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 taking inverses.


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 product 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:


G0: Closure

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

$\Box$


G1: Associativity

From Closed Substructure of Semigroup is Semigroup, associativity is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$.

$\Box$


G2: Identity

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 taking inverses

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

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

$\Box$


G3: Inverses

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 G0 (Closure) as applied to the group $\struct {H, \circ}$.
$(3): \quad a \in H \implies a^{-1} \in H$ follows from group axiom G3 (Inverses) 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.


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.


Also see


Sources