# Primitive action of wreath product

I say in advance that I am really new to Group Theory, so if my question is trivial I apologize in advance.

Let $$A$$ and $$H$$ be groups and $$\Omega$$ be a $$H$$-set. In this set-up, we can define the wreath product $$G: =A\,\text{Wr}_\Omega H$$. If $$A$$ acts on set $$\Lambda$$, then we get a canonical action of $$G$$ on the set $$\Lambda^\Omega=\lbrace f\colon \Omega \longrightarrow \Lambda| \: f \textit{ is a function}\rbrace$$. Wikipedia calls this action the primitive action of the wreath product. My question is the following:

Q: Is the action above indeed primitive? If it is not, do we have to assume something more on the action of $$A$$ on $$\Lambda$$ ( for example primitivity) to ensure the primitivity of the action of the wreath product?

Any help or reference is well accepted. Thank you.

• It may not be primitive, as $A$ could be imprimitive (e.g. $A=C_4$). What about checking the O'Nan-Scott theorem? – LeechLattice Sep 17 '19 at 1:13
• Even if $A$ and $H$ are primitive, the wreath product may not be primitive: Let $A=C_3$ and $H=S_3$, both on $3$ points. The wreath product is a subgroup of $AGL(3,3)$ and preserves subspaces of the form $(a,b,c)+(1,1,1)t$ for $t\in \mathbb{Z}_3$. – LeechLattice Sep 17 '19 at 3:43
• Thanks for the very helpful comments. If I take the wreath product of $S_n$ and $S_k$ do you think the action on $\lbrace 1,\ldots,n\rbrace ^k$ is still not primitive? – Vincenzo Zaccaro Sep 17 '19 at 3:52
• It's an $A$-action on $\Lambda$ (cf the Wikipedia page), not an $H$-action. I corrected accordingly. – YCor Sep 17 '19 at 6:09
• This kind of action of a wreath product is often called "product action". In the finite setting, precise conditions for a subgroup of such a permutation group to be primitive are given as type III(b) in Liebeck–Praeger–Saxl. – Colin Reid Sep 17 '19 at 8:23

Just a remark : if $$A$$ and $$B$$ are non trivial finite $$p$$-groups, each acting faithfully as transitive permutation groups, then the action of $$A \wr B$$ is never primitive as a permutation action. For the transitive action of the wreath product is on at least $$p^{2}$$ points, so the point stabilizer in the action is not a maximal subgroup.