subgroup definition $H \subseteq \text{group  } G $, $H$ is not empty.
If for any $a,b \in H$, we have $a^{-1}b^{-1} \in H$ , is it possible to deduce that $H$ is a subgroup of $G$ ?  
I feel like this is not necessarily true because the identity can't be proven to be in $H$, but I can't find an counterexample. 
 A: Hint: Consider $\mathbb{Z}/3$.
A: In general a counterexample: take a finite group $G$ with $3 \vert|G|$. By Cauchy' Theorem we can pick an element $g \in G$ of order 3. Now take $H=\{g\}$.
The story goes a little further.
Proposition Let $H$ be a subset of a group $G$ such that $\forall a,b \in H:  a^{-1}b^{-1} \in H$. Then $H$ is a subgroup of $G$ if and only if the identity element $e \in H$.
Proof. The only if part is obvious, so assume $e \in H$. We have to show that $H$ is closed under the group multiplication and that inverses lie in $H$. Let’s do the latter first: take $a \in H$, then by the property $a^{-1}e^{-1}= a^{-1} \in H$. Now take $ a,b \in H$, then $ b^{-1}a^{-1}=(ab) ^{-1} \in H$. But inverses lie in $H$, so $ab \in H \Box$.
We can in fact construct an infinite number of counterexamples.
Let $G$ be a group with $g \in G$ of order 6. Take $H=\{g,g^4\}$
Let $G$ be a group with $g \in G$ of order 9. Take $H=\{g,g^4,g^7\}$
Let $G$ be a group with $g \in G$ of order 12. Take $H=\{g,g^4,g^7,g^{10}\}$, and so on.
