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.

$\endgroup$
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. $\endgroup$ Commented Jun 11 at 2:32
  • $\begingroup$ I am confused about the case where only $H$ is the trivial group. Then $G\wr 1 \cong G\cong 1\wr G$. $\endgroup$ 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

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 $n=m^3$.

$\endgroup$
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. $\endgroup$ 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. $\endgroup$ Commented Jun 10 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.