4
$\begingroup$

(Cross-posted from MSE, with isomorphism replaced by conjugate: https://math.stackexchange.com/questions/4928697/commutativity-of-the-wreath-product?noredirect=1#comment10531931_4928697 )

Let $G$ be a subgroup of the symmetric group $\mathfrak{S}_n$ and $H$ be a subgroup of $\mathfrak{S}_m$. Recall that the wreath product $G \wr H$ is the semi-direct product $G^m \rtimes H$, where $H$ acts on the direct product $G^m$ by permuting components. We can also consider $H \wr G$, the semi-direct product $H^n \rtimes G$. Both can be seen as subgroups of $\mathfrak{S}_{nm}$. For example, $\mathfrak{S}_m \wr \mathfrak{S}_n$ is the stabilizer of the set partition

$$\{ \{ 1, …, m\}, \{m+1, …, 2m\},…, \{(n-1)m + 1, …, nm\}\}.$$

My question is : when are $G \wr H$ and $H \wr G$ conjugated? My suspicion is that it is only the case when $m = n$ and $G$, $H$ are conjugated, or when $G$ and $H$ are both the trivial group. But I can't come up with a proof, nor with a counterexample.

Attempt : First, if $|G|^m|H| \neq |H|^n|G|$, it is obvious, because it is the cardinality of each group. Suppose it is equal, and that there exists an isomorphism $f : G \wr H \to H \wr G$. I suspect that if $f$ is injective for both $G$ and $H$ (considered as subgroups of $G \wr H$), then $|H \wr G|$ must be much bigger than $|G \wr H|$, which is a contradiction. But i don't know if this intuition is good, nor how to formalize it.

Improve this question
$\endgroup$
6
  • $\begingroup$ "when $G$ and $H$ are both the trivial group" is redundant (it is just when $n=m=1$, in which case they're obviously conjugated). $\endgroup$
    – YCor
    Commented Jun 10 at 22:25
  • $\begingroup$ Conjugate in what? $\endgroup$
    – LSpice
    Commented Jun 10 at 23:10
  • $\begingroup$ @LSpice: as subgroups of $\mathfrak{S}_{mn}$, I would imagine based on the surrounding text. $\endgroup$
    – Sam Hopkins
    Commented Jun 11 at 2:32
  • $\begingroup$ I am confused about the case where only $H$ is the trivial group. Then $G\wr 1 \cong G\cong 1\wr G$. $\endgroup$
    – HenrikRüping
    Commented Jun 11 at 11:11
  • $\begingroup$ @HenrikRüping, re, $G \wr 1$ is isomorphic to $G^m$, not naturally to $G$. @‍SamHopkins, re, thanks; I seem to have skimmed right over that sentence. $\endgroup$
    – LSpice
    Commented Jun 11 at 15:22

1 Answer 1

Reset to default
13
$\begingroup$

If $G=\mathfrak{S}_m$ and $H=\mathfrak{S}_m\wr\mathfrak{S}_m$ then the associativity of the wreath product construction shows that $G\wr H$ and $H\wr G$ are isomorphic as permutation groups, and hence conjugate in $\mathfrak{S}_{n}$ with $n=m^3$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ You can obviously make examples like this with $G$ and $H$ any number of iterated wreath powers of the same group. I don't know whether these are the only examples. $\endgroup$
    – Dave Benson
    Commented Jun 10 at 21:57
  • 9
    $\begingroup$ I suspect that there is a uniqueness of factorisation theorem for finite permutation groups as an iterated wreath product of wreath indecomposable ones, in which case these would be the only examples. A five minute search did not reveal such a theorem in the literature, but it actually doesn't sound that hard to prove, so I suspect it's there somewhere. $\endgroup$
    – Dave Benson
    Commented Jun 10 at 22:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .