Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
en.wikipedia.org/wiki/Hopfian_group may be of interest to you. –  user1306 Apr 12 at 1:20
    
This question is a duplicate of math.stackexchange.com/questions/79852/… –  studiosus Apr 14 at 20:59
add comment

2 Answers 2

up vote 4 down vote accepted

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.

share|improve this answer
1  
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 at 1:25
    
@Olivier. I noticed the slip in my language. Corrected it. Thanks. –  P Vanchinathan Apr 12 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 at 1:30
    
"$S^1/C_n\simeq S^1$" –  Pedro Tamaroff Apr 12 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 at 1:46
add comment

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

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

share|improve this answer
    
Or residually finite. –  studiosus Apr 12 at 1:22
    
@studiosus Sorry, never heard about that. You can always add an answer of your own. =) –  Pedro Tamaroff Apr 12 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 at 1:31
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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