First Isomorphism Theorem (shorter version)
Given $\varphi: G \longrightarrow H$ a group homomorphism, then:

$\hspace{2cm}$ i.) $ker \varphi \unlhd G$
$\hspace{2cm}$ ii.) $G/ker\varphi \cong im\varphi$

Proof:
item(i): Let $g\in{G}$ and let $x\in{ker\varphi} \leq G$. Then, since $\varphi$ is a homomorphism, $\varphi(gxg^{-1})=\varphi(g)\varphi(x)\varphi(g^{-1})=\varphi(g)\cdot1_H\cdot\varphi(g^{-1})=\varphi(g)\varphi(g^{-1})=\varphi(gg^{-1})=\varphi(1_G)=1_H$.
item (ii): Let $K=ker\varphi$. Define $\Phi:G/K \longrightarrow H$ by $\Phi(aK)=\varphi(a)$. $aK=bK \Longleftrightarrow ab^{-1}\in{K} \Longleftrightarrow \varphi(ab^{-1})=1_H \Longleftrightarrow \varphi(a)=\varphi(b) \Longrightarrow \Phi$ is well-defined and 1-1. $\Phi$ is a homomorphism since $\Phi(aKbK)=\Phi(abK)=\varphi(ab)=\varphi(a)\varphi(b)=\Phi(aK)\Phi(bK)$. Also, $im\Phi=im\varphi$ $\therefore \Phi$ is an isomorphism. $\blacksquare$