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 20:21 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |
added 307 characters in body
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| May 25, 2015 at 2:16 | vote | accept | owb | ||
| May 25, 2015 at 2:00 | vote | accept | owb | ||
| May 25, 2015 at 2:03 | |||||
| May 25, 2015 at 1:53 | comment | added | owb | It suffices to prove 1, because it implies 2,3: put $x_i=1$ for all $i$. Let $k_i$ be the sum of exponents of $x_i$ in $w$. Then in abelian groups $\theta$ is equivalent to $$\phi:=(\exists x_1\dots x_m)(\forall y_1\dots y_n)x_1^{k_1}\dots x_m^{k_m}y_1^{l_1}\dots y_n^{l_n}=1.$$ In groups the sentence $\phi$ is equivalent to $$\psi:=(\forall y_1\dots y_n) y_1^{l_1}\dots y_n^{l_n}=1,$$ and hence to $\rho:=\bigwedge_i(\forall y) y^{l_i}=1$. The order of any finite cyclic group, in which $\rho$ holds, divides all $l_i$. Since there are infinitely many such cyclic groups, all $l_i=0$. | |
| May 25, 2015 at 1:35 | comment | added | owb | Your argument allows to show a more general statement. Suppose a sentence $$\theta:=(\exists x_1\dots x_m)(\forall y_1\dots y_n)w(x_1,\dots ,x_m,y_1,\dots ,y_n)=1$$ is true in infinitely many finite cyclic groups. Then (1) $l_i=0$ for all $i$, where $l_i$ is the sum of exponents of $y_i$ in $w$; (2) $\theta$ holds in all abelian groups; (3) if $n=1$ then $\theta$ holds in all groups. Note that if $n>1$, it is not clear whether `$\theta$ holds in all finite groups' implies '$\theta$ holds in all groups'. In particular, for $\theta$ of the form $(\exists x)(\forall yz)w(x,y,z)=1$. | |
| May 24, 2015 at 4:28 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |
added 241 characters in body
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| May 24, 2015 at 4:09 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |
added 41 characters in body
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| May 24, 2015 at 3:45 | history | answered | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |