Timeline for Cogroups in the category of groups are free
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2012 at 21:05 | vote | accept | Martin Brandenburg | ||
| Nov 13, 2012 at 19:24 | comment | added | Tyler Lawson | I wish I could say that the argument I've written down in the answer simplified things, but I'm not sure that it does. It does convert into the following: if you have a set of distinct nonidentity primitive elements in your cogroup G, then they generate a free subgroup (which is the last step in my comment-sketch). You still need to take a general element and use coassociativity to tell you about the height of terms appearing in its coproduct. | |
| Nov 13, 2012 at 17:40 | comment | added | Martin Brandenburg | Ok. We already know that the underlying group $G$ is free over some basis $S$; perhaps we can simplify the reasoning then? | |
| Nov 13, 2012 at 17:14 | comment | added | Tyler Lawson | @Martin: I doubt you're likely to get a better outline than what Bergman goes through in chapter 9.5 of that text, though some simplifications happen. To an element of G you assign a height corresponding to the length of a reduced word representing its coproduct; you use the unit to deduce that it provides some shuffling of two product decompositions; you show that any word is a product of words of degree two, which in this case are either primitive or the inverse of a primitive; you can then take a coproduct of a reduced word in primitive elements and show it's only primitive for length 1. | |
| Nov 13, 2012 at 15:39 | comment | added | Martin Brandenburg | In other words, I would like to know why every representable endofunctor of $\mathsf{Grp}$ is isomorphic to $G \mapsto G^S$ for some set $S$. This is Exercise 9.6:2 in Bergman's "An Invitation to General Algebra and Universal Constructions", where it follows from the more general classification for $\mathsf{Mon}$ - probably there is a more direct approach for groups. | |
| Nov 13, 2012 at 15:04 | comment | added | Martin Brandenburg | Thank you for the proof. I would like to know why for every cogroup $(G,\Delta,\epsilon)$ there is a set $S$ and an isomorphism of cogroups with $(F(S),\Delta_S,\epsilon_S)$, where $\Delta_S(s)=s's''$ and $\epsilon_S(s)=1$. This seems to come down to the statement that $G$ is free on the primitive elements. | |
| Nov 13, 2012 at 15:02 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Nov 13, 2012 at 14:23 | history | edited | Tyler Lawson | CC BY-SA 3.0 |
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| Nov 13, 2012 at 14:22 | comment | added | Tyler Lawson | Kan's result includes more: it tells you that, if you have a cogroup, the set of "primitive" elements $g$ whose coproduct is $g'g''$ form a set of generators for $G$ as a free group. I don't remember the proof of this. | |
| Nov 13, 2012 at 14:15 | history | answered | Tyler Lawson | CC BY-SA 3.0 |