homomorphism from $S_3$ to $\mathbb Z/3\mathbb Z$ 
TRUE/FALSE TEST:

*

*There is a non-trivial group homomorphism from $S_3$ to $\mathbb Z/3\mathbb Z.$

My Attempt:

True: Choose $a,b\in S_3$ such that $|a|=3,|b|=2.$
Then $S_3=\{1,a,a^2,b,ba,ba^2\}.$
Define $f:S_3\to\mathbb Z/3\mathbb Z:b^ia^j\mapsto j+3\mathbb Z$
Then $f$ is a nontrivial homomorphism.

Is the attempt correct?

There is some special tric to find homomorphism from a group G to another group G'. At first we have to collect all normal sub-groups of the group G , which are precisely given by the kernel of the mapping f:G-->G/N , by f(g)=gN. Then by fundamental theorem of group homomorphism we will have G/kerf is isomorphic to f(G) . Now if f(G) is a subgroup of G' , then the no. of homomorphism will be
Aut. (f(G)) × n , where n is the total no. of subgroups in G' of type f(G).
Here , we know that, {e}, A3 and S3 are the normal sub-groups of S3.
So, first take {e}
Then S3/{e} is isomorphic to S3 , but S3 is not a subgroup of Z3 . Hence , there is no one -one homomorphism.
Now, take A3
Then S3/A3 is isomorphic to Z2 . But , Z2 is not a subgroup of Z3 . So, in this case we will not have homomorphism.
Finally, take S3
Then S3/S3 , is isomorphic to {e} , and this is a subgroup of Z3 , so from that we will have trivial homomorphism.
So, we conclude that there is no non-trivial homomorphism from S3 to Z3.
