Nonempty subset $H$ of group $G$ is subgroup iff $a^{-1}b\in H $ for any $a,b \in H$. Subgroup test is given by
Nonempty subset $H$ of group $G$ is subgroup iff $ab^{-1} \in H $ for any $a,b \in H$
But why can't we say Nonempty subset $H$ of group $G$ is subgroup iff $a^{-1}b\in H $ for any $a,b \in H$
(Taking the inverse of the first element).
This is not the duplicate as I didn't ask about the proof, I ask why this particular statement is chosen over other.

I run through the proof; the same argument can prove this statement, but why almost all the book stated the first one but not second?
Is there any reason behind this?

 A: I suppose it is because $ab^{-1}$ is sort of like the "multiplicative difference" between $a$ and $b$, since if the operation is additive, it looks like $a-b\in H$.
So what you are checking is whether for any $a, b\in H$, their difference with respect to the group operation is in $H$.
Hence it makes sense to check $ab^{-1}\in H$, as otherwise it would not be so clear that it is the difference you're checking.
I hope this helps.
A: The logic behind the statement is that you need is to be able to apply the inverse operator to show that $H$ contains the identity, since it contains at least one element $x$ and $xx^{-1}=1$. Once you know this, it follows that $1x^{-1} = x^{-1}$ for any $x \in H$, and so $H$ is closed with respect to the inverse operator. Then it is immediate that it is also closed with respect to the group operation, as $ab=a(b^{-1})^{-1} \in H$.
Since left inverses and right inverses are the same thing in group theory, the two choices are equivalent and there is no particular reason to prefer one to the other.
