I was hoping someone could look over my solution to this exercise in Artin. I want my proof to be as rigorous as possible, so if I have skipped over anything and made any leaps of logic, please let me know.
a) In the definition of subgroup, the identity element in $H$ is required to be the identity of $G$. One might require only that $H$ have an identity element, not that it is the same as the identity in $G$. Show that if $H$ has an identity at all, then it is the identity in $G$, so this definition would be equivalent to the one given.
b) Show the analogous thing for inverses.
Proof. (a) Let $e_H$ be the identity element in $H$. Then, we have $$e_H e_H = e_H$$ by the definition of identity. We may, however, view this as an equation in $G$ (since all elements of $H$ are also in $G$), which has an identity element $e_G$. Since $H < G$, we of course have $H \subset G$, so $e_H \in G$. Hence, $e_H e_G = e_G e_H = e_H$. Hence, $$e_H e_H = e_H = e_H e_G = e_G e_H.$$ That is, $$e_H e_H = e_H e_G.$$ Since $e_H \in G$, $e_H^{-1}$ exists in $G$. Hence, we may cancel $e_H$ on the left: $$e_H = e_G.$$ (b) Let $h \in H$. Since $H$ is a group, it is closed under inverses, so $h^{-1}$ exists in $H$ with the property that $$hh^{-1} = h^{-1} h = e.$$ Using the result of part (a), I am simply referring to the result as $e$, having established that it is an identity in both $H$ and in $G$.
Since $H < G$ and hence $H \subset G$, we may view this equation as an equation in $G$. Hence, $h \in G$, so there exists an inverse in $G$. Call it $g$. Hence, $$hh^{-1} = e \iff g(hh^{-1}) = ge \iff (gh)h^{-1} = ge \iff h^{-1} = g.$$ Hence, the inverse in $H$ is equal to the inverse in $G$.