Two-Step Subgroup Test

Theorem
Let $\left({G, \circ}\right)$ be a group.

Let $H$ be a subset of $G$.

Then $\left({H, \circ}\right)$ is a subgroup of $\left({G, \circ}\right)$ iff:


 * $(1): \quad H \ne \varnothing$, that is, $H$ is not empty
 * $(2): \quad a, b \in H \implies a \circ b \in H$
 * $(3): \quad a \in H \implies a^{-1} \in H$.

That is, $\left({H, \circ}\right)$ is a subgroup of $\left({G, \circ}\right)$ iff $\left({H, \circ}\right)$ is a $H$ be a nonempty subset of $G$ which is closed under its operation and closed under taking inverses.

Proof

 * Let $H$ be a subset of $G$ that fulfils the conditions given.

It is noted that the fact that $H$ is nonempty is one of the conditions.

It is also noted that the group product of $\left({H, \circ}\right)$ is the same as that for $\left({G, \circ}\right)$, that is, $\circ$.

So it remains to show that $\left({H, \circ}\right)$ is a group.

We check the four group axioms:


 * G0: Closure: This is given by the definition.


 * G1: Associativity: From Subset Product of Associative is Associative, associativity is inherited by $\left({H, \circ}\right)$ from $\left({G, \circ}\right)$


 * G2: Identity: Let $e$ be the identity of $\left({G, \circ}\right)$.

Since $H$ is not empty, $\exists x \in H$.

Since $\left({H, \circ}\right)$ is closed under taking inverses, $x^{-1} \in H$.

Since $\left({H, \circ}\right)$ is closed under $\circ$, $x \circ x^{-1} = e = x^{-1} \circ x \in H$.


 * G3: Inverses: This is given by definition.

Therefore, $\left({H, \circ}\right)$ satisfies all the group axioms, and is therefore a group.

Therefore $\left({H, \circ}\right)$ is a subgroup of $\left({G, \circ}\right)$.


 * Now suppose $\left({H, \circ}\right)$ is a subgroup of $\left({G, \circ}\right)$.


 * $(1): \quad H \le G \implies H \ne \varnothing$ from the fact that $H$ is a group and therefore can not be empty.
 * $(2): \quad a, b \in H \implies a \circ b \in H$ follows from group axiom G0 (Closure) as applied to the group $\left({H, \circ}\right)$.
 * $(3): \quad a \in H \implies a^{-1} \in H$ follows from group axiom G3 (Inverses) as applied to the group $\left({H, \circ}\right)$.

Comment
This is called the two-step subgroup test although, on the face of it, there are three steps to the test. This is because the fact that $H$ must be non-empty is usually an unspoken assumption, and is not specifically included as one of the tests to be made.

Some sources, for example, use this property of subgroups as the definition of a subgroup, and from it deduce that a subgroup is a subset which is a group.

Also see

 * One-Step Subgroup Test