgroup condition

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