Any group can be embedded into a simple group hence no non-trivial injective groups. (Yes, you'll need the fact that any group can be embedded in a non-simple group.) For an infinite group G embed it first into the group of all permutations of G and then reduce by killing the subgroup of all permutations with support of cardinality less than that of G. The fact that any subgroup is an equalizer is an easy consequence of the pushout theorem for groups: to wit, the pushout of a pair of monics is a pullback [and all monic]. Hence the pushout of a monic with itself yields a pair of maps [indeed monic maps] for which it is an equalizer. The only proof I know for this lemma is via the construction of "free products with amalgamated subgroup" (that is, the construction of the pushouts). (On the top of page 36 of ftp://ftp.sam.math.ethz.ch/hg/EMIS/journals/TAC/reprints/articles/3/tr3.pdf it is mentioned in passing that all epics between groups are onto. Without a proof, alas.) Peter Quoting Michael Barr <barr@math.mcgill.ca>:
Googling around, I have come on several claims that there are no non-trivial injectives in the category of groups (e.g., Mac Lane in the 1950 Duality for groups paper credits Baer with an elegant proof, but gives no hint of what it might be and Baer's earlier paper on injectives doesn't mention it). I have not come on any proof of this, however.
Somewhere I have seen a proof that all monics in the category of groups are regular. I think it was in a paper by Eilenberg and ??? and it needed a special argument if there were elements of order 2. Can someone help me find this?
Michael
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