Two-Step Subgroup Test

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Theorem

Let $\struct {G, \circ}$ be a group.

Let $H$ be a subset of $G$.


Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ if and only if:

$(1): \quad H \ne \O$, 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, $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ if and only if $\struct {H, \circ}$ is a $H$ be a nonempty subset of $G$ which is:

closed under its operation

and:

closed under inversion.


Corollary

Let $G$ be a group.

Let $\O\subset H \subseteq G$ be a non-empty subset of $G$.


Then $H$ is a subgroup of $G$ if and only if:

$H H \subseteq H$
$H^{-1} \subseteq H$

where:

$H H$ is the product of $H$ with itself
$H^{-1}$ is the inverse of $H$.


Proof

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 operation of $\struct {H, \circ}$ is the same as that for $\struct {G, \circ}$, that is, $\circ$.

So it remains to show that $\struct {H, \circ}$ is a group.


We check the four group axioms:


Group Axiom $\text G 0$: Closure

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

$\Box$


Group Axiom $\text G 1$: Associativity

From Restriction of Associative Operation is Associative, associativity is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$.

$\Box$


Group Axiom $\text G 2$: Existence of Identity Element

Let $e$ be the identity of $\struct {G, \circ}$.

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

Therefore $\exists x \in H$.

From condition $(3)$, $\struct {H, \circ}$ is closed under inversion.

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

Since $\struct {H, \circ}$ is closed under $\circ$, $x \circ x^{-1} = e = x^{-1} \circ x \in H$.

$\Box$


Group Axiom $\text G 3$: Existence of Inverse Element

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

$\Box$


So $\struct {H, \circ}$ satisfies all the group axioms, and is therefore a group.

So by definition $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$.

$\Box$


Sufficient Condition

Now suppose $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$.

$(1): \quad H \le G \implies H \ne \O$ 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 $\text G 0$: Closure as applied to the group $\struct {H, \circ}$.
$(3): \quad a \in H \implies a^{-1} \in H$ follows from Group Axiom $\text G 3$: Existence of Inverse Element as applied to the group $\struct {H, \circ}$.

$\blacksquare$


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.


Some sources, for example 1965: J.A. Green: Sets and Groups, 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.

However, this definition is valid only if $\struct {G, \circ}$ is itself a group, as it is here so defined; it cannot be used to define a subgroup of a more general algebraic structure.


Also see


Linguistic Note

The Two-Step Subgroup Test is so called despite the fact that, on the face of it, there are three steps to the test.

This is because the fact that the subset must be non-empty is usually an unspoken assumption, and is not specifically included as one of the tests to be made.


Sources

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