# Commutativity of the wreath product

(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.

• "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).
– YCor
Commented Jun 10 at 22:25
• Conjugate in what? Commented Jun 10 at 23:10
• @LSpice: as subgroups of $\mathfrak{S}_{mn}$, I would imagine based on the surrounding text. Commented Jun 11 at 2:32
• I am confused about the case where only $H$ is the trivial group. Then $G\wr 1 \cong G\cong 1\wr G$. Commented Jun 11 at 11:11
• @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. Commented Jun 11 at 15:22

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 $$n=m^3$$.
• 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. Commented Jun 10 at 21:57