TITLE: Ways $S_3$ can act on a set of 4 elements.
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: We work instead in terms of the corresponding permutations representations (homomorphisms) $\phi: S_3\to S_4$. To avoid confusion we work with $S_3(\{1,2,3\})$ and $S_4(\{a,b,c,d\})$. Let $x = (1,2)$ and $y = (1,2,3)$, so that $$x^2=y^3=1,\text{ }y^2 = (1,3,2),\text{ }xy = (2,3) = yx^2,\text{ }xy^2 = (1,3) = yx.$$
By the First Isomorphism Theorem, $\text{Ker}\,\phi$ is normal in $S_3$ and $\text{Im}\,\phi \cong S_3/\text{Ker}\,\phi$, with isomorphism $\phi(g\text{Ker}\,\phi) = \phi(g)$. But$$yxy^{-1} = xy^2y^{-1} = xy\ne 1,\,x,$$so $S_3$ has no normal subgroups of order $2$, and $\text{Ker}\,\phi$ must be one of $\{1\}$, $A_3$, $S_3$.
If $\text{Ker}\,\phi = S_3$, then$$\text{Im}\,\phi\cong S_3/S_3\cong S_1,$$so $\phi$ is the trivial homomorphism, i.e. it always maps to $(a)(b)(c)(d)$.
If $\text{Ker}\,\phi = A_3$, then$$\text{Im}\,\phi\cong S_3/A_3\cong S_2,$$ $$\phi(1)=\phi(y)=\phi(y^2) = (a)(b)(c)(d),$$while$$\phi(x)=\phi(xy)=\phi(xy^2)$$ has order $2$ and takes the form $(p,q)$ or $(p,q)(r,s)$, for distinct $p,q,r,s\in\{a,b,c,d\}$.
If $\text{Ker}\,\phi = \{1\}$, then$$\text{Im}\,\phi\cong S_3/\{1\} \cong S_3.$$So $\phi(x)$ takes the form $(p,q)$ or $(p,q)(r,s)$, and $\phi(xy)$ takes the form $(p',q')$ or $(p',q')(r',s')$. Without loss of generality, $p' = p$.

If $\phi(x) = (p,q)$ and $\phi(xy) = (p,q')$, then $q\ne q'$, or else$$\phi(y) = \phi(x)\phi(xy) = (p,q)(p,q')$$becomes the identity, so $\phi(y) = (q,p,q')$. Because $\phi(x) = (q,p)$, $\phi$ simply maps $\pi\in S_3(\{1,2,3\})$ to the corresponding permutation with $1$, $2$, $3$ replaced by $q$, $p$, $q'$, respectively.
If $\phi(x) = (p,q)$ and $\phi(xy) = (p,q')(r',s')$ (or the other way around; these two cases are interchangeable by "switching" $1$ and $3$, since $x$, $xy$ are both transpositions), then $q\ne q'$, or else$$\phi(y) = \phi(x)\phi(xy) = (p,q)(p,q')(r',s')$$becomes a transposition and has order $2$ instead of $3$, so without loss of generality, assume $r' = q$. But then$$\phi(y) = (q,p,q')(q,s') = (q,s',p,q')$$has order $4$ instead of $3$, so this case is impossible.
If $\phi(x) = (p,q)(r,s)$ and $\phi(xy) = (p,q')(r',s')$, then $q\ne q'$, or else $\{r,s\} = \{r',s'\}$ and $\phi(x)=\phi(xy)$ forces $\phi(y)$ to be the identity. Without loss of generality, assume $q' = r$, so $\{r',s'\} = \{q,s\}$ means $\phi(xy) = (p,r)(q,s)$. Thus$$\phi(y) = (p,q)(r,s)(p,r)(q,s) = (p,s)(q,r)$$ has order $2$ instead of $3$, so this case is also impossible.

Remark. The only "unexpected" permutation here is $\phi$ taking transpositions to $(a, b)(c, d)$ (or $(a, c)(b, d)$ or $(a, d)(b, c)$) and other elements to the identity $(a)(b)(c)(d)$.