5
$\begingroup$

G is a group. For a subgroup H of G, note $[H]$ the class of subgroups which are conjugate to H.

Define the binary relation: $[H] \leq [K]$ iff $H_0 \subset K_0$ for some $H_0 \in [H]$ and $K_0 \in [K]$

It is easy to see that this relation is reflexive and transitive. But how to show that it is anti-symmetric?

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?

$\endgroup$
  • $\begingroup$ This preorder is Green's J-order on the idempotents of Scheins inverse semigroup of cosets of G. $\endgroup$ – Benjamin Steinberg Apr 26 '12 at 0:30
6
$\begingroup$

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 $K$.

$\endgroup$
5
$\begingroup$

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 $bKb^{-1} < H$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.