Timeline for On sentences true in all finite groups
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| May 25, 2015 at 16:26 | comment | added | Christian Remling | @owb: Thanks, you are right, of course. I've incorporated your correction now. | |
| May 25, 2015 at 16:25 | history | edited | Christian Remling | CC BY-SA 3.0 |
added 14 characters in body
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| May 25, 2015 at 2:03 | vote | accept | owb | ||
| May 25, 2015 at 2:16 | |||||
| May 25, 2015 at 1:31 | comment | added | owb | Here is a correct end of your argument: '$m+zn=0$ has no solution $z\in \mathbb Z$' implies that $n$ does not divide $m$, and hence $(\forall y) w(1,y)\ne 1$ and so $(\exists x)(\forall y) w(x,y)\ne 1$ hold in the cyclic group of order $n$. Thus if $(\forall x)(\exists y) w(x,y)=1$ holds in all cyclic groups then it holds in all groups. | |
| May 25, 2015 at 1:29 | comment | added | owb | @Cristian Remling: There is a small inaccuracy in your argument: '$m+zn=0$ has no solution $z\in \mathbb Z$' does not imply 'that $n$ has a prime factor that does not occur in $m$' (counterexample: $m=p$ and $n=p^2$, for a prime $p$). You claim: if $(\forall x)(\exists y) w(x,y)=1$ holds in all cyclic groups of prime orders then it holds in all groups. This is not true: for any prime $p$ and $w=x^py^{p^2}$ this sentence holds in every cyclic group of prime order but fails in the infinite cyclic group. | |
| May 24, 2015 at 5:12 | comment | added | Christian Remling | @BjørnKjos-Hanssen: I don't think this is really indirect, it's just the way I worded it, and you had already answered the second question. I just noticed you also added exactly the same argument to your answer before me, I didn't see this originally, sorry. | |
| May 24, 2015 at 5:00 | comment | added | Bjørn Kjos-Hanssen | This addresses only Question 1; also can you make it direct rather than indirect? +1 in any case | |
| May 24, 2015 at 4:41 | history | answered | Christian Remling | CC BY-SA 3.0 |