Can someone verify if my part solution is correct and help me complete it?
I found normal subgroups of S3: {e}, {S3}, {e, (123), (132)}
If kernel of homomorphism is {S3} then we have trivial homomorphism
If kernel is {e, (123), (132)} then we have two homomorphism because natural projection p from S3 to S3/kernel gives 2 elements group which is isomorphic to Z2 (g) and there are 2 homomorphisms from Z2 to Z24 (h1 and h2) and we get our desired homomorphism by composition h1gp and h2gp
If kernel is {e} then image of homomorphisms have order 6 and its a subgroup of Z24 so image is {0,4,8,12,16,20} and I do not know what to do next.
$$S_3=\langle a,b\mid a^3,b^2,aba^{-2}b\rangle. $$
There's two.
They can be specified using that $$S_3=\langle (12),(123)\rangle.$$
$$(123)\to e,(12)\to \bar{12},$$ and
$$(123)\to e,(12)\to e.$$
There's no embedding because any subgroup of an abelian group is abelian.
There's none with kernel $S_2$, because $S_2$ isn't normal.