Timeline for Frobenius complement/kernel of an infinite group
Current License: CC BY-SA 3.0
19 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Feb 8, 2016 at 13:06 | comment | added | HJRW | An infinite-group-theorist's perspective: Frobenius' theorem asserts that a malnormal subgroup of a finite group is necessarily a retract. This is a minor miracle, and no such phenomenon is true in the infinite case, as the many examples given here show. Most of them are far more complicated than necessary (Tarski monsters are massive overkill, as are super-rigidity style results). @DerekHolt's examples are good: the commutator of two generators in a free group generates a malnormal subgroup that isn't a retract, and if you want a finite order example, look at a hyperbolic triangle group. | |
| Jan 12, 2016 at 16:55 | comment | added | Derek Holt | I think there are simpler examples, such as any of the subgroups $\langle x \rangle$, $\langle y \rangle$ or $\langle xy \rangle$ in the Hurwitz gropup $\langle x,y \mid x^2=y^3=(xy)^7=1 \rangle$. | |
| Jan 12, 2016 at 16:42 | comment | added | Geoff Robinson | I think that the Tarski monster examples I gave still stand in the case that there is more than one conjugacy class of subgroups of order $p$. For if $G$ is a infinite group in which every non-identity proper subgroup has order $p$, then $G = \langle x,y \rangle$ whenever $\langle x \rangle \neq \langle y \rangle$ if $x$ and $y$ have order $p$. Taking $H = \langle x \rangle$ for any non-identity element of order $p$, then $\langle H,g^{-1}Hg \rangle = G$ for any $g \in G \backslash H$. | |
| Jan 12, 2016 at 16:41 | comment | added | Derek Holt | I think Geoff Robinson's example of a Tarski Monster is probably still a counterexample to the question with your new assumption that the conjugates of $H$ generate $G$ (which certainly holds in a Tarski Monster). But to provide a counterexample, we need one in which not all subgroups of order $p$ are conjugate. One easily finds statements of the form "there even exist Tarski Monsters in which all subgroups of order $p$ are conjugate" but frustratingly I could not find any definitve statement that there are some that do not have that property! But I expect an expert on the topic would know. | |
| Jan 12, 2016 at 16:25 | history | edited | Denis Serre | CC BY-SA 3.0 |
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| Jan 4, 2016 at 19:54 | answer | added | Derek Holt | timeline score: 5 | |
| Jan 4, 2016 at 16:25 | comment | added | Thomas | No conjugate of $ba$ is in $A$. For instance because it has infinite order. | |
| Jan 4, 2016 at 14:56 | answer | added | Venkataramana | timeline score: 3 | |
| Jan 4, 2016 at 13:48 | answer | added | Geoff Robinson | timeline score: 2 | |
| Jan 4, 2016 at 13:07 | comment | added | Denis Serre | @Thomas. A sentence such as "$c$ is not conjugate in $A$" is meaningless. Only a pair of elements can be conjugate. | |
| Jan 4, 2016 at 11:16 | comment | added | Thomas | Sorry, let $b\in B, a\in A$ two non trivial elements. Then $ba$ is not conjugate in $A$, neither $b^{-1}$, but the product is $a$. | |
| Jan 4, 2016 at 10:19 | comment | added | Denis Serre | @Thomas. What do you make of this ? Which conclusion can you draw about $N$ ? | |
| Jan 4, 2016 at 9:13 | comment | added | Thomas | Take the free product of two finite groups $A*B$ . Certainly $gAg^{-1}\cap A$ is reduced to $1$. But $abab^{-1}$ is of infinite order and is therefore not conjugate in $A$. | |
| Jan 4, 2016 at 8:57 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 122 characters in body; edited title
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| Jan 4, 2016 at 8:05 | comment | added | Yemon Choi | Might this arxiv.org/abs/1104.3065 be relevant? If I recall correctly it has an example of a malnormal H where the Frobenius complement is not a subgroup (these examples predate the paper) | |
| Jan 4, 2016 at 7:47 | comment | added | abx | If $G$ is commutative the hypothesis implies $H=\{0\} $ or $G$, hence the conclusion is trivially satisfied. | |
| Jan 4, 2016 at 7:36 | comment | added | Włodzimierz Holsztyński | I must be something missing. Let $(F\ +\ 0)$ be an arbitrary commutative group such that $|F|>1$. Let $G:=F\times F$ and $H:=\{(x\ x)\in G: x\in F\}$. Then for $0\ne a\in F$ we get: $(a\ 0)\ (0\ a)\in N$ but $(a\ 0)+(0\ a)=(a\ a)\notin N$ hence $N$ is not a group. | |
| Jan 4, 2016 at 7:03 | history | asked | Denis Serre | CC BY-SA 3.0 |