Tour:Nonempty finite subsemigroup of group is subgroup

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). See below for the proof and more details.
There is a general condition called the subgroup condition for arbitrary groups, as we'll see in the next article.
Proceed to Guided tour for beginners:Sufficiency of subgroup condition OR return to Guided tour for beginners:Subgroup

Statement

Verbal statement

Any nonempty multiplicatively closed subset (or equivalently, nonempty subsemigroup) of a finite group is a subgroup.

Symbolic statement

Let $G$ be a finite group and $H$ be a nonempty subset such that $x,y\in H\implies xy\in H$. Then, $H$ is a subgroup of $G$.

Proof

Finite order of every element

We use finiteness to observe that $x$ has finite order, or equivalently, for any $x\in G$ that there exists a $n$ such that $x^{n}=e$. This can be shown as follows:

For any $x\in G$, the set $x,x^{2},x^{3},\ldots$ is a finite set and hence there are positive integers $k$ and $l$ such that $x^{k}=x^{l}$. This gives $x^{k-l}=e$.

The proof

Since $H$ is nonempty, there exists $x\in H$. We now demonstrate the three conditions for $H$ to be a subgroup:

• Binary operation: The product of two elements in $H$ is in $H$, by assumption
• Identity element: There exists $x\in H$ and from the previous section we know that there exists $n$ such that $x^{n}=e$. Since $H$ is closed under multiplication, $e\in H$
• Inverse element: For any $x\in H$, there exists $n$ such that $x^{n}=e$. Hence, $x^{n-1}=x^{-1}$. Since $H$ is closed under multiplication, $x^{n-1}\in H$ and hence $x^{-1}\in H$.

Related results

• Sufficiency of subgroup condition

External links

Links to related riders

Mathlinks discussion forum page for Subgroup of a finite group