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.

  1. 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., J. Integer Seq. 16, No. 5, Article 13.5.8, 23 p. (2013). ZBL1288.20002.

  • $\begingroup$ I am not completely sure what you mean by a direct product of two actions. Certainly, if groups $G$ and $H$ act on sets $X$ and $Y$, then there is an induced action of $G \times H$ and $X \times Y$. But in your situation does that mean that if $H$ was the direct product of actions of $H_1$ and $H_2$, then $H = H_1 \times H_2$? (We can assume that the actions of $G$ and $H$ are faithful.) $\endgroup$
    – Derek Holt
    Commented Apr 25, 2019 at 10:27
  • $\begingroup$ @DerekHolt: Yes, that is what I meant to write. The wording is from oeis.org/A005226. $\endgroup$ Commented Apr 25, 2019 at 11:08
  • $\begingroup$ I think it must be very unusual for a wreath product to decompose nontrivially as a direct product, but of course that would need proof. $\endgroup$
    – Derek Holt
    Commented Apr 25, 2019 at 12:17
  • $\begingroup$ @DerekHolt: well, it evidently decomposes whenever $H$ is a direct product. I am quite sure it does not decompose otherwise (provided $H$ is nontrivial). $\endgroup$ Commented Apr 25, 2019 at 13:19
  • 2
    $\begingroup$ I don't believe it does decompose even when $H$ does. Let $G = \langle (1,2) \rangle$, and $H = \langle (1,2)(3,4),(1,3)(2,4) \rangle$. Then $G \wr H$ is a group of order $2^6$ that does not deompose as a direct product. $\endgroup$
    – Derek Holt
    Commented Apr 25, 2019 at 14:08

1 Answer 1


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 $Y_i$.

  • $\begingroup$ Follow up question: is "connected permutation group" standard terminology, or is this a concept which doesn't arise too frequently? $\endgroup$ Commented Apr 26, 2019 at 21:41

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