No injective groups with more than one element? There are several claims in the literature that there are no injective groups (with more than one element), but I have not found a proof.  For example, Mac Lane claims in his Duality from groups paper in the 1950 Bull. A.M.S. that Baer showed him an elegant proof, but gives no hint of what it might be.  Does anyone know of an actual published proof?
 A: The argument in the slides by Maria Nogin (née Voloshina) linked to by a-fortiori in a comment was published as:
Maria Nogin, A short proof of Eilenberg and Moore’s theorem, Central European Journal Of Mathematics Volume 5, Number 1 (2007), 201–204. Also available on her homepage.

Added: The above paper was also mentioned in Jonas Meyer's answer to the same question on math.SE. As Steve D. points out in a comment there, the result  appears as Exercise 7 on page 9 of D.L. Johnson's Topics in the Theory of Group Presentations, Cambridge University Press 1980:


The argument by Eilenberg and Moore appears on pages 21/22 of Foundations of Relative Homological Algebra, Memoirs of the AMS, Volume 55 (1965). Here's a scan of the relevant passage for the convenience of the readers:


A: Does this work? I want to use the following statement, which I believe is true. 
Let $S$ be an infinite set. Let $\Pi(S)$ be the group of bijections $S\to S$. Let $N$ be the normal subgroup consisting of those bijections whose support is smaller (in cardinality) than $S$. Then every proper normal subgroup of $\Pi(S)$ is contained in $N$. 
Now suppose that $G$ is a nontrivial injective group. Certainly it is infinite. Let $S$ be the set of elements of $G$. $S$ is isomorphic to a subgroup of $\Pi(S)$. Being injective, it is therefore isomorphic to a quotient of $\Pi(S)$, and is therefore at least as big as $\Pi(S)/N$, which is as big as $2^S$, contradiction.
A: Let $G$ be a non-trivial injective group and $g \in G$ non-trivial. From category theory, $G \times G$ is injective as well. But, now $G \times G$ embeds into a group $H$, where $(g,e)$ and $(e,g)$ are conjugate using an HNN-extension, or just embedding into the group $H:=(G \times G) \rtimes \mathbb Z/2\mathbb Z$. Since $(g,e)$ and $(e,g)$ are not conjugate in $G \times G$, $H$ cannot split back to $G \times G$. This is a contradiction.
