Let $\phi: G_1 \to G_2$ be a group homomorphism.
Let $\ker \left({\phi}\right)$ be the kernel of $\phi$.
Then: $\operatorname {Im} \left({\phi}\right) \cong G_1 / \ker \left({\phi}\right)$ where $\cong$ denotes group isomorphism.
Proof:
Let $K = \ker \left({\phi}\right)$.
$G_1 / K$ exists.
We need to establish that the mapping $\theta: G_1 / K \to G_2$ defined as: :$\forall x \in G_1: \theta \left({x K}\right) = \phi \left({x}\right)$ is well-defined.
That is, we need to ensure that: $\forall x, y \in G: x K = y K \implies \theta \left({x K}\right) = \theta \left({y K}\right)$
Let $x, y \in G: x K = y K$. Then:
$xK = yK$
$\iff x^{-1} y \in K = \ker \left({\phi}\right)$
$\iff \phi \left({x^{-1} y}\right) = e_{G_2}$
$\iff \left({\phi \left({x}\right)}\right)^{-1} \phi \left({y}\right) = e_{G_2}$
$\iff \phi \left({x}\right) = \phi \left({y}\right)$
Thus we see that $\theta$ is well-defined.
I don't understand the two lines here:
$\iff \phi \left({x^{-1} y}\right) = e_{G_2}$
$\iff \left({\phi \left({x}\right)}\right)^{-1} \phi \left({y}\right) = e_{G_2}$
I know that $\phi \left({x^{-1} y}\right) = {\phi \left({x^{-1}}\right)} \phi \left({y}\right)$, but I don't understand how to get from $\phi \left({x^{-1}}\right)\phi \left({y}\right) = \left({\phi \left({x}\right)}\right)^{-1}\phi \left({y}\right)$.
$\phi(x)\phi(x)^{-1}=e_2=\phi(e_1)=\phi(xx^{-1})=\phi(x)\phi(x^{-1})$