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 $G$ be a group and let $H$ be a nonempty subset of $G$. If for all $a$ and $b$ in $H$, $ab^{-1}$ 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. If we take a = x and b = x, then ab⁻¹ = xx⁻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 ab⁻¹ = 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.