Proof of Group Homomorphisms Preserve the Identity Element and Inverses

We write both operations multiplicatively, as permitted by Group and Abelian Group and Group Homomorphism and Isomorphism.

The identity element. By condition 2 of Group and Abelian Group applied in $G$ we have $e_Ge_G=e_G$. Applying $\varphi$ and using the homomorphism property from Group Homomorphism and Isomorphism gives

On the other hand, condition 2 of Group and Abelian Group applied in $H$ to the element $\varphi(e_G)$ gives

Combining the two displays yields $\varphi(e_G)\varphi(e_G)=e_H\varphi(e_G)$, and the right-hand cancellation law, claim 1 of Cancellation Laws and Basic Inverse Identities in a Group applied in $H$, gives $\varphi(e_G)=e_H$.

Inverses. Let $a\in G$. By Uniqueness of the Identity Element and of Inverses in a Group we have $aa^{-1}=e_G$ and $a^{-1}a=e_G$. Applying $\varphi$, using the homomorphism property, and using $\varphi(e_G)=e_H$ just proved, we obtain

and

Thus $\varphi(a^{-1})$ satisfies both equations required of an inverse of $\varphi(a)$ in $H$, and the uniqueness of inverses in Uniqueness of the Identity Element and of Inverses in a Group, applied in $H$, gives $\varphi(a^{-1})=\varphi(a)^{-1}$.