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

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) $$H \le G \implies H \ne \varnothing$$ from the fact that $$H$$ is a group and therefore can not be empty.
 * 2) $$a, b \in H \implies a \circ b \in H$$ follows from group axiom G0 (Closure) as applied to the group $$\left({H, \circ}\right)$$.
 * 3) $$a \in H \implies a^{-1} \in H$$ 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 frequently assumed as one of the "givens", and is then not specifically included as one of the tests to be made.

Also see

 * One-Step Subgroup Test