2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

$G/N = 1$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.