Notes for CMU-1504/0:26:15: Difference between revisions
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So, by renaming the nails, you get a bijection from $[[x,y],z]$ to e.g. $[[y,z],x]$. Is this obvious without considering the visual? [[User:Isomorphismes|Isomorphismes]] ([[User talk:Isomorphismes|talk]]) 19:39, 19 October 2015 (EDT) |
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I don't think I said that. "Dropping a nail" is the same as setting $x=e$ or $y=e$ or $z=e$, and it is easy to convince oneself that any one of these substitutions maps $[[x,y],z]$ to $e$. (And the same is true for further-iterated commutators, such as $[[[w,x],y],z]$, etc. --[[User:Drorbn|Drorbn]] ([[User talk:Drorbn|talk]]) 19:55, 19 October 2015 (EDT) |
- I don't think I said that. "Dropping a nail" is the same as setting $x=e$ or $y=e$ or $z=e$, and it is easy to convince oneself that any one of these substitutions maps $[[x,y],z]$ to $e$. (And the same is true for further-iterated commutators, such as $[[[w,x],y],z]$, etc. --[[User:Drorbn|Drorbn]] ([[User talk:Drorbn|talk]]) 19:55, 19 October 2015 (EDT) |
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Latest revision as of 19:01, 19 October 2015
So, by renaming the nails, you get a bijection from $[[x,y],z]$ to e.g. $[[y,z],x]$. Is this obvious without considering the visual? Isomorphismes (talk) 19:39, 19 October 2015 (EDT)
- I don't think I said that. "Dropping a nail" is the same as setting $x=e$ or $y=e$ or $z=e$, and it is easy to convince oneself that any one of these substitutions maps $[[x,y],z]$ to $e$. (And the same is true for further-iterated commutators, such as $[[[w,x],y],z]$, etc. --Drorbn (talk) 19:55, 19 October 2015 (EDT)
