What is wrong with the followig proof that $\mathbb{Z}$ is isomorphic to $\mathbb{Z} \times \mathbb{Z}$ Since 
$ G =( \prod_{1}^{\infty} \mathbb{Z} )\times  \mathbb{Z}  \times \mathbb{Z}  \cong (\prod_{1}^{\infty} \mathbb{Z}) \times \mathbb{Z}  $, 
by taking quotients we get
$\mathbb{Z \times Z} \cong G/ \prod_{1}^{\infty} \mathbb{Z} \cong \mathbb{Z}$.
Therefore 
$\mathbb{Z} \cong \mathbb{Z \times Z}$.
But $\mathbb{Z}$ is indecomposable!  What's wrong with above proof?
 A: Quotienting by isomorphic copy need not preserve isomorphisms.  For a simpler related example: $C_n=\mathbb{Z}/n\mathbb{Z}$ and $m\mathbb{Z}\cong n\mathbb{Z}$ for all $m,n>0$, but obviously the cyclic groups $C_m$ and $C_n$ are not isomorphic if $m\neq n$.
A: Your argument is based on the assumption that $G/N_1\cong G/N_2$ follows from $N_1\cong N_2$. Since $\mathbb Z\not\cong\mathbb Z\times\mathbb Z$, your example shows that the assumption is incorrect. Here is a simpler example.
The non-isomorphic groups $\mathbb Z_4$ and $\mathbb Z_2\times\mathbb Z_2$ are both homomorphic images of the group $G=\mathbb Z_2\times\mathbb Z_4$. Thus there are subgroups $N_1,N_2$ of $G$ such that $G/N_1\cong\mathbb Z_4$ and $G/N_2\cong\mathbb Z_2\times\mathbb Z_2$. The subgroups $N_1$ and $N_2$ are isomorphic, as they are groups of order $2$;
so $N_1\cong N_2$ but $G/N_1\not\cong G/N_2$.
P.S. In other words (considering your comment on another answer, "If I have $G\cong H\times K$, can i not say $G/H\cong K$?"): here we have $G\cong N_2\times\mathbb Z_4$, but $G/N_2\not\cong\mathbb Z_4$.
