# Are there injective groups?

One exercise in Robinson's A Course in the Theory of Groups is to prove that the groups which have the projective property are necessarily free. I'm not able to prove that because I haven't gotten to that chapter yet, but I tried to find in the book any mention of groups that have the injective property. I know that the injective abelian groups are exactly the divisible groups. But nothing is said about injectivity in the class of all groups. Why is that?

-
The non-existence of non-trivial injective groups appeared today on MO: mathoverflow.net/questions/100245 – t.b. Jun 21 '12 at 17:27

"The only injective object in the category of groups is the trivial group," is the statement of the theorem in M. Nogin's "A short proof of Eilenberg and Moore's theorem," 2007. The cited work of S. Eilenberg and J.C. Moore is "Foundations of relative homological algebra," 1965.

-
Thank you very much! – user23211 Jan 27 '12 at 0:24
Or, as Mac Lane put it, "every fascist group consists only of the identity element." [1950, Duality for Groups]. – Zhen Lin Jan 27 '12 at 0:31
Was that paper actually published? It basically follows exactly a hint to an exercise in D.L. Johnson's "Topics in the Theory of Group Presentations", published over 30 years ago! And the exercise is in the first 10 pages of the book! – user641 Jan 27 '12 at 2:05
That's true: tinyurl.com/7rpvjrc (link to the Google Books) -- exercise 7. – Damian Sobota Jan 27 '12 at 2:58

Another approach is to consider again the free group $F$ on two letters $x,y$ and the automorphism $x\leftrightarrow y$ that gives a semidirect product $F\rtimes C_2$ where the generator $\sigma$ of $C_2$ acts by $\sigma x\sigma =y$ and $\sigma y\sigma =x$. Then consider the canonical injection $\iota : F\to F\rtimes C_2$ and given $g\in G$, where $G$ is injective, the morphism $F\to G$ with $x\to 1$ and $y\to g$. An extension $\psi: F\rtimes C_2\to G$ gives that $\psi z g\psi z=1$ so that $g=1$ since $(\psi z)^2=1$.

-