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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 $G$ 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, ab^-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. Then the identity is in H since we can write it as e = x x^-1 which is in H by the initial assumption.
• Let x be an element of H. Since the identity e is in H it follows that ex^-1 = x^-1 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.