Subgroup Test
Theorem
One-step subgroup test
Let the group $G$ and let $H$ be a nonempty subset of $G$ (empty set가 아닌 부분집합). If whenever $a$ and $b$ are elements of $H$ then $ab^{-1}$ is also an element of $H$, then $H$ is a subgroup of $G$. Equivalently, if whenever $a$ and $b$ are in $H$ then $a-b$ is also in $H$, then $H$ is a subgroup.
$$ (a, b \in H \implies ab^{-1} \in H) \implies H \le G $$
Two-step subgroup test
For a nonempty subset $H$ of the group $G$, if the following two conditions are satisfied, then $H$ is a subgroup of $G$.
- $a$, $b \in H \implies ab \in H$
- $a \in H \implies a^{-1} \in H$
$$ (a, b \in H \implies ab \in H) \land (a \in H \implies a^{-1} \in H) \implies H \le G $$
Explanation
In brief, the two-step test asks you to check whether the subset is closed under the group operation and under taking inverses.
Proof
One-step subgroup test
Assume that whenever $a,\ b$ is an element of $H$, then $ab^{-1}$ is also an element of $H$. We then verify the three conditions for $H$ to be a group.
[1] Since the operation on $H$ is the same as that of the group $G$, associativity holds trivially.
[2] Let $a=x,\ b=x$. Then $ab^{-1}=xx^{-1}=e$, and by the assumption it belongs to $H$, so $H$ has an identity element.
[3] Let $a=e,\ b=x$. Then $ex^{-1}=x^{-1}$, and by the assumption it is an element of $H$, so any element $b$ of $H$ has an inverse.
[4] From [3] we have that every element has an inverse; let $a=x,\ b=-y$. Then $x(y^{-1})^{-1}=xy$ and by the assumption it belongs to $H$, hence $H$ is closed under the operation.
From [1]–[4], $H$ is closed under the operation of $G$, satisfies associativity, and has an identity and inverses, so it is a group. Therefore the subset $H$ is a subgroup of $G$.
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Two-step subgroup test
Assume 1. and 2.. Then, by the one-step test, it suffices to show that whenever $a, b \in H$ then $ab^{-1} \in H$. Let $a$ and $b \in H$. Then by 2. we have $b^{-1} \in H$, and by 1. we have $ab^{-1} \in H$. Therefore $H$ is a subgroup of $G$.
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