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 $G \to C_{m}$.
Yes, you can have nontrivial homomorphisms. For instance, from
$$(\mathbb Z, +)$$
to
$$
(\mathbb Z/3\mathbb Z, +)
$$
you can send $n$ to $n \bmod 3$.
Consider the matrix
$$
A = \begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 1 \\
1 & 0 & 0
\end{bmatrix}
$$
Since $A^3 = I$, the map that sends $n \in (\mathbb Z, +)$ to $A^n$ in $P(3; \mathbb R)$ is essentially the same map as in the first example, except that this time, the $\mathbb Z / 3 \mathbb Z$ is a subgroup of the matrix group.
As for maps from $\mathbb R$ to $P(n; \mathbb R)$: the homomorphic image of a connected topological group will still be connected, and the only connected subgroup of $P(n; \mathbb R)$ is the trivial one. (Indeed, this remark applies to any codomain that's discrete, as @Clement notes .)
Best Answer
There is an infinite group which contains every single finite group! Just take the direct sum of all finite groups...You get a group, and this is infinite. However, this is boring because the group is so wild. For example, there is no finite set which will generate this group (note that the same problem exists for Christian Blatter's group - in fact, his group is uncountable while this one is countable). Often the challenge in group theory is to give finitely presentable groups with wild properties, such as containing all finite groups. For example, in the group we have just constructed every element has finite order, but it is an open problem if there exists a finitely presentable group in which every element has finite order.
In order to overcome this problem of non-finite generation/presentation we need quite a big theorem, called Higman's Embedding Theorem. This states that a group is "recursively presentable" if and only if it can be embedded into a finitely presentable group. Rotman's book An introduction to the theory of groups contains an accessible proof of Higman's Embedding Theorem.
Thereom: There is a finitely presented group which contains every single finite group. Specifically, this group is finitely generated and countable.
Proof: Start by enumerating all finite groups and then take their direct sum using this enumerations. (Basically, do what I said in the first paragraph but be careful when doing it). This gives you a recursively presented group, and so you can use Higman's Embedding Theorem to obtain a finitely presented group containing your group. Done.
(Note that the above proof actually works for all finitely presented groups but not all finitely generated groups...but to do that properly I'd have to convince you that there are only countably many finitely presented groups but uncountably many finitely generated groups. These facts both hold, but are stories for a different day...)