Two-Step Subgroup Test

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subset of $G$.
Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ :
:$(1): \quad H \ne \O$, 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 $\struct {G, \circ}$ $\struct {H, \circ}$ is a $H$ be a nonempty subset of $G$ which is:
:closed under its operation
and:
:closed under inversion.
=== Corollary ===
== 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:
==== ====
The closure condition is given by condition $(2)$.
==== ====
From Restriction of Associative Operation is Associative, associativity is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$.
==== ====
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 $\circ$, $x \circ x^{-1} = e = x^{-1} \circ x \in H$.
==== ====
From condition $(3)$, every element of $H$ has an inverse.
So $\struct {H, \circ}$ satisfies all the group axioms, and is therefore a group.
So by definition $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$.
=== 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 as applied to the group $\struct {H, \circ}$.
:$(3): \quad a \in H \implies a^{-1} \in H$ follows from as applied to the group $\struct {H, \circ}$.
== 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 , 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 ==
* Two-Step Subgroup Test using Subset Product
* One-Step Subgroup Test
== Linguistic Note ==
== Sources ==
* : $\S 5.2$. Subgroups
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Theorem $8.4: \ 1^\circ - 2^\circ$
* : $\S 1.9$: Subgroups: Lemma $6$
* : $1$: Subgroups: $1.\text{T}.6$
* : Chapter $\text{II}$: Groups: Subgroups
* : Chapter $2$: Subgroups and Cosets: $\S 35$
* : $\S 36.3$: Subgroups
* : $\S 3.2$: Groups; the axioms
* : Chapter $4$: Subgroups: Proposition $4.2$
Category:Named Theorems
Category:Subgroups