Working on homology and completion a question has arisen in my head. I know that $R$-mod as a category has enough projectives in it, and as such the category of abelian groups has it as they are in $\mathbb{Z}$-mod. But if we expand it and don't assume that that groups are abelian, does the category still have enough projectives in it? If not what is a counter example as a group that cannot get it to work?

Any references on this would be appriciated as well to read up on.


Free groups are projective in the category of groups (the exact same argument works as for modules), and for any group $G$, you can take the free group $F$ on the underlying set of $G$ and there is a canonical epimorphism $F\to G$. This works much more generally for pretty much any sort of algebraic structure.

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  • $\begingroup$ got any source on that free groups are projective? While I understand the clear connection to free module I'd still like to see a source. $\endgroup$ – Zelos Malum Feb 13 '16 at 11:31
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    $\begingroup$ @ZelosMalum it's straightforward from the universal property of free groups: if you have an epimorphism $f: S \rightarrow T \rightarrow 0$ and a morphism $g : F(S) \rightarrow T$ from the free group on the set $S$, you can define a morphism $F(S) \rightarrow S$ making the triangle commute by sending each $s \in S$ to some preimage via $f$ of $g(s)$. $\endgroup$ – Abel Feb 13 '16 at 15:46
  • $\begingroup$ I have checked it but I feel fairly uncertain as we're dealing with a set and then groups which makes it feel not quite compelte $\endgroup$ – Zelos Malum Feb 13 '16 at 15:47

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