Counterexample to homomorphism problem
Here it is:
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.
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
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...
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 hn ≠ e, in which case:
φ(e) = φ(xn) = φ(x)n = hn ≠ 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.