In his Algebra Artin gives an example of a map from $S_4$ to $S_3$, which is realised by variously partitioning the former's set of four indices $\{1,2,3,4\}$ into pairs
$$\Pi_1:\{1,2\}\cup\{3,4\}, \;\Pi_2:\{1,3\}\cup\{2,4\},\; \Pi_3:\{1,4\}\cup\{2,3\}$$
and the set $\{\Pi_1, \Pi_2, \Pi_3\}$ of order 3 is permuted along with $S_4$.
so far, so good, but the author then claims that it is a homomorphism. explaining that:
If $p$ and $q$ are elements of $S_4$, the product $pq$ is the composed
permutation $p \circ q$, and the action of $pq$ on the set $\{\Pi_1, \Pi_2, \Pi_3\}$is the composition of the actions of $q$ and $p$.
Therefore $\varphi(pq) = \varphi(p)\varphi(q)$, and $\varphi$ is a
homomorphism.
And yet I struggle to come up with a formal proof, only to end up with examples: take two elements from $S_4$, say (1234) and (24), then
$$\varphi((1234))=(\Pi_1\Pi_3), \varphi((24))=(\Pi_1\Pi_3)$$
$$\varphi((1234))\varphi((24))=(\Pi_1\Pi_3)(\Pi_1\Pi_3)=e$$
on the other hand:
$$(1234)(24)=(12)(34)$$
$$\varphi((1234)(24))=\varphi((12)(34))=e$$
hence
$$\varphi((1234))\varphi((24))=\varphi((1234)(24))$$
therefore $\varphi$ is etc.
which I find rather unsound. I'd like to know if Artin's intuitive proof could be made formal, thanks!
Best Answer
As pointed out in the comments, I'll rather take $\Pi_1:=\{\{1,2\},\{3,4\}\}$, etc., and $X:=\{\Pi_i,i=1,2,3\}$. The map $S_4\times X\to X$, defined by: $$\sigma\cdot\{\{i,j\},\{k,l\}\}:=\{\{\sigma(i),\sigma(j)\},\{\sigma(k),\sigma(l)\}\}$$ fulfils the axioms of a group action, and this action is equivalent to the homomorphism $\varphi\colon S_4\to S_X$ defined by: $$\varphi_\sigma(\Pi_i):=\sigma\cdot\Pi_i$$ But $|X|=3$, and hence $S_X\cong S_3$.