Let G be a finite group with subgroups H and K. Then the set of not necessarily equivariant maps from G/H to G/K is itself a finite G-set under the conjugation action. Is there a good description of it?
2
-
2$\begingroup$ The question is a bit vague. What kind of 'good description' do you want / expect? $\endgroup$– Gerrit BegherCommented Jan 25, 2017 at 11:52
-
$\begingroup$ A reasonable description of a finite $G$-set is its decomposition into transitive $G$-sets? Still the question is quite broad (e.g., the case $H=K=1$, i.e. describing $G^G$ as $G$-set, is not clear-cut, at least for me). $\endgroup$– YCorCommented Jan 25, 2017 at 17:47
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Even though the following gives only a minor generalisation, here's one perspective:
Given any two sets with an action of a given group (the sets and the group don't even need to be finite) there is an induced action on the set of maps:
$$G\times \mathrm{Set}(X,Y)\to \mathrm{Set}(X,Y)$$ $$(g, f)\mapsto (x\mapsto (g^{-1}).f(g.x))$$
The fixpoints of this action are then precisely the equivariant maps.
answered Jan 25, 2017 at 12:01