Two-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

$(2): \quad a, b \in H \implies a \circ b \in H$

$(3): \quad a \in H \implies a^{-1} \in H$.

That is, $\left({H, \circ}\right)$ is a subgroup of $\left({G, \circ}\right)$ iff $\left({H, \circ}\right)$ is a $H$ be a nonempty subset of $G$ which is closed under its operation and closed under taking inverses.

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:

• G0: Closure: This is given by the definition.

• 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 not empty, $\exists x \in H$.

Since $\left({H, \circ}\right)$ is closed under taking inverses, $x^{-1} \in H$.

Since $\left({H, \circ}\right)$ is closed under $\circ$, $x \circ x^{-1} = e = x^{-1} \circ x \in H$.

• G3: Inverses: This is given by definition.

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.

$(2): \quad a, b \in H \implies a \circ b \in H$ follows from group axiom G0 (Closure) as applied to the group $\left({H, \circ}\right)$.

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

$\blacksquare$

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.

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

Also see

• One-Step Subgroup Test

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 5.2$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 8$: Theorem $8.4: \ 1^\circ - 2^\circ$
• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.9$: Lemma $6$
• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): Chapter $\text{II}$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 35$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 36.3$
• John F. Humphreys: A Course in Group Theory (1996): $\S 4$: Proposition $4.2$