Can a subgroup be mapped onto its parent group as a homomorphism?

[majutsu]

Consider the cyclic group G={a,a^2,a^3,...a^12=u} and its subgroup G`={a^2,a^4,...,a^12}. My book says that the mapping
a^n ---> a^2n is an homomorphism of G onto G` (this seems true)

and that X: a^n ---> a^n is homomorphism of G` onto G (this seems to be false to me, a misprint)

A homomorphism of G` onto G would have
1)every g` in G` has a unique image g in G (true)
2)if X(a`)=a and X(b`)=b then X(a` o b`)=X(a x b) with operator o for G` and x for G (true)
3)every g in G is an image (false) e.g. a^3 in G is not the image of any g` in G`.

In fact, it would seem to me that you can have a homomorphism of a group onto a subgroup, but you could not have a homomorphism of a subgroup onto a group ever. Is this right, or am I missing something?

 

[Muzza]

3)every g in G is an image (false) e.g. a^3 in G is not the image of any g` in G`.

Have you confused homomorphism with isomorphism?

 

[Hurkyl]

you could not have a homomorphism of a subgroup onto a group ever.

Try looking at infinite groups.

 

[majutsu]

I see the problem. It is not a misprint, just poor language. The original text says (paraphrased) a homomorphism of G into H as
F(G,*g) ---> (H,*h)
and
F(g1 *g g2)=F(g1) *h F(g2)

Then the book reads, "if (iii) every h in H is an image we have a homomorphism of G onto H." This is actually a definition of the term ONTO, that is to say epimorphism, or surjective. But if you are not used to all this, it almost makes it sound like a requirement of homomorphism, which I realize now it is not. This is why it sounded like I was confusing isomorphism with homomorphism, as isomorphisms are injective and surjective, that is one-to-one.

You all have really helped clear this up. Carrying this misunderstanding forward could have been deadly.