Subgroup test

In abstract algebra, the one-step subgroup test is a theorem that states that for any group, a nonempty subset of that group is itself a group if the inverse of any element in the subset multiplied with any other element in the subset is also in the subset. The two-step subgroup test is a similar theorem which requires the subset to be closed under the operation and taking of inverses.

One-step subgroup test

Let ${\displaystyle G}$  be a group and let ${\displaystyle H}$  be a nonempty subset of ${\displaystyle G}$ . If for all ${\displaystyle a}$  and ${\displaystyle b}$  in ${\displaystyle H}$ , ${\displaystyle ab^{-1}}$  is in ${\displaystyle H}$ , then ${\displaystyle H}$  is a subgroup of ${\displaystyle G}$ .

Proof

Let ${\displaystyle G}$  be a group, let ${\displaystyle H}$  be a nonempty subset of ${\displaystyle G}$  and assume that for all ${\displaystyle a}$  and ${\displaystyle b}$  in ${\displaystyle H}$ , ${\displaystyle ab^{-1}}$  is in ${\displaystyle H}$ . To prove that ${\displaystyle H}$  is a subgroup of ${\displaystyle G}$  we must show that ${\displaystyle H}$  is associative, has an identity, has an inverse for every element and is closed under the operation. So,

• Since the operation of ${\displaystyle H}$  is the same as the operation of ${\displaystyle G}$ , the operation is associative since ${\displaystyle G}$  is a group.
• Since ${\displaystyle H}$  is not empty there exists an element ${\displaystyle x}$  in ${\displaystyle H}$ . If we take ${\displaystyle a=x}$  and ${\displaystyle b=x}$ , then ${\displaystyle ab^{-1}=xx^{-1}=e}$ , where ${\displaystyle e}$  is the identity element. Therefore ${\displaystyle e}$  is in ${\displaystyle H}$ .
• Let ${\displaystyle x}$  be an element in ${\displaystyle H}$  and we have just shown the identity element, ${\displaystyle e}$ , is in ${\displaystyle H}$ . Then let ${\displaystyle a=e}$  and ${\displaystyle b=x}$ , it follows that ${\displaystyle ab^{-1}=ex^{-1}=x^{-1}}$  in ${\displaystyle H}$ . So the inverse of an element in ${\displaystyle H}$  is in ${\displaystyle H}$ .
• Finally, let ${\displaystyle x}$  and ${\displaystyle y}$  be elements in ${\displaystyle H}$ , then since ${\displaystyle y}$  is in ${\displaystyle H}$  it follows that ${\displaystyle y^{-1}}$  is in ${\displaystyle H}$ . Hence ${\displaystyle x(y^{-1})^{-1}=xy}$  is in ${\displaystyle H}$  and so ${\displaystyle H}$  is closed under the operation.

Thus ${\displaystyle H}$  is a subgroup of ${\displaystyle G}$ .

Two-step subgroup test

A corollary of this theorem is the two-step subgroup test which states that a nonempty subset of a group is itself a group if the subset is closed under the operation as well as under the taking of inverses.