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.

Also see

 * One-Step Subgroup Test