If $\space o(\ker\phi) = n$, then $\varphi$ is an $n$-to-$1$ mapping from $G$ onto $\varphi(G)$ If $G,H$ are groups and $\varphi\colon G\to H$ a group homomorphism, how can I prove that if the order of $\ker\phi$ is $n$ then $\varphi$ is an $n$-to-$1$ mapping from G onto $\varphi(G)$

If $\phi$ is a homomorphism from $G$ into $H$.
Let $\phi(a)=\phi(b)\Rightarrow \phi(a)(\phi(b))^{-1}=e_H\Rightarrow \phi(ab^{-1})=e_H\Rightarrow ab^{-1}\in \ker \phi$
If $a,b\in G$ and $ab^{-1}\in \ker \phi$, then $\phi(ab^{-1})=e_H\Rightarrow \phi(a)=\phi(b)$.
Hence if $\ker \phi=n$, for every $t\in H$ $\exists a\in G, $ such that $\phi(a)=t$; therefore there are $n$ distinct elements in the set $\{ah:h\in \ker \phi\}$, that are preimages of $t$.
