Nonempty finite subsemigroup of group is subgroup

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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 $G$ 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 $x\in H$ :

All the positive powers of $x$ are in $H$
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 $g\in H\implies g^{-1}\in H$ $g\in H\implies g^{-1}\in H$ : Set $x=g$ $x=g$ 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, $g^{-1}\in H$ .