TITLE: Cogroups in the category of groups are free
QUESTION [10 upvotes]: 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.

REPLY [3 votes]: 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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