From the chinese remainder theorem to products of transitive G-sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:07:31Z http://mathoverflow.net/feeds/question/81018 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81018/from-the-chinese-remainder-theorem-to-products-of-transitive-g-sets From the chinese remainder theorem to products of transitive G-sets David Roberts 2011-11-15T23:53:24Z 2011-11-16T00:07:56Z <p>Note: I am aware of the question <a href="http://mathoverflow.net/questions/18893/analog-to-the-chinese-remainder-theorem-in-groups-other-than-z-n" rel="nofollow">http://mathoverflow.net/questions/18893/analog-to-the-chinese-remainder-theorem-in-groups-other-than-z-n</a></p> <p>For an abelian group $A$, every transitive $A$-set $M$ is of course isomorphic, as an $A$-set, to a quotient group $A/H$, by picking a point $m\in M$ and letting $H = Stab(m)$. Note that stabiliser groups of different points are conjugate, hence equal.</p> <p>For a pair of transitive $A$-sets, $N,M$, their product $M\times N$ is an $A$-set by the diagonal action $(m,n) \stackrel{a}{\mapsto} (am,an)$. This is not in general a transitive $A$-set, but is the disjoint union of transitive $A$-sets. An easy result is that </p> <p>$$Stab(m,n) = Stab(m)\cap Stab(n)$$</p> <p>The Chinese remainder theorem is precisely the statement</p> <p>$$\mathbb{Z}/(k)\times\mathbb{Z}/(l) \simeq \mathbb{Z}/((k)\cap (l)) \simeq \mathbb{Z}/(kl)$$</p> <p>for coprime $k$ and $l$ (and generalised to more than two factors) and so the product of transitive $\mathbb{Z}$-sets <em>is</em> a transitive $\mathbb{Z}$-set. There is also the version where one has to consider the gcd of the factors, and this is when things get a bit more interesting, and break away from the ring-theoretic approach - the disjoint union of rings is not a ring!</p> <p>Describing the structure of $A/H \times A/K$ for subgroups $H, K \lt A$ is only mildly interesting - it is a disjoint union of a number of copies of isomorphic transitive $A$-sets. This is not what my question is about, but there may be some combinatorial interest in the case of finite $A$. Consider instead a finite nonabelian group $G$ - not necessarily nilpotent! - and a pair of subgroups $H, K \lt G$. Fairly elementary observations show that </p> <p>$$G/H\cap K \hookrightarrow G/H\times H/K$$</p> <p>and that generally the orbits look like $G/(H\cap gKg^{-1})$. This seems to me to be an interesting combinatorial/group-theoretic problem, enumerating/classifying the various subgroups $H\cap gKg^{-1} \lt G$, and the number of orbits in the product.</p> <p>My question is: has anyone done any work on something like this?</p> <hr> <p>Postscript: people know know me might wonder why I was thinking about this. Well, the general problem of determining the structure of the product $G/H\times H/K$ came up thinking about proper-class-sized $G$ with set-sized $G/H$, $G/K$. It quickly became apparent that this would be nontrivial even for finite $G$!</p> http://mathoverflow.net/questions/81018/from-the-chinese-remainder-theorem-to-products-of-transitive-g-sets/81020#81020 Answer by Andreas Blass for From the chinese remainder theorem to products of transitive G-sets Andreas Blass 2011-11-16T00:07:56Z 2011-11-16T00:07:56Z <p>There has been a good deal of work on this, under the heading of "Burnside rings". The Burnside ring of a finite group $G$ is the Grothendieck ring obtained from the finite $G$-sets with the operations of disjoint union and cartesian product. One of the easy ingredients of the theory is Burnside's notion of the <em>mark</em> of a subgroup $H$ of $G$ in a $G$-set $X$; this is just the number of points in $X$ fixed by the action of all elements of $H$. The mark of any fixed $H$ gives a ring-homomorphism from the Burnside ring of $G$ to $\mathbb Z$, and these homomorphisms, for varying $H$, are jointly monic. These facts let you decompose products of transitive $G$-sets into their transitive parts easily, once you've calculated the <em>table of marks</em>, i.e., the matrix, with rows and columns indexed by (conjugacy classes of) subgroups of $G$, with the $(H,K)$-entry being the mark of $H$ in $G/K$. (For easy calculation, you also want the inverse of this matrix, but that's easy to find because the table of marks is triangular if you order the subgroups by size.)</p>