$G/N = 1$ iff $N = G$ and $G/N = G$ iff $N = 1$ Suppose that $G$ is a finite group and $N \trianglelefteq G$. Then
$$G/N \cong \{1_G\} \iff |G/N| = |\{1_G\}| \iff \frac{|G|}{|N|}= 1 \iff |G| = |N| \iff N=G \text, $$
which answers this question. Likewise, $G/N \cong G$ if and only if $N$ is trivial.
I thought these facts were true for infinite groups as well, but now I'm having doubts. If $N=G$, then $a^{-1}b \in N$ for all $a, b \in G$, so there is only one coset and $G/N \cong 1$. If $N=1$, then $a^{-1}b \in N \iff b = a$, so every coset is a singleton and $G/N \cong G$. Is the statement below true or false?

If $G$ is a group with a normal subgroup $N$, then $G/N \cong 1$ implies $N=G$, and $G/N \cong G$ implies $N=1$.

I'd appreciate seeing lots of counterexamples if it's false. And I'd like to know if the statement generalizes to rings and ideals.

Here are some ideas. Suppose that $G/N \cong G$. The isomorphism $\phi : G/N \to G$ composes with the usual map $\pi : G \to G/N$ to produce a surjective homomorphism $\phi \circ \pi : G \to G$. Notice that $\ker \phi \circ \pi = \ker \pi = N$ since $\phi(\pi(g)) = 1_G \iff \pi(g) \in \ker \phi = \{ N \} \iff \pi(g)=N \iff g \in \ker \pi$. By the correspondence theorem, there is a bijection between subgroups of $G$ containing $\ker \phi \circ \pi =N$ and subgroups of $G$. I'd like to say "hence every subgroup of $G$ contains $N$, so $N$ must be trivial," but I'm not sure if that's true.
If $G/N$ is trivial, then $N = \ker \pi = G$. I expected the two claims to be equally difficult to prove...

EDIT: Quotient ring being isomorphic to the initial ring is relevant.
 A: $G/N = 1$ implies $N = G$, because $G/N$ has only only element (which then must be $N$).
$G/N = G$ does not imply $N = 1$. You can take $G$ to be $H \times H \times \dots$, for some group $H$, and $N = H \times 1 \times 1 \times \dots$. Then $G/N$ kills one of the copies of $H$, but $G/N = 1 \times H \times H \times \dots$ is still isomorphic to $H \times H \times \dots = G$. So for infinite groups this can indeed fail.
A: That $G/N$ is trivial if and only if $N=G$ holds for all groups, finite or infinite groups. For if $N\neq G$, then let $x\in G$, $x\notin N$; we will have $xN\neq eN$, so $G/N$ contains at least one nontrivial element.
That $G/N\cong G$ implies $N$ is trivial is true for finite groups, but need not be true for infinite groups. For example, take the group $G\cong \prod_{n=1}^{\infty} C_2$, the product of infinitely many copies of the cyclic group of order $2$. If $$N=C_2\times\{e\}\times\{e\}\times\cdots\times \{e\}\times\cdots,$$
then $N$ is not trivial, but $G/N\cong G$.
For another example, the Prüfer $p$-group $C_{p^{\infty}}$ has the property that for every proper normal subgroup $N$, $C_{p^{\infty}}/N\cong C_{p^{\infty}}$. And there are infinitely many proper normal subgroups.
The magic word is “Hopfian group”. A group $G$ is Hopfian if and only if every surjective morphism $f\colon G\to G$ is a bijection; in other words, if $G/N\cong G$, then $N=\{e\}$. Every finite group is Hopfian, as are many infinite groups, but not all.
The same holds for rings: $R/I$ is trivial if and only if $I=R$, by the same argument. A ring is Hopfian if and only if $R/I\cong R$ implies $I=\{e\}$. Every finite ring is Hopfian, as are other types of rings, but not every ring is Hopfian: $\prod_{i=1}^{\infty}\mathbb{Z}_2$ is an example.
In general, in any class of algebraic objects, an object is Hopfian if every surjection $\mathscr{O}\to\mathscr{O}$ must be a bijection; and it is co-Hopfian if every injection $\mathscr{O}\to\mathscr{O}$ must be a bijection.
