How many homomorphism from $S_3$ to $S_4$? Please find these using fundamental theorem.
I think if $f\colon G \rightarrow G'$ is a group homomorphism then $G/\ker f$ is isomorphic to a subgroup of $G'$. For one choice of $\ker f$, the order of $G/\ker f$ is $k$ and $G'$ has $n$ subgroups of order $k$ . Hence there are $n$ homomorphisms.
But in case of $S_3$ to $S_4$, if $S_3/\ker f$ is a subgroup of $S_4$ then we have three cases:
$\ker f = \{\mathrm{id}\}$: then $S_3$ is isomorphic to a subgroup of $S_4$. There are 4 subgroups in $S_4$ isomorphic to $S_3$ so in this case we have 4 homomorphisms.
$\ker f = A_3$ then $S_3/A_3$ is a subgroup of order 2 in $S_4$. Again, 9 subgroups in $S_4$ so 9 homomorphisms.
$\ker f = S_3$ which gives 0 homomorphism.
So in total 14 homomorphisms. Am I right ?