Timeline for Is a free group a product of f.g subgroups of infinite index?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 19, 2014 at 14:47 | vote | accept | Pablo | ||
| Dec 18, 2014 at 20:33 | answer | added | HJRW | timeline score: 7 | |
| Dec 18, 2014 at 18:00 | comment | added | Benjamin Steinberg | Can you not use Marshall Hall's theorem to reduce to the case where one of the subgroups is a free factor? If one of the subgroups is a proper free factor then you can use the finiteness of the Stallings graph of the other to show you can't reach every vertex in the Schreier graph using generators from the first factor. | |
| Dec 18, 2014 at 16:53 | comment | added | Ashot Minasyan | There are lots of ways to prove that $HK$ is a proper subset of $F$. For example, Theorem 1.1 in arxiv.org/abs/1308.3192 claims that there is a finite index subgroup $K'$ of $K$, s.t. $M:=\langle H,K'\rangle$ still has infinite index in $F$. Then, clearly $HK=\cup_{i=1}^k HK'g_i \subseteq \cup_{i=1}^k Mg_i \neq F$ (where $K=\sqcup_{i=1}^k K'g_i$). | |
| Dec 18, 2014 at 10:52 | comment | added | HJRW | I'm not sure about the votes to close - this seems a reasonable question to me. Anyway, the answer is 'no'. I'll try to write down an answer when I have time. | |
| Dec 18, 2014 at 8:15 | review | Close votes | |||
| Dec 18, 2014 at 11:45 | |||||
| Dec 17, 2014 at 21:40 | history | asked | Pablo | CC BY-SA 3.0 |