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 $\left({G, \circ}\right)$ :


 * $(1): \quad H \ne \varnothing$, that is, $H$ is non-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 $\left({G, \circ}\right)$ $\left({H, \circ}\right)$ is a $H$ be a nonempty subset of $G$ which is:
 * closed under its operation

and:
 * closed under taking inverses.

Necessary Condition
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
The closure condition is given by condition $(2)$.

G1: Associativity
From Closed Substructure of Semigroup is Semigroup, associativity is inherited by $\left({H, \circ}\right)$ from $\left({G, \circ}\right)$.

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

From condition $(1)$, $H$ is non-empty.

Therefore $\exists x \in H$.

From condition $(3)$, $\left({H, \circ}\right)$ is closed under taking inverses

Therefore $x^{-1} \in H$.

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

G3: Inverses
From condition $(3)$, every element of $H$ has an inverse.

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

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

Sufficient Condition
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 $\left({H, \circ}\right)$.

Also defined as
Some sources specify condition $(1)$ as being $e \in H$ in order to satisfy the non-emptiness condition, but from Identity of Subgroup it can be seen that this is not necessary, as this follows automatically by conditions $(2)$ and $(3)$.

Some sources completely omit to state the fact that $H$ needs to be non-empty.

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

 * Two-Step Subgroup Test using Subset Product
 * One-Step Subgroup Test