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 $a, b \in H ⟹ a b \in H$ . 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 $x, x^{2}, x^{3}, \dots$ 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∈H⟹g−1∈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

External links

Links to related riders

Mathlinks discussion forum page for Subgroup of a finite group