One-Step Subgroup Test
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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;
- $(2): \quad \forall a, b \in H: a \circ b^{-1} \in H$.
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:
- G1: Associativity: From Subset Product of Associative is Associative, associativity is inherited by $\left({H, \circ}\right)$ from $\left({G, \circ}\right)$
Since $H$ is nonempty, $\exists x \in H$.
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.
- G3: Inverses: If we take $a = e$ and $b = x$, then $a \circ b^{-1} = e \circ x^{-1} = x^{-1} \in H$.
Thus, $H$ is closed under taking inverses, i.e. every element of $H$ has an inverse.
- G0: Closure: Let $x, y \in H$.
Then $y^{-1} \in H$, so we may take $a = x$ and $b = y^{-1}$.
So, $a \circ b^{-1} = x \circ (y^{-1})^{-1} = x \circ y \in H$.
Thus, $H$ is closed.
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 $As $\left({H, \circ}\right)$ 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
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 5.2$
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 8$: Theorem $8.4: \ 1^\circ - 3^\circ$
- Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.9$: Lemma $8$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 35 \alpha$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 36.4$
- John F. Humphreys: A Course in Group Theory (1996): $\S 4$: Proposition $4.2$
