# When does a non-trivial Homomorphism exist?

Let $$G$$ and $$H$$ be groups and $$p \in \mathbb{N}$$ be a prime such that $$p|o(G)$$ and $$p|o(H)$$. Then, we know that there exist $$g$$ and $$h$$ in $$G$$ and $$H$$ respectively which generate cyclic subgroups of order $$p$$. Then, given an arbitrary $$x \in G$$, $$x=yg^m$$ for some $$y \in G$$ and $$0 \leq m < p$$. Define $$\phi : G \to H,$$ $$\phi(yg^m)=h^m.$$ Then, this map is a homomorphism so far as $$\phi$$ is well-defined. This basically boils down to whether different representations of $$x \in G$$ affect the image of $$x$$ under $$\phi$$. I don't know how to proceed if I want to prove or disprove the well-definedness of this map.

• $g=g\cdot g^0=g^0\cdot g^1$. What should $\phi(g)$ be? Feb 25 '20 at 14:32

Your $$\phi$$ is never well-defined, the representation $$x=yg^m$$ is never unique or valid. Your "some $$y\in G$$" is simply $$xg^{-m}$$ and can be taken for any $$m$$. Thus, by your definition, we have $$\phi(x)=h^m$$ for any $$m$$, which makes sense only when $$h$$ is the neutral element.
In fact two groups having orders with a non-trivial common divisor is not enough for a non-trivial homomorphism to exist. Even when the orders are equal. Consider any non-abelian simple group $$G$$ (e.g. the alternating group $$A_n$$) and any abelian group $$H$$ such that $$|H|=|G|$$, e.g. $$H=\mathbb{Z}_{|G|}$$. Note that no non-trivial homomorphism $$G\to H$$ exists.
This is true, however, if $$G$$ is assumed to be abelian, or more generally when $$G$$ has a normal subgroup of index $$p$$. That's because in this situation there's a quotient map $$G\to\mathbb{Z}_p$$ which composed with an embedding $$\mathbb{Z}_p\to H$$ gives a non-trivial homomorphism.
Here, checking that the map $$\phi$$ is well defined amounts to the following task :
Check if $$h^m$$ depends on the choice of $$y, m$$ such that $$x = yg^m$$.
In other words, you have to check whether it is possible to write $$x = y g^m$$ and $$x = y'g^{m'}$$ with $$m \neq m'$$ or not. If it is possible, $$\phi$$ isn't well defined. If it isn't possible, $$\phi$$ is well defined.