Timeline for A non-trivial property of all groups
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2011 at 20:37 | comment | added | user6976 | Good! I guess, the question mathoverflow.net/questions/54921/… by Andreas Thom is what I really had to ask. | |
| Feb 9, 2011 at 20:35 | vote | accept | CommunityBot | moved from User.Id=6976 by developer User.Id=69903 | |
| Feb 9, 2011 at 20:17 | comment | added | Kevin Buzzard | Perhaps what is really going on is that if you allow $G$ to be finite, but let $S$ be a multi-set, then the criterion fails for $G=(Z/2Z)^2$ and $\epsilon=1/4$ say, because at least $1/4$ of the elements in $S$ will just be one element of $G$, and $G$ is not generated by one element. Now $G$ is finite so this isn't allowed---but now replace $(Z/2Z)^2$ with any infinite finitely-generated group which admits a surjection onto $(Z/2Z)^2$ and now you have a real counterexample via the same argument. | |
| Feb 9, 2011 at 19:43 | comment | added | Nick S | I think that the same idea works for $Z \times Z_n$ and $\epsilon < \frac{1}{n}$. Any set $S$ will have a subset with $\epsilon |S|$ elements which lies entirely in some $\Z \times {k}$. Then if $k=0$ clearly the set cannot generate, while if $k \neq 0$, the set cannot generate elements of the form $(Z \backslash \frac{n}{gcd (n,k)} Z )\times \{ 0 \}$. | |
| Feb 9, 2011 at 19:22 | history | answered | Stephen S | CC BY-SA 2.5 |