Timeline for Isomorphic free groups have bijective generating sets
Current License: CC BY-SA 4.0
28 events
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| May 4, 2020 at 10:29 | comment | added | Asaf Karagila♦ | @YCor: Good question. I feel like the answer should be yes. But I have nothing off hand to give you as an example. | |
| May 4, 2020 at 9:31 | comment | added | YCor | OK thanks, so I should restate my question replacing $\mathbf{R}$ and $\mathbf{R}/\mathbf{Q}$ with $X$ and $X^2$ for $X$ such that there is no injection $X^2\to X$. Namely: is it consistent (for some such $X$, and say with ZF+DC) that there is an group isomorphism $F^2_X\to F^2_{X^2}$ (or $F_X\to F_{X^2}$)? | |
| May 4, 2020 at 9:30 | comment | added | Asaf Karagila♦ | @YCor: Of course not. Since that would imply every infinite set has the same cardinality as its finite subsets, which implies $X\times X$ is equipotent with $X$, which implies choice. | |
| May 4, 2020 at 9:28 | comment | added | YCor | @AsafKaragila thanks, indeed. Does there always exist (say in ZF+DC) a bijection $X\to F_X^2$ for every infinite $X$? (oh,probably not) | |
| May 4, 2020 at 9:25 | comment | added | Asaf Karagila♦ | @YCor: Well, in that case, you can easily see that the cardinality of $F^2_{\Bbb R}$ is $2^{\aleph_0}$, while that of $F^2_{\Bbb{R/Q}}$ is at least $|\Bbb{R/Q}|$ (and probably equal, but it's too early in the morning for this. So if there is no bijection between $\Bbb R$ and $\Bbb{R/Q}$, the groups will not be isomorphic. | |
| May 4, 2020 at 9:23 | comment | added | YCor | @AsafKaragila Yes, this is one realization of 'the' free 2-elementary abelian group over $X$. | |
| May 4, 2020 at 9:23 | comment | added | Asaf Karagila♦ | @YCor: So free 2-elementary abelian = the finite subsets of $X$ (with symmetric difference as addition)? (Also, I had meant that maybe the groups are too elementary... :-P) | |
| May 4, 2020 at 9:21 | comment | added | YCor | @AsafKaragila a $p$-elementary abelian group, for prime $p$, is a synonym of vector space over the field on $p$ elements (I don't think a terminology question can be too elementary!). Purely group-wise, it means abelian group in which $px=0$ for all $p$. | |
| May 4, 2020 at 7:55 | comment | added | Asaf Karagila♦ | @YCor: I feel like I asked this before, what's a 2-elementary abelian group? (Too elementary?) | |
| May 3, 2020 at 19:46 | comment | added | YCor | Is the case of vector spaces already known? For instance, let $F^2_X$ be the free 2-elementary abelian group on $X$. Is ZF(+DC?) + "there's an isomorphism $F^2_{\mathbf{R}}\to F^2_{\mathbf{R}/\mathbf{Q}}$" + "there is no bijection $\mathbf{R}\to\mathbf{R}/\mathbf{Q}$" consistent (assuming ZF consistent)? (I allow middle choice here, at least to understand my own question) | |
| May 3, 2020 at 19:17 | comment | added | Ali Caglayan | @LSpice I've changed it to "or" as that sounds better. | |
| May 3, 2020 at 19:17 | history | edited | Ali Caglayan | CC BY-SA 4.0 |
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| May 3, 2020 at 18:54 | comment | added | LSpice | It's not a big deal, but I just can't parse it, even understanding what you mean. Is 'more' meant to be 'or'? | |
| May 3, 2020 at 18:52 | comment | added | Ali Caglayan | @LSpice Maybe that wasn't the best way to phrase it. BPIT is strictly weaker than AC so the proof I am looking for would use ZF + something weaker than choice. | |
| May 3, 2020 at 14:46 | comment | added | LSpice | Your last parenthesis ("more rather statements weaker than choice") doesn't make sense to me. Is it what was intended? | |
| May 3, 2020 at 14:46 | history | edited | LSpice | CC BY-SA 4.0 |
Name of "this paper"
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| May 3, 2020 at 14:29 | history | edited | Ali Caglayan |
edited tags
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| May 3, 2020 at 3:25 | comment | added | user44143 | I’d expect that most constructive uses focus on the countable case, where countable choice or dependent choice might be enough. | |
| May 2, 2020 at 23:51 | comment | added | Ali Caglayan | @LSpice This is Kleppmann's dissertation which I think Paul was referring to. | |
| May 2, 2020 at 23:47 | comment | added | LSpice | @PaulPlummer's reference: Läuchli - Auswahlaxiom in der Algebra (MSN). I think the Kleppman reference is actually to Kleppmann - Generating sets of free groups and the axiom of choice (MSN) (two 'n's). | |
| May 2, 2020 at 23:26 | history | edited | Ali Caglayan | CC BY-SA 4.0 |
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| May 2, 2020 at 23:15 | comment | added | Ali Caglayan | @YCor Hopefully things are more clear now. | |
| May 2, 2020 at 23:15 | history | edited | Ali Caglayan | CC BY-SA 4.0 |
made statement clearer
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| May 2, 2020 at 22:29 | comment | added | user35370 | In Kleppmans dissertation (which includes results from that paper you cite) it is asked if any choice is need. I guess there is a similar question you can ask for vector spaces and that is not a theorem of ZF (found in L ̈auchli. Auswahlaxiom in der Algebra): that is there is a vector space, in a suitable model, with two bases of different cardinalities. | |
| May 2, 2020 at 21:23 | comment | added | YCor | @MikeShulman when I first read it, I thought of it as "there exists a free generating subset of the first one and...etc." I converged to the wording I'm writing just by elimination. Actually "free" is hopelessly ambiguous since mathematicians use the same word for "free over some given subset" and "free over some subset"; of course often this doesn't matter, but precisely here it matters. | |
| May 2, 2020 at 21:22 | comment | added | Mike Shulman | @YCor What else could it mean? | |
| May 2, 2020 at 21:21 | comment | added | YCor | The statement is way too vague and has several interpretations. I guess you mean the statement "if there is an isomorphism from the free group on $X$ and the free group on $Y$, then there is a bijection from $X$ to $Y$". | |
| May 2, 2020 at 20:55 | history | asked | Ali Caglayan | CC BY-SA 4.0 |