Counterexample to homomorphism problem

Here it is:

Quote:

Show that if H is any group and h is an element of H with

, then there is a unique homomorphism from

to H such that

.

I have been able to prove such a homomorphism exists with a cyclic subgroup of H anyway. How do I extend this to they whole group H? For example when I do this using I end up with a "homomorphism" that is a relation, not a function.

And as far as uniqueness goes the homomorphism maps x to h. Isn't that the only function that will do that? I mean if we have how could we have a distinct function ? If they both map x to h wouldn't they have to be the same function automatically? I'm missing something here...

-Dan

Re: Counterexample to homomorphism problem

There is no requirement that the homomorphism be surjective....

The whole point of the problem is that: o(h)|n. If this were NOT so, we would have h^{n} ≠ e, in which case:

φ(e) = φ(x^{n}) = φ(x)^{n} = h^{n} ≠ e, so φ is not a homomorphism (which must map identity to identity).

Homomorphisms are just certain functions between groups, the image of a homomorphism does NOT have to be the entire target group.