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:


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

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


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

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.

Also see

 * Two-Step Subgroup Test