Timeline for "Big" groups $G$ with trivial $Out(G)$
Current License: CC BY-SA 3.0
24 events
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| Sep 22, 2017 at 14:37 | history | edited | Peter Heinig | CC BY-SA 3.0 |
I was requested to review an edit to this. While considering it, someone else approved this edit. Being interested in torsion in groups, I improved a few things: there seems to have been a non-working link in the post, which I deleted; also, someone added 'requires $n>1$', which I clarified further.
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| S Sep 22, 2017 at 14:25 | history | suggested | user1729 | CC BY-SA 3.0 |
Logan's result requires $n>1$
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| Sep 22, 2017 at 14:01 | review | Suggested edits | |||
| S Sep 22, 2017 at 14:25 | |||||
| Apr 10, 2017 at 14:21 | comment | added | HJRW | @ADL, thanks for the reference. It was clear to me that this should be true, but I got bogged down in the details of the proof. | |
| Apr 10, 2017 at 11:07 | comment | added | ADL | @HJRW Fischer, Karrass and Solitar proved (On one-relator groups having elements of finite order, PAMS 1972) that $G=\langle X; R^n\rangle$, $n>1$, is virtually free if and only if $G$ is the free product of a free group and a finite cyclic group. In fact, they prove that if $G$ has more than one end then it has infinitely many and is a free product of a nontrivial free group and an indecomposable one-relator group (which is decidable). It follows (perhaps not "clearly", but it does follow!) that if $G=\langle X; R\rangle$, $R$ not a proper power, is virtually free then it is free. | |
| Mar 22, 2017 at 12:19 | comment | added | YCor | @HJRW But you don't tell me why a non-free infinite (finite cyclic)-by-free group can't be a 1-relator group. | |
| Mar 22, 2017 at 8:13 | comment | added | HJRW | @YCor, yes, one also needs to invoke Stallings' theorem to say that every virtually free group is a graph of finite groups. The upshot of all this is that the only 2-generator, one-relator virtually free groups are free products of cyclic groups (one finite). | |
| Mar 22, 2017 at 1:16 | comment | added | Igor Rivin | @YCor The first statement is Theorem 4.13 in Magnus-Karrass-Solitar. | |
| Mar 21, 2017 at 23:57 | comment | added | YCor | OK thanks (that all torsion elements of 1-relator groups are conjugate to a power of the relator is probably standard but nontrivial; at least it's not a fact I know by heart!). On the other hand, to conclude you need to sort of use Stallings's theorem or at least some easier version thereof. Then there's slightly more than what you claim: you can have a semidirect product $C\rtimes F$ of a cyclic and a free group, which also has this property of having a single maximal finite subgroup up to conjugacy (but I don't know if this can be 1-related). | |
| Mar 21, 2017 at 21:54 | comment | added | HJRW | My reasoning was quite simple: the only torsion in a one-relator group comes from powers of the relator; in particular, there is only one conjugacy class of (maximal) finite subgroup, which is cyclic. Therefore, if it is virtually free, it must be a free product of an infinite cyclic group and a finite cyclic group. | |
| Mar 20, 2017 at 21:14 | comment | added | Igor Rivin | @YCor sorry typo in my long comment "every two generator subgroup of the torsion-free subgroup is free' | |
| Mar 20, 2017 at 21:07 | comment | added | Igor Rivin | By the way, @YiftachBarnea , this group IS word-hyperbolic. | |
| Mar 20, 2017 at 21:04 | comment | added | Igor Rivin | @YCor Magnus-Karrass and Solitar says that every one-relator group with torsion is virtually torsion-free. Pride (1977) says that a two generator subgroup of a one relator torsion group is either free or has torsion (actually, closer to Henry's statement, it is a free product of cyclic groups, or a one relator group with torsion). Which tells us that the torsion-free subgroup has the obvious property (that every two generator subgroup thereof has torsion), but since the finite index subgroup I conjecture need not be two-generator, I am not sure what Henry has in mind. | |
| Mar 20, 2017 at 20:44 | comment | added | YCor | Thanks Igor but this is not my question. I refer to HJRW's claim "if it were virtually free, it would be a free product of cyclic groups" | |
| Mar 20, 2017 at 20:39 | comment | added | Igor Rivin | @YCor It appears to be a long-standing conjecture that every one-relator group of the form $\langle a, b \left| R^n\right. \rangle$ is virtually free by cyclic, so my is not so far off the wisdom of the ancients (in this case, of the late lamented G Baumslag. | |
| Mar 20, 2017 at 20:15 | comment | added | YCor | @HJRW are you using some result on 1-relators that are virtually free? they're always free products of cyclic groups? | |
| Mar 20, 2017 at 20:12 | comment | added | YCor | to suspect is not conjecture, and this "suspicion" was already discarded. And indeed a non-abelian free product of cyclic groups has non-inner automorphisms (if at least one has $\ge 3$ elements, take inversion on it and identity on others; if all have 2 elements, take a non-trivial permutation of factors). | |
| Mar 20, 2017 at 19:52 | history | edited | Yiftach Barnea | CC BY-SA 3.0 |
added 2 characters in body
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| Mar 20, 2017 at 16:33 | history | edited | Igor Rivin | CC BY-SA 3.0 |
added info on Dyer's automorphism.
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| Mar 20, 2017 at 13:39 | comment | added | Igor Rivin | @HJRW ah, ok (although in the other comment thread you conjecture that a free product of cyclic groups has a trivial outer automorphism group...) | |
| Mar 20, 2017 at 6:24 | comment | added | HJRW | It's not. If it were virtually free, it would be a free product of cyclic groups, and have a non-trivial outer automorphism. | |
| Mar 20, 2017 at 0:58 | comment | added | Igor Rivin | @YiftachBarnea I am suspecting that yes, but can't do better than that at the moment. | |
| Mar 19, 2017 at 23:42 | comment | added | Yiftach Barnea | is this group virtually (non-abelian free)? | |
| Mar 19, 2017 at 23:28 | history | answered | Igor Rivin | CC BY-SA 3.0 |