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 [math]\displaystyle{ G }[/math] be a group and let [math]\displaystyle{ H }[/math] be a nonempty subset of [math]\displaystyle{ G }[/math]. If for all [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] in [math]\displaystyle{ H }[/math], [math]\displaystyle{ a b ^{-1} }[/math] is in [math]\displaystyle{ H }[/math], then [math]\displaystyle{ H }[/math] is a subgroup of [math]\displaystyle{ G }[/math].

Proof

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

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

Thus [math]\displaystyle{ H }[/math] is a subgroup of [math]\displaystyle{ G }[/math].

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.