Nonempty finite subsemigroup of group is subgroup

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