Let $G$ and $H$ be subgroups of the symmetric groups $\mathfrak S\_m$ and $\mathfrak S\_n$. Assume that $n>1$ and that $H$ is a 'connected' permutation group, that is, there is no non-trivial $H$-stable subset $X\subset [n]$ (with $[n]=\{1,\dots,n\}$) such that $H$ is isomorphic to the direct product of the image of its action on $X$ and the image of its action on $[n]\setminus X$.


For example, the subgroup of $\mathfrak S\_5$ generated by $(1,2)(3,4,5)$ is not connected, because it is the direct product of $\langle(1,2)\rangle\lt\mathfrak S\_2$ and $\langle(3,4,5)\rangle\lt\mathfrak S\_{\{3,4,5\}}$.


This terminology is from <http://oeis.org/A005226> and [1].



> 
> 1. Question: Is there a more standard name for this property?
> 
> 
> 


My original question concerns the observation that, apparently, $G\wr H = G^n \rtimes H$ is a connected subgroup of $\mathfrak S\_{mn}$, provided that $H$ is connected. Note that, evidently, if $H$ is not connected, $G\wr H$ is not connected either.


Perhaps this appears more natural when using the language of combinatorial species: suppose that $\mathcal G$ is a molecular species (that is, cannot be written as a sum) and $\mathcal H$ is an atomic species (so it is molecular and additionally cannot be written as a product), and $\mathcal H\neq \mathcal X$. Then $\mathcal H\circ \mathcal G$ is atomic.


It seems to me that this has been overlooked in the literature on species. I admit that I did not try to prove it yet - I am mostly interested in a reference.



> 
> 2. Question: Is this known in terms of group actions?
> 
> 
> 


[1] *Naughton, L.; Pfeiffer, G.*, [**Integer sequences realized by the subgroup pattern of the symmetric group.**](http://www.emis.de/journals/JIS/VOL16/Naughton/naughton2.html), J. Integer Seq. 16, No. 5, Article 13.5.8, 23 p. (2013). [ZBL1288.20002](https://zbmath.org/?q=an:1288.20002).



Connected permutation groups and wreath product
I am afraid that I would have no idea where to look for a reference for this statement, but here is a very rough sketch proof. I can fill in details if necessary.


Let $A\_1,\ldots,A\_s$ be the orbits of $G$ on $[m]$ and $B\_1,\ldots,B\_t$ the orbits of $H$ on $[n]$. We can assume that each $|B\_j| > 1$, since otherwise the action of $H$ would be disconnected.


Then the orbits of $W := G \wr H$ on $[mn]$ can be labelled $C\_{ij}$, $1 \le i \le s$, $1 \le j \le t$, where each $C\_{ij}$ is a union of $|B\_j|$ orbits of size $|A\_i|$ of the base group $G^n$ of $W$, and the components of $G^n$ are acting as in the action of $G$ on $A\_i$ on these orbits, and the induced action of $H$ on the set of orbits is the same as its action on $B\_j$.


Suppose that the action of $W$ is disconnected. Then $[mn] = X\_1 \cup X\_2$ with $W \cong W\_1 \times W\_2$, and the action induced from product action sof $W\_i$ on $X\_i$.


Then $X\_1$ and $X\_2$ are unions of some of the orbits $C\_{ij}$. Suppose that $C\_{11} \in X\_1$. The key to the proof is that $C\_{i1}$ must lie in $X\_1$ for all $i$. To see this, note that if say $C\_{2j} \in X\_2$, then the pointwise stabilizer of $X\_1$ and hence of $C\_{11}$ must act transitively on $C\_{21}$. But we are assuming that $|B\_1|>1$, and the action of $H$ on the base group orbits in $C\_{11}$ is the same as its action on those of $C\_{21}$ (i.e. action of $H$ on $B\_1$) and so the pointwise stabilizer of $C\_{11}$ must stabilize each orbit of the base group within $C\_{21}$ and hence it cannot act transitively $C\_{21}$.


So there is a partition of $[n]$ into two sets $Y\_1$ and $Y\_2$ such that, for $k=1,2$, $X\_k$ is the union of all orbits $C\_{ij}$ with $j \in Y\_k$. Now it is clear by looking at the induced action of $H$ on the orbits of the base group that the decomposition $W = W\_1 \times W\_2$ induces a decomposition $H = H\_1 \times H\_2$ with $H\_i$ acting on $Y\_i$.


