If H is a subset of G, prove that H is also a subgroup.

Question

$G$ is a group, and $H$ is a nonempty subset of $G$. We know that, for any two values $a$ and $b$ in $H$, $ab^{-1}$ is also in $H$. Given this, how do we know that $H$ is also a subgroup of $G$? More specifically, I need help proving that, for every $a$ in $H$, $a^{-1}$ is also in $H$. Thank you.

Answer 1

To show $h^{-1}$ is in $H$ for all $h\in H$, note that the given property implies: $hh^{-1} =e \in H$. But now letting $e$ take the role of $a$ and $h$ take the role of $b$ above we get that:

$eh^{-1} =h^{-1}\in H$

Answer 2

Since $H$ is nonempty, there is some $a \in H$. Now, $aa^{-1} = 1$ is also an element of $H$. Also, since $1 \in H$ and $a \in H$, so is $1\cdot a^{-1} = a^{-1}$. Lastly, if $a,b \in H$, then $b^{-1} \in H$ and thus $a(b^{-1})^{-1} = ab \in H$.