I am having a hard time understanding when to use each different subgroup test and if given at any case, its sufficient to use either to proof that the given subgroup is truly a subgroup.
One-Step Subgroup Test
Let $G$ be a group and $H$ a nonempty subset of $G$. Then, $H$ is a subgroup of $G$ if $ab^{-1}$ is in $H$, whenever $a$ and $b$ are in $H$.
Two-Step Subgroup Test
Let $G$ be a group and $H$ a nonempty subset of $G$. Then, $H$ is a subgroup of $G$ if $ab \in H$ whenever $a,b \in H$ (closed under multiplication), and $a^{-1} \in H$ whenever $a \in H$ (closed under taking inverses).
Finite Subgroup Test
Let $H$ be a nonempty finite subset of a group $G$. Then, $H$ is a subgroup of $G$ if $H$ is closed under the operation of $G$.
I would really appreciate your help 🙂
Best Answer
They are all equivalent. They are really just restatements of the the definition of a subgroup. The definition says that the subset must form a group on its own. That means the subset $H\subseteq G$ contains the identity, the inverses and is closed.
All these properties can be encompassed with each of the three tests, the "trickiest" being the finite case as it is not as evidently true but it can be shown. But in all cases they are equivalent and which you use is up to you and is usually determined by which is easiest to see and prove.
Though in most cases it obvious if it is a subgroup or not.