a non trivial example: $G/H$ is isomorphic to $G$ I need a non trivial  examples of group isomorphism that maps $G\to H$ such that 
 $G/H \cong G$.
 A: For both the title and the text of the question, you can take $G=\mathbb Z\times\mathbb Z\times\mathbb Z\times\mathbb Z\times\dots$ (an infinite product) and 
$H=\mathbb Z\times\{0\}\times\mathbb Z\times\{0\}\times\dots$.  Then all three of $G$, $H$, and $G/H$ are isomorphic.  (You can replace all the $\mathbb Z$'s by your favorite other group to get more examples.)
A: The first finitely generated example found was the Baumslag-Solitar group $G = \langle x,y \mid y^{-1}x^2y = x^3 \rangle$, with $H$ equal to the normal closure in $G$ of $w^2x^{-1}$, where $w=x^{-1}y^{-1}xy$.
So $G/H \cong \langle x,y,w \mid y^{-1}x^2y = x^3, w = x^{-1}y^{-1}xy, x=w^2 \rangle$.
Eliminating $x$, using $x=w^2$, gives
$G/H \cong \langle y,w \mid y^{-1}w^4y = w^6, w = w^{-2}y^{-1}w^2y \rangle = \langle y,w \mid y^{-1}w^2y=w^3 \rangle \cong G$
(because $y^{-1}w^2y=w^3$ implies $y^{-1}w^4y = w^6$, so that relation is redundant).
That's all reasonably straightforward but, although it seems very plausible that $H$ is a nontrivial subgroup of $G$, that does need to be proved, and you need the theory of HNN extensions to prove it.
A: One additional example, as the OP asked for a group defined without direct products: fix a prime $p$ and let $T_p=\{z\in\mathbb{C}:\exists n\hspace{3pt}z^{p^n}=1\}$. Let $\varphi:T_p\to T_p$ be defined by $\varphi(z)=z^p$. Then $\varphi$ is a epimorphism (surjective homomorphism) which is not injective, i.e. $\ker(\varphi)\neq\{1\}$, and $T_p/\ker(\varphi)\cong T_p$.
