One-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) $$\forall a, b \in H: a \circ b^{-1} \in H$$.

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:


 * 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 nonempty, $$\exists x \in H$$.

If we take $$a = x$$ and $$b = x$$, then $$a \circ b^{-1} = x \circ x^{-1} = e \in H$$, where $$e$$ is the identity element.


 * G3: Inverses: If we take $$a = e$$ and $$b = x$$, then $$a \circ b^{-1} = e \circ x^{-1} = x^{-1} \in H$$.

Thus, $$H$$ is closed under taking inverses, i.e. every element of $$H$$ has an inverse.


 * G0: Closure: Let $$x, y \in H$$.

Then $$y^{-1} \in H$$, so we may take $$a = x$$ and $$b = y^{-1}$$.

So, $$a \circ b^{-1} = x \circ (y^{-1})^{-1} = x \circ y \in H$$.

Thus, $$H$$ is closed.

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) $$H \le G \Longrightarrow H \ne \varnothing$$ from the fact that $$H$$ is a group and therefore can not be empty.
 * 2) As $$\left({H, \circ}\right)$$ is a group, it is closed and every element has an inverse. So it follows that $$\forall a, b \in H: a \circ b^{-1} \in H$$.

QED

Comment
This is called the "one-step subgroup test" although, on the face of it, there are two 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.