Actual Question is to check if there is :
Non trivial group homomorphism $\eta : S_3 \rightarrow \mathbb{Z}/3\mathbb{Z}$.
What I have tried so far is :
I take $(1 2)\in S_3$, this has order $2$.
So, order of $\eta(1 2)$ should divide $2$ but then, I have no element of order $2$ in $\mathbb{Z}/3\mathbb{Z}$
which would give me only possibility of $\eta (1 2)$ as $\bar{0}$
with similar reasoning I would see only possibility for images of $(2 3)$ and $(1 3)$ is $\bar{0}$.
But then, $\{(1 2), (2 3), (1 3)\}$ generates $S_3$ So, if each generator goes to $\bar{0}$, so should be the images of all elements
i.e., all elements must be mapped to $\bar{0}\in \mathbb{Z}/3\mathbb{Z}$.
So, I do not have non trivial homomorphism $\eta : S_3 \rightarrow \mathbb{Z}/3\mathbb{Z}$.
I like this idea and i would like to see for generalization of this idea.
i.e., Can I say something about non existence of a non trivial homomorphism from $G$ to $H$ if i know orders of all elements of $G$ and $H$.
I would be thankful if some one can give me an exciting example (though it is subjective) which is not very trivial which follows same idea as above question.
Thank you.
Best Answer
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 $G \to C_{m}$.