Isomorphisms of two subgroups in $S_6$ Is there any nice group action to see why groups $S_2 \times S_4$ and $S_2 \wr S_3$ of order 48 are isomorphic? Or is this "just" an abstract property which becomes invisible when we switch to permutation groups? 
 A: It's normal that it's not so visible since they are not isomorphic as permutation groups (one is transitive and not the other). 
You can still view it as follows: first, you view $S_2\wr S_3$ as the stabilizer of some unordered partition $\mathcal{P}$ of the 6-element set $X_6$ into 3 2-element blocks. Then you consider the set $K$ of unordered partitions $\mathcal{Q}$ of $X_6$ into 2 3-elements sets that are "orthogonal" to $\mathcal{P}$, in the sense that for every $(P,Q)\in\mathcal{P}\times\mathcal{Q}$, $P\cap Q$ is a singleton. Then you can see that $K$ has cardinal 4. Since $S_2\wr S_3$ acts on $K$, this yields a homomorphism $S_2\wr S_3$ into $\mathfrak{S}(K)\simeq S_4$. On the other hand, the signature map yields a homomorphism $S_2\wr S_3\to S_2$. Finally, combining these two, you can check that the resulting homomophism $S_2\wr S_3\to S_2\times S_4$ is injective; hence by cardinality it's an isomorphism. 
A: If $G = S_2 \wr S_3$, then $Z(G) = \langle (1,2)(3,4)(5,6) \rangle$, which clearly does not intersect $A_6$, so $G = Z(G) \times (G \cap A_6)$. It is not too hard to show that $G \cap A_6 \cong S_4$ - for example, it must have $4$ Sylow $43$-subgroups, and you could consider the conjugation action on them.
