Suppose $\emptyset\neq H\subseteq G$ and if $a, b\in H$, then $a^{-1}b^{-1}\in H$. Prove or disprove that $H \leq G$. I want to answer the following:

Suppose that $H$ is a nonempty subset of a group $G$ with the property that if $a$ and $b$ belong to $H$ then $a^{-1}b^{-1}$ belongs to $H$. Prove or disprove that this is enough to guarantee that $H$ is a subgroup of $G$.

After looking at this question on someone else's stack exchange question, I saw that the answer is no. However, I was wondering if there is any sort of disproof for this, or if literally the only way to disprove the statement is by showing a specific counterexample. 
I would appreciate more insight into how we can tell that the statement fails. Specifically, if there is any way to show that the statement fails in general terms or to show that it fails using a one-step subgroup test or something of this nature, it would be greatly appreciated. 
 A: It seems like you're asking about the statement:

For all groups $G$ and all non-empty subsets $H$ of $G$ satisfying the condition that, for all pairs $a,b\in H$ we have $a^{-1}b^{-1}\in H$, it is the case that $H$ is a subgroup of $G$.

However you phrase it, this statement is saying "for all examples satisfying these conditions, we have some outcome." The negation of this is that there is a counterexample, which satisfies the condition, but not the outcome. That is to say: a disproof of this statement is, almost by definition, a counterexample (although, technically, it also suffices to show that a counterexample exists without actually constructing one). So, there isn't really any formal way in which one can ask for more than a counterexample.
That said, maybe you want a more intuitive way to find a counterexample than just "here's an example" - and there really is a good general way to approach this kind of question: let's say that a set $H$ satisfying the condition that $a,b\in H$ implies $a^{-1}b^{-1}\in H$ is good. A reasonable approach is to look at the smallest good sed containing a given set - or, otherwise said, the set which is generated by the operation that takes a pair $a,b\in H$ to $a^{-1}b^{-1}$.
For instance, suppose we wanted to find the smallest good set $H$ in $\mathbb Z$ containing $1$. We would notice that $(-1)+(-1)\in H$ by the operation. Then, we could use the operation on $1$ and $-2$ to get that $2+(-1)\in H$ - though we already knew that - and that $2+2\in H$, which is new. We'd then use the operation on $1$ and $4$ to get $-5\in H$ and on $4$ and $4$ to get $-8\in H$. We can keep combining elements this way by the axioms to create a bigger set - and then the question is: what set do we end up with? 
Well, it's a bit annoying to prove explicitly, but if you just run this process for a while, you'll realize that $H$ looks to be the set of numbers that are equal to $1$ mod $3$ - and you can figure this out by a completely rote process in which you just iteratively find elements in $H$, put them in a list, and look at that list carefully. This leaves you with an explicit counterexample - but also can be extended, by mapping $\mathbb Z$ into any other group, that if $a\in H$ then $a^{3k+1}\in H$ for every $k\in \mathbb Z$ (and that these elements form a good set) - which suddenly gives you a lot of counterexamples: the set $\{a^{3k+1} : k\in\mathbb Z\}$ suffices for any group in which the order of $a$ is either infinite or divisible by $3$, since it is not a subgroup under these conditions (though it is always a good set). 
Abstracting a bit, you can find out that the question is really asking: suppose that $H$ is good and $x\in H$ and $y\in H$. Does it follow that $xy\in H$ and $x^{-1}\in H$ and $e\in H$? This sort of structure is modeled really well by free groups, and leads to a fairly powerful fact:

It suffices to check this conjecture for a single good set: let $\langle x,y\rangle$ be the free group on two generators and $H$ be the smallest good set containing $x$ and $y$. Then, the earlier highlighted statement is true if and only if $H$ is a subgroup.

Note that this applies to basically any variant of the question. The reason for this is that if this $H$ were a subgroup, then we would have essentially proven that the relations $e\in H$ and $x^{-1}\in H$ and $xy\in H$ all follow from the axioms of a good set and the conditions $x\in H$ and $y\in H$ - but since we worked in a free group, this would hold for all groups.
It's difficult to describe exactly what $H$ is in this case, but it shows an important point: there is a counterexample which is essentially the most general possible. It is often a viable strategy to approach questions like this by trying to construct a "free" counterexample where you either find out that a specific example is a counterexample or show that a specific example implies the statement in general. That is, instead of going for a wild goose chase, you narrow the issue to one example that contains within it everything relevant to the question in the most free possible way*. This often leads to very difficult computations if one wants to work out what something like $H$ is explicitly, but often allows one to become inspired mid-computation and find a counterexample based on their work on the universal example.
(*This is an example of a "universal property" in case you ever encounter that term elsewhere - though it's a strange example of a universal property, to be sure)
A: Take $\{6\} \subset \mathbb{Z}_9$ so $6^{-1} =3, \text {and } 3+3=6 \in \{6\}$
