If there is a homomorphism $\phi: G \to H$ then $G / \ker \phi \cong \phi(H)$. If $\phi$ is not trivial then $\phi(H) \neq 1$.
In your example $G=S_3$ and $H=C_3$ so if a nontrivial homomorphism exists, then $\phi(C_3) = C_3$ because $C_3$ has no nontrivial subgroups. Therefore $|\ker \phi| = 2$. But $S_3$ has no normal subgroups of order 2, so such a homomorphism cannot exist. As the groups get bigger, arguments like this based on the order are in general easier than arguments based on generators.
In general, if $G$ is generated by $\{g_1, g_2, \dots, g_n\}$, and the set map $\phi : \{g_1, g_2, \dots, g_n\} \to H$ maps
$g_i \mapsto h_i$, then it can be extended uniquely to a homomorphism as long as all the relations satisfied by the $g_i$ are also satisfied by the corresponding $h_i$ in $H$.
In your example, you choose three generators of $G$ and in your case $H$ was cyclic of prime order. Since the generators satisfied $g_i^2=1$ but none of the nonidentity elements of $H$ can satisfy $h_i^2=1$ you knew that all the $h_i$ were 1, and so the homomorphism was trivial. As the groups get more complicated there will be more to check. In general this is a slick method only in special cases where $H$ has nice structure and the presentation of $G$ is especially easy.
One generalization that you can prove is this: If $G$ is generated by
$\{g_1, g_2, \dots, g_n\}$ and the order of $g_i$ is relatively prime to $m$ for each $i$, then there can be no nontrivial homomorphism $G \to C_{m}$.
By the first isomorphism theorem, $A/\ker\varphi \cong \varphi(A)$. In particular: \begin{equation}\frac{29}{|\ker\varphi|}=|\varphi(A)|.\end{equation} Since 29 is a prime number, either $\ker\varphi=\{0\}$ (which implies $\varphi$ is injective) or $\ker\varphi=A$ (i.e. $\varphi$ is trivial).
If $|B|=80$ then since $\varphi(A)$ is a subgroup of $B$, $|\varphi(A)|$ divides $80$ by Lagrange's theorem. But $|\varphi(A)|$ also divides 29, so $|\varphi(A)|=1$ and $|\ker\varphi|=29$, which implies $\varphi$ is trivial.
Best Answer
This is not a homomorphism at all. According to how you defined it $\varphi(1.5+1.5)=\varphi(3)=3$. However, $\varphi(1.5)+\varphi(1.5)=1+1=2$.
Anyway, a group homomorphism is injective if and only if its kernel is trivial.