5

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?

Henry
  • 3,125

2 Answers2

4

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.

P Vanchinathan
  • 19,504
  • 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 '14 at 01:25
  • @Olivier. I noticed the slip in my language. Corrected it. Thanks. – P Vanchinathan Apr 12 '14 at 01:29
  • 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 01:30
  • "$S^1/C_n\simeq S^1$" – Pedro Apr 12 '14 at 01:40
  • 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 01:46
2

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

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

Pedro
  • 122,002