tag:blogger.com,1999:blog-2492990990386446733.post1797422156884562121..comments2015-12-21T10:56:28.061+00:00Comments on Proof Roof: Commuting probability of compact groupsSean Eberhardnoreply@blogger.comBlogger5125tag:blogger.com,1999:blog-2492990990386446733.post-5743065745025984482015-12-21T10:56:28.061+00:002015-12-21T10:56:28.061+00:00Being an open subgroup (and therefore in particula...Being an open subgroup (and therefore in particular a closed subgroup), $F$ contains the connected component of the identity, but $F$ itself need not be connected. The connected component of the identity might even be trivial.<br /><br />That being said, yes $F$ is an open normal subgroup and so $G/F$ is a finite group. Also if $G_0\subset F$ is the connected component of the identity in $G$ then $G_0$ is a normal subgroup and so $G/G_0$ is a (not necessarily finite) group.<br /><br />The finiteness of $[F,F]$ is equivalent to the BFC-ness of $F$. The left-to-right direction of this is easy: if $e,f\in F$ then $e^{-1} e^f \in [F,F]$, so $e$ has at most $|[F,F]|$ conjugates. The other direction is a theorem of Bernhard Neumann. In the compact case one can prove it as follows: Take a commutator $c = [e,f]=e^{-1}f^{-1}ef$. If one replaces $e$ with any element $e'$ of $C_F(f)e$, and then $f$ with any element $f'$ of $C_F(e')f$, then still $[e',f']=c$. If $F$ is $n$-BFC then $C_F(f)$ and $C_F(e')$ both have index at most $n$, so $\mu(\{(e,f)\in F^2 : [e,f]=c\}) \geq 1/n^2$, so there are at most $n^2$ commutators $c$. Now the task is reduced to showing that if $F$ has finitely many commutators then $[F,F]$ is finite, which is a classical theorem of Schur.Sean Eberhardhttps://www.blogger.com/profile/13157396039399316710noreply@blogger.comtag:blogger.com,1999:blog-2492990990386446733.post-51781718817952673372015-12-21T08:28:44.176+00:002015-12-21T08:28:44.176+00:00Again, thank you so much for your time and help!Again, thank you so much for your time and help!Dr. Avraham Goldsteinhttps://www.blogger.com/profile/11007802930054466285noreply@blogger.comtag:blogger.com,1999:blog-2492990990386446733.post-77239979082741744582015-12-21T08:28:21.096+00:002015-12-21T08:28:21.096+00:00Hello and thank you for your prompt and clear repl...Hello and thank you for your prompt and clear reply! Indeed, a very nice and useful Lemma.<br /><br />Actually, after carefully working-out all the steps of the proof, I realized that there are two other points there, which I wanted to understand. Could you, please, help me with them?<br /><br />First one: Since $X_n\cdot X_m$ is a subset of $X_{nm}$, it follows that $F$ is equal to some $X_k$ for some large-enough $k$. This is actually explicitly stated in your proof, when you say that $F$, which is FC-center, is also BFC-subgroup. But that implies that $F$ is also closed (again, this is stated explicitly in the proof, when you say that $F$ is compact). Thus, $F$ is the connected component of the identity in $G$ (which is a normal subgroup of G). Thus, the index $t$ of $F$ in $G$ is just the number of connected components of $G$. Moreover, the multiplication in $G$ defines a binary operation on the set of connected components of $G$, making it a finite group $B=G/F$. Am I right?<br /><br />Second one: why is $[F,F]$ finite? How does it follow from the facts, that $F$ is a BFC-group and that $F$ is compact?Dr. Avraham Goldsteinhttps://www.blogger.com/profile/11007802930054466285noreply@blogger.comtag:blogger.com,1999:blog-2492990990386446733.post-6219671368208626312015-12-20T22:01:43.199+00:002015-12-20T22:01:43.199+00:00
Hi, thanks for your comment! This is a useful le... <br />Hi, thanks for your comment! This is a useful lemma. Suppose $X$ is a symmetric subset of a group $G$ containing the identity, and suppose $X^{k+1}\neq X^k$. Pick $x\in X^{k+1}\setminus X^k$. Then $xX\subset X^{k+2}\setminus X^{k-1}$. Thus if $G$ has a Haar measure $\mu$ then $\mu(X^{k+2}\setminus X^{k-1})\geq \mu(X)$ whenever $X^{k+1}\neq X^k$. Thus if $X^{3m}\neq\langle X\rangle$ then $$\mu(G) \geq \sum_{k=1}^m \mu(X^{3k+2}\setminus X^{3k-1}) \geq m \mu(X).$$Sean Eberhardhttps://www.blogger.com/profile/13157396039399316710noreply@blogger.comtag:blogger.com,1999:blog-2492990990386446733.post-72321450565691886382015-12-20T18:37:17.520+00:002015-12-20T18:37:17.520+00:00Beautiful proof! One point, which is not clear to ...Beautiful proof! One point, which is not clear to me:<br /><br />"This implies that the group $H_n$ generated by $X_n$ is generated in at most $6\div\epsilon$ steps" - why?Dr. Avraham Goldsteinhttps://www.blogger.com/profile/11007802930054466285noreply@blogger.com