Suppose $H\subset G$, where $G$ is a group and H is nonempty and closed under the binary operation of $G$.
Additionally, assume $H$ has the property that if $a$ is not in $H$ then $a^{-1}$ is not in $H$.
It seems rather trivial to see the given condition: $$\left(a\notin H\implies a^{-1}\notin H\right)\iff \left(a^{-1}\in H \implies a\in H\right)$$
by contraposition. The latter of the two above is simply one of the hallmark conditions of a subgroup (among closure under the binary operation of $G$) i.e. there's a theorem we've proved previously in class:
Theorem: If $ab\in H$ when $a\in H$ $\land$ $b\in H$ and if $a^{-1}\in H$ when $a\in H$, then $H$ is a subgroup of $G$ assuming $G$ is a group with $\emptyset \neq H\subset G$.
Thus since in the problem we're given $H$ is closed under the operation of $G$ and is a nonempty subset of $G$, then doesn't seeing the given property's equivalence with its contrapositive as I've shown finish the proof by showing $a^{-1}\in $ when $a\in H$?
Seems too easy.
Best Answer
You are in the right track, but it is not just the contrapositive of the statement, so perhaps you missed a small detail.
Note that the contrapositive of "$a \not\in H$ implies $a^{-1} \not\in H$" is the following statement:
With this, use $a = (a^{-1})^{-1}$ to conclude that $H$ is closed under inverses.