Reidemeister-Schreier Method for Finding Stabilizer of an Element in a Group Action - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T11:51:40Z http://mathoverflow.net/feeds/question/64572 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64572/reidemeister-schreier-method-for-finding-stabilizer-of-an-element-in-a-group-acti Reidemeister-Schreier Method for Finding Stabilizer of an Element in a Group Action Larry 2011-05-11T06:24:55Z 2011-06-06T17:51:19Z <p>Hi, Everyone:</p> <p>I would appreciate some references for the version of Reidemeister-Schreier that is used to find the stabilizer of a point under a group action. The only refs. I have found are about Schreier-Sims method, but I have not been able to find anything on it.</p> <p>The version of R-S I know of allows us to find a presentation of a subgroup H of a group G, by using transversals, etc. </p> <p>I think there is a connection between the two, but I am not sure.</p> <p>Thanks in Advance.</p> <p>Sorry, I forgot to ask something important: I would like to know how the following process --the adaptation of R-S to group actions ( or maybe a version of Cayley graphs) produces a set of generators for the stabilizer of a fixed element sk , under a group action:</p> <p>We start with a group action HxS-->S (could also be a left action), and we are given the (finite) set {$h_1$,..,$h_n$} of generators for H; S is a finite set. We then define a graph G by:</p> <p>1)The vertices are the elements sj of S</p> <p>2)We join $s_i$ with $s_k$ with an edge labeled $h_j$ , if $h_j$.$s_i$=$s_k$ , i.e., if the action of $h_j$ on $s_i$ results in sk.</p> <p>3) We construct a spanning-tree T for G, rooted at $s_k$ (the element of S being stabilized); I think it is clear that G is connected --|n|-connected, actually, where n is the size of the generating set for H (tho we mayhave loops) , to guarantee the existence of a spanning- tree. </p> <hr> <p>Claim: the edges in G-T generate the stabilizer Stb{{$s_k$})of $s_k$ under this action.</p> <p>Anyone have a suggestion for showing this? </p> <p>I don't remember the place where I read this, but I remember some related results:</p> <p>The background/context is a generalization of the fact that , given a group H and any subgroup</p> <p>H' of H , there is an action by H for which H' is the stabilizer. Specifically, this</p> <p>action is the "standard" action of H on H/H' (standard group quotient); we just define, for</p> <p>any h1H' on H/H' and h in H:</p> <p>h.(h1H' ) --> (h.h1)H'</p> <p>Then H' is the stabilizer of the coset eH'=H' .</p> <p>I think this is also related to the method for finding the fundamental group of a rooted</p> <p>connected graph G: we find a spanning-tree T. Then each edge e=(gi,gj) in G-T defines a non-</p> <p>trivial element of $Pi_1$(G): we start at , say, $g_i$ (which is in T, since T spans) , then we</p> <p>find the (unique; any two paths would form a loop in T) path $P_i$ in T from $g_i$ to the root g,</p> <p>and from g we find the unique path $P_j$ to $g_j$; the other vertex in e. Then the composition</p> <p>$P_i$$P_j$e forms a non-trivial loop in G. It is just a little more work to show that these</p> <p>edges freely generate the fundamental group.</p> <p>These are the results that were related to the issue of the stabilizer.</p> <p>Thanks for any Suggestions, Refs.</p> http://mathoverflow.net/questions/64572/reidemeister-schreier-method-for-finding-stabilizer-of-an-element-in-a-group-acti/67066#67066 Answer by Robert Bell for Reidemeister-Schreier Method for Finding Stabilizer of an Element in a Group Action Robert Bell 2011-06-06T17:51:19Z 2011-06-06T17:51:19Z <p>This is an answer to the OP's second question. Let $H$ be a group acting, say on the right, on a set $S$. Suppose that $H$ is generated by $X$, and let $G$ be the graph with vertex set $S$ and edges of the form $(s,sx)$, where $s \in S$ and $x \in X$. Then $G$ is connected if and only if $H$ acts transitively on $H$: a path from $s$ to $s'$ yields a sequence $x_1^{\epsilon_1}, \dots, x_n^{\epsilon_n}$ and the product of these represents an element $h$ such that $sh = s'$. </p> <p>Fix $s \in S$. Suppose $P$ is a collection of closed paths in $G$ based at $s$ such that every closed path based at $s$ is expressible as the concatenation of finitely many paths in $P$. To each path $p \in P$ you can associate a word $w_p$ over $X$ and, hence, a element of the group $H$. We have that $H_s = gp\langle \ w_p : p \in P \ \rangle$: every such word stabilizes $s$, and every element of the stabilizer, when expressed as a word in $X$, defines a closed path at $s$; the latter closed path is a concatenation of the subpaths which define closed paths at $s$ and so the word is expressible as a product of the $w_p$'s and their inverses.</p> <p>To find such a generating set for $H_s$, choose a spanning tree $T$ in $G$. For each edge $e$ in $G-T$, let $p_e$ be the unique reduced path in $T$ from $p$ to the initial vertex of $e$ followed by $e$ followed by the unique reduced path in $T$ from the terminal vertex of $e$ to $s$. Each such path $p_e$ defines a word over $X$ and these words generate $H_s$ as in the above proof. </p> <p>If $S$ is the set of right cosets of a subgroup $H' &lt; H$, then $T$ is essentially a Schreier transversal and the elements defined by the paths $p_e$ are the generators which are commonly denoted $\gamma(t,x) = tx\overline{tx}^{-1}$, where $F$ is free with basis $X$, $F'$ is the preimage of $H'$ under a map $F \to H$ which realizes $X$ as a generating set for $H$, $T$ is a Schreier transversal for $F' &lt; F$, and $\bar{w}F' = wF'$ is the unique coset representative $\overline{w} \in T$ for $w$.</p> <p>There is a detailed treatment in the textbook by Lyndon and Schupp.</p>