The equivalence of two one-step subgroup tests. When proving a nonempty set $H$ of $(G;\circ)$ (a group) is a subgroup of $G$. There are some equivalent conditions.
One of them goes as

*

*$\forall a, b \in H, a \circ b^{-1} \in H$,

but I can also show that


*$\forall a, b \in H, a^{-1} \circ b \in H$
is an equivalent condition, too.
But these two conditions are different. '1.' operates $a$ (an element of $H$) and $b^{-1}$ (maybe not in $H$), '2.' is obviously different when it is not a commutative group.
My question is:

Why are they stating the same thing?

Appreciate your helping hands.
 A: A condition on a subset $H$ being a subgroup is that it is a group on its own (with respect to same group operation, $\circ$). This means that if $a \in H$ then $a^{-1} \in H$. It also means that $\forall a, b \in H$ we have $a \circ b \in H$. The condition that $\forall a,b \in H, a^{-1} \circ b \in H$ is just combining these two conditions.
You can get the condition about inverses by letting one of the elements be $1$ and you get the closure condition by using the inverses.
Once you have gained the condition that every element in $H$ also has it's inverse in $H$ from $$\forall a,b \in H, a^{-1} \circ b \in H$$ (by letting $b=1$). Then you can see that $$\forall a,b \in H, a \circ b^{-1} \in H$$ is only different in switching $a$ for $a^{-1}$ and $b$ for $b^{-1}$.
A: You have to understand the essence of these conditions. Subgroup is a subset of the group which itself is a group. It means it should have three properties.

*

*Identity $1 \in H$.

*Every element has an inverse i.e. for $x\in H$, $x^{-1}\in H$.

*Closure i.e. for $x, y\in H$ we have $xy \in H$.

The significance of these one-step subgroup tests is that they verify all these three conditions in one go.
For example if you prove that for all $x, y\in H$, $xy^{-1}\in H$ then taking same element say $a$ at the place of $x$ and $y$ will show you that the identity belongs to $H$. Now that you have identity in it, take $x = 1$ and $y$ equal to any element and that will give you the inverse in $H$. You can verify last one too. Once you understand this it will be trivial to understand why those two are equivalent conditions.
