# $G/N = 1$ 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.

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.

$$G/N = 1$$ 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.