Homomorphism $\psi$ from $S_3$ to $S_4$ [duplicate]
Solution 1:
There are 34 homomorphisms from $S_3$ to $S_4$.
Let's counting homomorphisms by analysis of its kernel.
Case 1. $S_3$ is the kernel: As Kaj Hansen commented, there is the trivial homomorphism and this is the only choice.
Case 2. $A_3$ is the kernel: There are 9 homomorphisms.
Case 3. $1$ is the kernel: Your argument is wrong; as you've already noticed, there are 4 subgroups (the stabilizers of a single letter) that isomorphic to $S_3$. But you also have to take permutations (rename of letters) into account. Hence there are $4 \times 3! = 24$ homomorphisms.
After all, we have $1 + 9 + 24 = 34$ homomorphisms from $S_3$ to $S_4$.
Generally speaking, let $G, G'$ be finite groups and $N$ a normal subgroup of $G$. Suppose $G'$ has $n$ subgroups that isomorphic to (not just the orders are same) $G/N$. What we can say is the number of homomorphisms from $G$ to $G'$ with its kernel $N$ equals $n \times \vert \operatorname{Aut}(G/N) \vert$.