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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.

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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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