# Nonempty finite subsemigroup of group is subgroup

## Contents

This article gives the statement, and possibly proof, of a basic fact in group theory.
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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 $x \in H$:

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