Show that if a finite group, $G$, contains a proper subgroup of index $2$ in $G$, then $G$ is not simple.
Proof
Let $H$ be a proper subgroup of index $2$ in $G$. We know that $H$ is normal because it is of index $2$. Now, assume that $H = \{e\}$. Thus, $G/H \simeq G$. However, $G/H$ is of order $2$ because $H$ is of index $2$. Since $H$ is a proper subgroup, the order of $H$ must be less than the order of $G$. This means that the order of $G$ is greater than $G/H$. Thus, $G$ is not isomorphic to $G/H$ and $H \neq \{e\}$ and $G$ is not simple.
Now, I have a problem with the statement "Since $H$ is a proper subgroup, the order of $H$ must be less than the order of $G$. This means that the order of $G$ is greater than $G/H$." If $|G|=2$, then the proof seemingly falls apart. How can I rectify this problem?