When does a non-trivial Homomorphism exist?

Question

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.

Answer 1

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.

Answer 2

Here, checking that the map $\phi$ is well defined amounts to the following task :

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.