Actions of $Z_n$ and actions of $Z_{n-1}$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:16:50Z http://mathoverflow.net/feeds/question/61081 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61081/actions-of-z-n-and-actions-of-z-n-1 Actions of $Z_n$ and actions of $Z_{n-1}$ Gejza Jenča 2011-04-08T16:53:05Z 2011-04-20T11:59:30Z <p>I am playing with some questions concerning connections between certain poset partitions and their linear extensions. This is not my usual playground, I just happened to stumble upon something.</p> <p>When I was playing with these things, I came up with a very simple construction that I cannot find anywhere.I suspect that this construction is a minor particular case of something well known and much more general. It reminds of an equivariant map, but the group is not fixed here. </p> <p>Let $n>2$, let $X$ be a finite set of cardinality $n$. Let $\oplus_n$ be an action of $Z_n$ on $X$ such that the action of $1$ is a cycle of length $n$. Pick $x\in X$ and let $\pi$ be a partition of the set $X$ such that $\{x,x\oplus_n 1\}$ is the only nonsingleton block of $\pi$.</p> <p>In a very simple way, this gives rise to an action $\oplus_{n-1}$ of $Z_{n-1}$ on $\pi$ by ,,squeezing the action'' at $x$:</p> <p>Put</p> <ul> <li>$\{x,x\oplus_n 1\}\oplus_{n-1} 1=\{x\oplus_n 2\}$</li> <li>$\{x\oplus_n(n-1)\}\oplus_{n-1} 1=\{x,x\oplus_n 1\}$</li> <li>$\{y\}\oplus_{n-1} 1=\{y\oplus_n 1\}$ for any $y\neq x\oplus_n(n-1)$.</li> </ul> <p>It can be visualized in a simple way by a digraph construction: if we identify the action of $1$ on $X$ with an oriented cycle, this construction corresponds to a contraction of an edge. </p> <p>Has anyone seen this construction before? Is there any name for it?</p> http://mathoverflow.net/questions/61081/actions-of-z-n-and-actions-of-z-n-1/62399#62399 Answer by Tom De Medts for Actions of $Z_n$ and actions of $Z_{n-1}$ Tom De Medts 2011-04-20T11:59:30Z 2011-04-20T11:59:30Z <p>I might be missing something, but it seems to me that there is not much going on in your construction. In fact, your original action of $Z_n$ on $X$ does nothing more than putting a cyclic ordering on your set $X$, and your complicated-looking construction creates a new cyclically ordered set $Y$ obtained from $X$ by removing one element (and retaining the ordering).</p> <p>In any case, the construction is rather artificial from a group-theoretical point of view, so I don't expect a more general construction (e.g. for arbitrary group actions) in that sense. It is faintly reminiscent of the construction of a primitive action from an imprimitive permutation group by considering the induced action on the blocks of imprimitivity.</p>