2
$\begingroup$

Suppose that $G$ and $G'$ are two groups isomorphic to each other. Is it true that any onto homomorphism from $G$ to $G'$ is an isomorphism, i.e. any surjective homomorphism has a trivial kernel?

If it was not the case, then $G≈G'$ also Image of homomorphism $ =G'≈G/K$. Hence $G≈G/K$. This implies K is trivial. Am I right ?

$\endgroup$
  • 1
    $\begingroup$ It is true for finite groups, for trivial reasons. $\endgroup$ – Bernard Jan 10 '17 at 19:46
  • $\begingroup$ Is it also true for groups with finitely countable elements ? @ Bernard $\endgroup$ – Shreedhar Bhat Jan 10 '17 at 19:49
  • $\begingroup$ @ShreedharBhat no, my example works if you change $\mathbb R$ with a finite group. $\endgroup$ – Jorge Fernández Hidalgo Jan 10 '17 at 21:38
5
$\begingroup$

clearly false, consider $\mathbb R ^\mathbb N$ and the "scoot to the left" function. (in other words $(x_1,x_2,\dots)\rightarrow (x_2,x_3,\dots)$)

$\endgroup$
0
$\begingroup$

The additive groups $(\mathbb C,+)$ and $(\mathbb R,+)$ are isomorphic; the map $z\mapsto\Re z$ is a surjective homomorphism but not an isomorphism.

Let $F$ be the free group generated by a countably infinite set $X.$ Any non-injective surjection from $X$ to $X$ extends to a surjective homomorphism from $F$ to $F$ which is not an isomorphism.

$\endgroup$

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.