Describe all ways in which $S_3$ can operate on a set of four elements.
My approach: This question can be broken down into: How many homomorphisms exist from $S_3$ to $S_4$. Say $\varphi : S_3 \to S_4$ is a homomorphism. Then we have three possibilities for $\text{ker }\varphi$: $\{1\}, \{1, (1\ 2\ 3), (1\ 3\ 2)\}$, and $S_3$.
The case in which $\text{ker }\varphi = S_3$ is the trivial homomorphism that maps everything to the identity.
Now, the case in which $\text{ker }\varphi = \{1\}$ is the same as saying that the mappings are injective. This comes down to picking three of the four elements and permutating them and leaving the fourth one fixed. There are $\binom43 = 4$ ways of doing this.
Say $\text{ker }\varphi = \{1, (1\ 2\ 3), (1\ 3\ 2)\}$. This means that $\varphi((1)) = \varphi((1\ 2\ 3)) = \varphi((1\ 3\ 2)) = (1)$. Morover we can observe the following two properties immediately:
- $\varphi((1\ 2\ 3)) = \varphi((1\ 3))\varphi((1\ 2)) = (1)$. Equivalently $\varphi((1\ 2)) = \varphi((1\ 3))$.
- $\varphi((1\ 3\ 2)) = \varphi((1\ 3))\varphi((2\ 3)) = (1)$. Equivalently $\varphi((1\ 3)) = \varphi((2\ 3))$.
and thus $\varphi((1\ 2)) = \varphi((1\ 3)) = \varphi((2\ 3))$. But we know by properties of homomorphisms that $\vert \varphi((1\ 3)) \vert \mid \vert (1\ 3) \vert = 2$. So $\vert \varphi((1\ 3)) \vert$ is 1 or 2. But if the order of $\varphi((1\ 3))$ were 1 it would be in the kernel, which would be a contradiction to the kernel we chose, so it must be 2. We can map $(1\ 3)$ to any 2-cycle in $S_4$, of which there are 6, as well as any product of disjoint 2-cycles, of which there are 3. Hence we have 9 possible homomorphic mappings given this kernel.
Adding up all the possible homomorphisms from $S_3$ to $S_4$ that we counted, we get 14 different ways in which $S_3$ can act on four elements, as described above.
Is this correct? Are there 14 homomorphisms from $S_3$ to $S_4$? Is my reasoning correct? Or are there any hidden assumptions I made that I shouldn't have made?