An equivalence relation on group actions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:40:19Z http://mathoverflow.net/feeds/question/110703 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110703/an-equivalence-relation-on-group-actions An equivalence relation on group actions David Feldman 2012-10-25T21:30:30Z 2012-10-27T12:28:04Z <p>Suppose a group $G$ acts faithfully on a set $X$, or equivalently, $G$ is a subgroup of ${\rm Sym}(X)$. </p> <p>By functoriality, $G$ acts on $P(X), P(P(X)), P(P(P(X))),$ etc. ($P(\cdot)$ means powerset.) Henceforth, I'll omit parentheses.</p> <p>One can recover $G$ from ${\rm Fix}_G(PPPX)$ because, for example, one can encode a well-ordering of $X$ as an element of $PPX$.</p> <p>Generally, one cannot recover $G$ from ${\rm Fix}_G(PPX)$. For example, the alternating group and symmetric groups on a finite set will give the same set of fixed elements.</p> <p>Write $G\sim H$ if both groups act on $X$ and ${\rm Fix}_G(PPX) = {\rm Fix}_H(PPX)$. Equivalently, $$\forall u,v \subset X (\exists g\in G, gu=v \leftrightarrow \exists h\in H, hu=v) \ .$$</p> <p><strong>Questions</strong> </p> <p>When does $G\sim H$ imply $G=H$? </p> <p>Are there nontrivial examples of $G\sim H$ for infinite $X$? for $X$ of any infinite cardinality? </p> <p>Is there a classification of such pairs for finite $X$? </p> <p>Does this phenomenon have a name? </p> <p>Can one always recover $G$ from ${\rm Fix}_G(PPPX)$ in <em>ZF</em>?</p> http://mathoverflow.net/questions/110703/an-equivalence-relation-on-group-actions/110786#110786 Answer by David Feldman for An equivalence relation on group actions David Feldman 2012-10-26T21:24:33Z 2012-10-26T21:24:33Z <p>Since posting my question, I've thought of a general way to get examples, which I'll merely illustrate. Take $X={\Bbb N}\times {\Bbb N}$, say. Consider for the moment the group of permutations, a wreath product, each having the form $$p((x,y))=(f(x),g(x,y))\ .$$ For $G$ further restrict allowing any permutation $f$ but making $y\mapsto g(x,y)$ always finitary alternating; for $H$ merely insist on having $y\mapsto g(x,y)$ finitary.</p> <p>Then $G\subset H$ properly, but they fix the same elements of $PPX$. $|G|=|H|=c$ and neither group is transitive on sets of a given cardinality and cocardinality.</p> <p>One could enlarge $G$ and $H$ by including permutations that agree with these up to finitely many elements or up to finitely many rows (first coordinate determines the row). Thus we get examples where the action doesn't preserve an equivalence relation.</p> <p>For a different example, we could make $f$ finitary alternating (for $G$) or merely finitary (for $H$) but put no restrictions on $g$.</p> http://mathoverflow.net/questions/110703/an-equivalence-relation-on-group-actions/110809#110809 Answer by Dima Pasechnik for An equivalence relation on group actions Dima Pasechnik 2012-10-27T05:02:52Z 2012-10-27T05:15:56Z <p>In the case of a finite group $G$, a lot can be said by looking at $Fix_G(PPX)$, as it includes information on primitivity of the action on $X$, etc. E.g. if $G$ is doubly transitive on $X$, one can list all the possible examples of $H$ equivalent to $G$ using the classification of finite simple groups. </p> <p>More generally, for primitive groups, one can use <a href="http://www.maths.qmul.ac.uk/~raw/talks_files/ONStalk.pdf" rel="nofollow">O'Nan-Scott theorem</a> to partition such groups into few relatively well-understood classes, and, hopefully, derive the list you are looking for. As a toy example, consider $G\cong S_5$ acting on the set $X$ of pairs of {1,...,5}. There are just two nontrivial invariant graphs on $X$, the Petersen graph, and its complement. $G$ is the automorphism group of the Petersen graph, thus $H$ must be a subgroup of $G$. It follows by inspection that the only $H\neq G$, $G$~$H$ is (<strong>EDIT</strong>: actually, it could be that this $H$ is distinguished from $G$ by other orbits on sets, this still needs to be checked!) the index 2 subgroup in $G$, isomorphic to $A_5$.</p> <p>For imprimitive groups, probably there is a reduction to the primitive case. (And needless to say, intranisitive case reduces to the transitive.) </p> http://mathoverflow.net/questions/110703/an-equivalence-relation-on-group-actions/110825#110825 Answer by Yves Cornulier for An equivalence relation on group actions Yves Cornulier 2012-10-27T12:28:04Z 2012-10-27T12:28:04Z <p>The group of pwop (piecewise order-preserving) permutations of $\mathbf{N}$ is equivalent to the full symmetric group (i.e., has the same orbits on the power set $2^\mathbf{N}$. These are permutations for which there's a finite partition such that on every component, the permutation is order-preserving. It is transitive on moieties (infinite subsets with infinite complement) and obviously is also transitive on finite subsets of given cardinality. You can google "transitive on moieties" to find more.</p>