partial order on conjugate classes of subgroups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T10:40:37Z http://mathoverflow.net/feeds/question/95203 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95203/partial-order-on-conjugate-classes-of-subgroups partial order on conjugate classes of subgroups Linxiao 2012-04-25T22:00:38Z 2012-04-26T01:32:27Z <p>G is a group. For a subgroup H of G, note $[H]$ the class of subgroups which are conjugate to H.</p> <p>Define the binary relation: $[H] \leq [K]$ iff $H_0 \subset K_0$ for some $H_0 \in [H]$ and $K_0 \in [K]$</p> <p>It is easy to see that this relation is reflexive and transitive. But how to show that it is anti-symmetric?</p> <p>P.S. In a book, the author claims that this relation defines a partial order on the classes of conjugate subgroups in a context where G is a compact Lie group. But I don't think the compact Lie group condition be essential, right?</p> http://mathoverflow.net/questions/95203/partial-order-on-conjugate-classes-of-subgroups/95206#95206 Answer by Tom Goodwillie for partial order on conjugate classes of subgroups Tom Goodwillie 2012-04-25T22:12:11Z 2012-04-25T22:12:11Z <p>The compactness is essential. Let $G$ be the group of conformal automorphisms of $\mathbb R^2$. Let $H$ be the group of translations $(x,y)\mapsto (x+m,x+n)$, $m$ and $n$ integers. Let $K$ be the group of translations $(x,y)\mapsto (x+m,x+2n)$, $m$ and $n$ integers. $K$ is a subgroup of $H$, and a conjugate of $H$ is a subgroup of $K$, but $H$ is not conjugate to $K$.</p> http://mathoverflow.net/questions/95203/partial-order-on-conjugate-classes-of-subgroups/95214#95214 Answer by Mark Sapir for partial order on conjugate classes of subgroups Mark Sapir 2012-04-26T01:32:27Z 2012-04-26T01:32:27Z <p>Here is a countable example. Take the Baumslag-Solitar group $\langle a, b \mid bab^{-1}=a^{4}\rangle$. Let $K=\langle a\rangle$, $H=\langle a^2\rangle$. Then $[K]\ne [H]$ (the standard properties of HNN extensions, see the book of Lyndon and Schupp), but $K > H$ while $bKb^{-1} &lt; H$. </p>