One-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 $\struct {G, \circ}$ if and only if:
- $(1): \quad H \ne \O$, that is, $H$ is non-empty
- $(2): \quad \forall a, b \in H: a \circ b^{-1} \in H$.
Proof
Necessary Condition
Let $H$ be a subset of $G$ that fulfils the conditions given.
It is noted that the fact that $H$ is non-empty 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:
$\text G 1$: Associativity:
From Subset Product within Semigroup is Associative, associativity is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$.
$\text G 2$: Identity
Let $e$ be the identity of $\struct {G, \circ}$.
Since $H$ is non-empty, $\exists x \in H$.
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.
$\text G 3$: Inverses
If we take $a = e$ and $b = x$, then $a \circ b^{-1} = e \circ x^{-1} = x^{-1} \in H$.
Thus every element of $H$ has an inverse also in $H$.
$\text G 0$: Closure
Let $x, y \in H$.
Then $y^{-1} \in H$, so we may take $a = x$ and $b = y^{-1}$.
So:
- $a \circ b^{-1} = x \circ \paren {y^{-1} }^{-1} = x \circ y \in H$
Thus, $H$ is closed.
Therefore, $\struct {H, \circ}$ satisfies all the group axioms, and is therefore a group.
Therefore $\struct {H, \circ}$ 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.
- $(2): \quad$ As $\struct {H, \circ}$ is a group, it is closed and every element has an inverse. So it follows that $\forall a, b \in H: a \circ b^{-1} \in H$.
$\blacksquare$
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
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.2$. Subgroups
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 8$: Theorem $8.4: \ 1^\circ - 3^\circ$
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Subgroups: Lemma $8$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 35 \alpha$
- 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\S 1.2$
- 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 36.4$: Subgroups
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $4$: Subgroups: Proposition $4.2$