Timeline for Proof of an 'easy' exercise in a book of Tits
Current License: CC BY-SA 4.0
11 events
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| Feb 13, 2023 at 14:53 | history | edited | Max Horn | CC BY-SA 4.0 |
added 195 characters in body
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| Apr 20, 2011 at 7:35 | comment | added | Thomas Kalinowski | Finally, I also see the inversion argument: (3) is applied to $G_1$, $y^{-1}G_2$ and $z^{-1}G_3$, giving an element $g_1\in G_1\cap y^{-1}G_2\cap z^{-1}G_3$, and then $hg_1\in G_2\cap G_3$. | |
| Apr 20, 2011 at 7:12 | comment | added | Thomas Kalinowski | The last $v$ should be $v^{-1}$. | |
| Apr 20, 2011 at 7:09 | comment | added | Thomas Kalinowski | (2) => (3) can be done without the second "wlog": $g_1^{-1}\tilde g_1=uv$ with $u\in G_1\cap G_2$ and $v\in G_1\cap G_3$, and then [yg_2^{-1}u=g_1u=1\cdot g_1u=zg_3^{-1}\tilde g_1^{-1}g_1u=zg_3^{-1}v ] is in $yG_2\cap G_1\cap zG_3$. | |
| Apr 19, 2011 at 22:26 | comment | added | Thomas Kalinowski | @Johannes: I'm stuck at the same points as Darij. The second "wlog" in the answer is not clear to me, and inverting an element from $G_1\cap yG_2\cap z G_3$ (whose existence follows from (3)) gives something in $G_1\cap G_2y^{-1}\cap G_3 z^{-1}$, not the claimed $g_1\in G_1\cap G_2y\cap G_3 z$. | |
| Apr 19, 2011 at 16:58 | comment | added | darij grinberg | Sorry for my typo, but it should be $G_1\cap yG_2\cap zG_3$, and that's quite a difference to $G_1\cap G_2y\cap G_3z$. | |
| Apr 19, 2011 at 16:57 | comment | added | darij grinberg | @Johannes: sorry, there are two "wlogs" in that proof and I meant the second one: "wlog we may assume $g_1\in G_1\cap G_2$ and $\tilde{g}_1\in G_1\cap G_3$." | |
| Apr 19, 2011 at 16:01 | vote | accept | Thomas Connor | ||
| Apr 19, 2011 at 15:28 | comment | added | Johannes Hahn | @darij: wlog $x=1$ is just the usual symmetry: If $xG_1,yG_2,zG_3$ are pairwise non-disjoint than $G_1,x^{-1}yG_2,x^{-1}zG_3$ are as well. By the above proof $G_1\cap x^{-1}yG_2\cap x^{-1}zG_3$ is non-empty and hence $xG_1\cap yG_2\cap zG_3$ is non-empty. I don't see what's the other problem. If there is an element of $G_1\cap G_2 y \cap G_3 z$ (you have $yG_2$ there, that's just a typo, isn't it? But that doesn't matter, you can always change sides by inverting) and you can name it $g_1$. What's unclear about it? | |
| Apr 19, 2011 at 14:17 | comment | added | darij grinberg | Looks nice, but I'm not comfortable with two things: (a) In the proof of (2) => (3), why can we "wlog assume"? (b) In the proof of (3) => (1), why does there exist $g_1\in G_1\cap G_2y\cap G_3z$ ? I only see that there exists some element of $G_1\cap yG_2\cap G_3z$. | |
| Apr 19, 2011 at 11:01 | history | answered | Max Horn | CC BY-SA 3.0 |