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The free group $F(S)$ on a set $S$ is a cogroup in the category of groups since $\hom(F(S),G) \cong G^S$ carries a natural group structure for every group $G$. I have read that these are the only cogroups in the category of groups. This result is attributed to Kan:

Daniel M. Kan, On monoids and their dual, Bol. Soc. Mat. Mexicana (2) 3 (1958), 52–61. MR 0111035 (22 #1900)

However I have no access to this paper, and could not find it online either. Perhaps someone knows the paper and can give me a hint how to prove the result? Thanks a lot.

Edit: Tyler's answer explains why the underlying group of any cogroup is free. I accept it because meanwhile I've found Kan's paper.

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    $\begingroup$ There is discussion of Kan's result in Section 13 of the following paper -- dropbox.com/s/pnzlq6ken5vpafk/Bergman%20Colimits.pdf -- I don't know if that is useful to you... $\endgroup$
    – Nick Gill
    Commented Nov 13, 2012 at 13:52
  • $\begingroup$ Thank you, this is interesting. It doesn't contain Kan's proof, but lots of other related observations. $\endgroup$
    – Martin Brandenburg
    Commented Nov 13, 2012 at 14:30
  • $\begingroup$ To claim that you have no access to this paper is a bit of overstatement; Münster's library lists it in stock: superfix.uni-muenster.de/Katalog/start.do?Query=0=%22864164%22 $\endgroup$
    – Dmitri Pavlov
    Commented Nov 14, 2012 at 21:15
  • $\begingroup$ Oh, you're right! Thanks. I found it in the archive. $\endgroup$
    – Martin Brandenburg
    Commented Nov 15, 2012 at 21:06
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    $\begingroup$ May I suggest that you scan it and make it available on your homepage (or elsewhere)? Scanning a 10 page paper takes a relatively small amount of time compared to borrowing it from the library archive. For example, I maintain my own archive of scanned papers: dmitripavlov.org/scans $\endgroup$
    – Dmitri Pavlov
    Commented Nov 24, 2012 at 9:55

2 Answers 2

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For any group $G$, there are two maps $G \ast G \to G$ given by projection onto the individual factors. For an element $g \in G$, I'll write $g'$ for its image in the first factor and $g''$ for its image in the second factor.

Claim: The equalizer of the diagram $G \ast G \rightrightarrows G$ is always free on the elements $g' g''$ as $g$ ranges over the non-identity elements.

Once you know this, it automatically follows that cogroup objects are free: they are groups $G$ equipped with a map $G \to G \ast G$ which equalizes these two arrows, and which is injective, and this exhibits them as a subgroup of a free group.

To prove the claim, take the free group $F$ on symbols $s_g$ for non-identity elements of $G$, and take the group homomorphism $F \to G \ast G$, $s_g \mapsto g' g''$.

The kernel is trivial: if you take a reduced word in the $s_g$ and look at its image, you get a word in $G \ast G$. You can take this word and write down the associated reduced word in $G \ast G$ by collecting adjacent terms, and there cannot be any cancellation unless the original word was unreduced.

The image is the equalizer: if you have a word in $G \ast G$ of the form $a' b'' c' d'' \cdots$, you can rewrite it as $$ a' a'' (a^{-1} b)'' (a^{-1} b)' (b^{-1} a c)' (b^{-1} a c)'' (c^{-1} a^{-1} b d)'' \cdots $$ which is a product of elements in the image of $F$; the term left at the end will be $$ (\cdots e^{-1} c^{-1} a^{-1} b d f \cdots)'' $$ which is trivial precisely when $ace\cdots = bdf\cdots$, or equivalently the two projections are equal.

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  • $\begingroup$ 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. $\endgroup$
    – Tyler Lawson
    Commented Nov 13, 2012 at 14:22
  • $\begingroup$ 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. $\endgroup$
    – Martin Brandenburg
    Commented Nov 13, 2012 at 15:04
  • $\begingroup$ 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. $\endgroup$
    – Martin Brandenburg
    Commented Nov 13, 2012 at 15:39
  • $\begingroup$ @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. $\endgroup$
    – Tyler Lawson
    Commented Nov 13, 2012 at 17:14
  • $\begingroup$ Ok. We already know that the underlying group $G$ is free over some basis $S$; perhaps we can simplify the reasoning then? $\endgroup$
    – Martin Brandenburg
    Commented Nov 13, 2012 at 17:40
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This is an extended comment to Tyler's answer. The Kurosh theorem implies the equalizer is a free product of a free group with conjugates of subgroups of the free factors. But each non-trivial subgroup of a free factor is killed by one projection and not the other. Thus the equalizer is free.

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