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

Proof

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

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

Thus $H$ is a subgroup of $G$.

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.

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