Homomorphisms preserve identity and inverses

Statement

Let $G$ and $H$ be groups with identity elements $e_G$ and $e_H$, and let $\varphi \colon G \to H$ be a group homomorphism. Then $\varphi(e_G) = e_H$, and $\varphi(a^{-1}) = \varphi(a)^{-1}$ for each $a \in G$.

Remarks

The first equation follows from $\varphi(e_G) \varphi(e_G) = \varphi(e_G) = \varphi(e_G) e_H$ and cancellation. The second follows from $\varphi(a) \varphi(a^{-1}) = \varphi(e_G) = e_H$ and the uniqueness of inverses. A consequence is that $\varphi(a^n) = \varphi(a)^n$ for each integer $n$.

Explanations

L1 human

A map that respects products automatically respects the identity and the inverses. You do not need to ask for these two properties in the definition of a homomorphism. For example, the exponential map sends $0$ to $1$ and $-x$ to $1/e^x$.