When taking the fixed points commutes with taking the orbits

Let $G$ and $H$ be groups, both acting on a set $X$ on the left, in such a way that the two actions commute. (Equivalently, let $G \times H$ act on $X$.)

The set $\text{Fix}_H(X)$ of $H$-fixed points carries a $G$-action, and we can then take the set $\text{Fix}_H(X)/G$ of orbits. But also, the set $X/G$ of $G$-orbits carries an $H$-action, and we can take the set $\text{Fix}_H(X/G)$ of fixed points. There is a canonical map $$\lambda\colon \text{Fix}_H(X)/G \to \text{Fix}_H(X/G),$$ which is easily seen to be injective.

It's a fact that if $G$ and $H$ are finite groups with coprime orders, this map $\lambda$ is bijective. So then, $$\text{Fix}_H(X)/G \cong \text{Fix}_H(X/G).$$ (A bit more generally, this is true whenever $G$ and $H$ are possibly-infinite groups with finite coprime exponents.) The proof isn't hard.

I only discovered this yesterday, but I guess it's well-known — maybe as a special case of something more general. Can anyone give me a reference?

Update Perhaps I should have explained the context. In category theory, limits commute with limits and colimits commute with colimits, but limits do not usually commute with colimits. There are various theorems giving restricted conditions under which limits do commute with colimits. This result about group actions, which Peter Johnstone and I came upon, gives a new (?) such theorem. That's why I'm after a reference. For more explanation, see this $n$-Category Café post.

• This is true for all actions if and only if no nontrivial subquotient of $H$ is isomorphic to a subgroup of $G$. So there are not many more cases. No more for finite groups. For infinite groups it is as you say, except I guess one can have infinite exponents if the other is trivial. Dec 18 '13 at 2:50
• I would say that the proof is sufficiently straightforward as not to require a reference. Dec 18 '13 at 8:41
• Thanks, both. Derek, I agree that the proof is short, but I'd like to find somewhere that this is mentioned in print. E.g. you sound like you already knew this... Dec 18 '13 at 9:30
• perhaps some thing is missing in your question? assume that G=H act on a set with only one fixed point. then the left side of your equality is a single point while the right side is the whole set! Am I missing some thing? Dec 18 '13 at 9:43
• Assume that $G=H=\mathbb{R}$ in my previous comment. Dec 18 '13 at 9:53

To see when this is possible, we can consider each orbit of $G \times H$ separately. An orbit of $G \times H$ corresponds to a subgroup of $G \times H$.

A subgroup of $G \times H$ corresponds to a triple of $A,B,C,D$ where $A$ is a normal subgroup of $B$ a subgroup of $G$, $C$ is a normal subgroup of $D$ a subgroup of $H$, plus an isomorphism $B/A \cong D/C$. The subgroup corresponding to $A, B, C,D,\cong$ consists of pairs $(g,h) \in G \times H$ where $g \in B$, $h \in D$, and the class of $g$ in $B/A$ is sent by the isomorphism to the class of $h$ in $D/C$. One can also reverse this process.

$X/G$ is a single orbit of $H$. It is the orbit corresponding to the subgroup $D$, that is, the action of $H$ on $H/D$. So $X/G$ has an $H$-fixed point if $D = H$.

On the other hand $Fix_H(X)$ is nonempty if and only if $C= G$. In this case, the map is surjective. So my comment was off slightly - $H$ and $G$ have this property if and only if no nontrivial quotient of $H$ is a subquotient of $G$. This condition is quite messy in general, and there are lots of examples, like $G$ finite and $H$ not residually finite.

But this is certainly a general result that your statements follow from.

• Thanks very much, Will. I'm formally accepting this as an answer because it's so satisfying. On the other hand, my original question still isn't answered: where can I find any mention of the coprime-order result (or a generalization thereof) in print? Dec 21 '13 at 2:05
• Dear Will, Instead of "not residually finite", do you perhaps mean "not having any finite quotients"? (e.g. H infinite simple) Dec 21 '13 at 4:41
• I like your DNA-like "a triple of $A$, $B$, $C$, $D$". ("Our 3 subgroups are $A$, $B$, $C$, and $D$. I'll come in again.") It looks like you really mean "a quintuple $(A, B, C, D, i)$"? Aug 18 '17 at 21:46
• @LSpice I think I meant to write "tuple". Aug 19 '17 at 8:34