Is an onto homomorphism from G to itself an automorphism

A homomorphism from $G$ to itself is an automorphism if it is bijective.

I am trying to make the condition of bijectiveness weaker. 1-1 is not enough because there is a 1-1 homomorphism from $\mathbb{Z}$ to $\mathbb{2Z}$. What about onto? If a homomorphism from $G$ to itself is onto, then is it an automorphism? Or, similarly, if $H$ is a nontrivial normal subgroup of $G$, can $G$ and $G/H$ be isomorphic?

-
en.wikipedia.org/wiki/Hopfian_group may be of interest to you. – user1306 Apr 12 '14 at 1:20
This question is a duplicate of math.stackexchange.com/questions/79852/… – studiosus Apr 14 '14 at 20:59

Consider $z\mapsto z^n$ from the group of nonzero complex numbers to itself. By fundamental theorem of algebra it is onto. That is, any complex number has $n$th roots. But it takes the same value on all $n$th roots of unity. So an infinite group quotiented by a finite subgroup CAN BE isomorphic to itself.

-
Did you mean to write that the circle is an example of an infinite group with a quotient by a finite subgroup isomorphic to itself? – Olivier Bégassat Apr 12 '14 at 1:25
@Olivier. I noticed the slip in my language. Corrected it. Thanks. – P Vanchinathan Apr 12 '14 at 1:29
FTA is a little excessive for this purpose; it is onto because $\mathbb{R}$ is complete and because of complex polar coordinates. – Ryan Reich Apr 12 '14 at 1:30
"$S^1/C_n\simeq S^1$" – Pedro Tamaroff Apr 12 '14 at 1:40
You are right. Shortest path for this proof is the polar route. As nth roots exists for [positive reals. However, I feel that mathematician in their quest for brevity or economy, state FTA in that way. In my class I formulate it as surjectivity of polynomial functions. As zero is a special number I did not want my students to get the idea that zero value is always achieved as opposed to others. (The proof by appealing to Liouville's theorem exploits non-vanishing of the entire function and obscures the surjectivity). – P Vanchinathan Apr 12 '14 at 1:46

If $G$ is finite, yes, for one-one and onto are equivalent.

For you last question, $S^1/C_n\simeq S^1$.

-
Or residually finite. – studiosus Apr 12 '14 at 1:22
@studiosus Sorry, never heard about that. You can always add an answer of your own. =) – Pedro Tamaroff Apr 12 '14 at 1:26
This is a wonderful concept meaning that intersection of all finite index subgroups is trivial. Malcev proved that finitely generated matrix groups are residually finite. – studiosus Apr 12 '14 at 1:31