Timeline for Is the free product of arbitrarily many copies of `${\mathbb{Z}}$` and `${\mathbb{Z}}/2$` residually nilpotent?
Current License: CC BY-SA 3.0
11 events
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| Oct 11, 2011 at 3:09 | vote | accept | CommunityBot | ||
| Oct 10, 2011 at 10:07 | comment | added | user6976 | Yes, the free product of $F*G$ where $F$ is free and $G$ is a free product of ${\mathbb Z}/2{\mathbb Z}$ is residually 2-group, hence residually nilpotent. For finitely many factors it follows from Gruenberg's result I mentioned in my answer. For infinitely many factors it follows from the standard nonsense. Take any word in generators, then apply Gruenberg's result to the subgroup generated by only the letters that appear in that word. | |
| Oct 10, 2011 at 8:47 | history | edited | user2529 | CC BY-SA 3.0 |
added 502 characters in body; edited title
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| Oct 10, 2011 at 6:36 | answer | added | James | timeline score: 2 | |
| Oct 9, 2011 at 21:21 | comment | added | user6976 | In fact the free product of abelian groups is always residually solvable because the derived subgroup is free. | |
| Oct 9, 2011 at 21:14 | comment | added | user6976 | They are probably residually solvable. | |
| Oct 9, 2011 at 20:23 | comment | added | Andreas Thom | Maybe the right thing to expect is that any such free product is residually elementary amenable. | |
| Oct 9, 2011 at 9:53 | answer | added | user6976 | timeline score: 17 | |
| Oct 9, 2011 at 9:46 | comment | added | Torsten Ekedahl | Indeed, let $G$ be the free quotient of $\mathbb Z/2$ and $\mathbb Z/3$. Any nilpotent quotient of $G$ is torsion and hence finite thus a product of $p$-groups for primes $p$. It is also generated by one element of order dividing $2$ and one of dividing $3$. This forces the quotient to be cyclic. | |
| Oct 9, 2011 at 6:51 | comment | added | Will Sawin | Intuitively, I wouldn't think it would be true for, say, 2 cyclic groups. Is it, in fact, true for such simple examples? | |
| Oct 9, 2011 at 6:20 | history | asked | user2529 | CC BY-SA 3.0 |