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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?

flag	asked Jun 21 at 13:45 Michael Barr 545●1●7

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Google gives zimmer.csufresno.edu/~mnogin/talks/… which may not be formally published, but is definitely short. – a-fortiori Jun 21 at 13:52

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And sweet, definitely short and sweet. – Lee Mosher Jun 21 at 14:06

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That's a nice proof! Since it is short, it could be copied as an answer below (with proper attribution) so it is easier to find and cite... – François G. Dorais♦ Jun 21 at 14:16

About Maria Voloshina proof, that injectivity of $f: F[a, b]\to F[c, d]:a \mapsto c,\ b\mapsto d\cdot c\cdot d^{-1} $ follow from $\forall n:P(n)$, where $P(n)$:: the (reducted) word f(w) end with $c$ or by $d^{−1}$ for each (reducted) word $w\in F[a,b]$ of lenght $n$ ($n>0$). And $\forall n>0:P(n)$ follow easly by induction. – Buschi Sergio Jun 21 at 15:48

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This question has appeared in MSE a few months ago: math.stackexchange.com/questions/102820/… – Bruno Stonek Jun 21 at 16:03

Theo Buehler · Accepted Answer · 2012-06-21 17:04:02Z UTC

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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:

Exercise 7

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:

Original argument part 1 Original argument part 2

edited Jun 21 at 17:04

Theo Buehler

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	I forgot: Eilenberg-Moore refer to the paper by R. Baer, Die Kompositionsreihe der Gruppe aller eineindeutigen Abbildungen einer unendlichen Menge auf sich, Studia Math. 5 (1934), 15-17 which is available on the home page of the Polish virtual library of science: pldml.icm.edu.pl. Direct link to the paper: matwbn.icm.edu.pl/ksiazki/sm/sm5/sm512.pdf – Theo Buehler Jun 21 at 19:22
	I just wanted to say that the proof attributed to Maria Nogin uses more about homotopy than I wanted my readers to have to know. And there is something wrong with the inductive argument that that map is 1-1 since clearly the image of the length 1 word $a^{-1}$ does not end in either $c$ or $d^{-1}$. I guess the argument of Eilenberg-Moore may be the easiest. Now I have to track down the reference to Baer. Thanks to all who answered. – Michael Barr Jun 21 at 20:41
	@Michael: I'm not sure what inductive argument you are referring to here. Nogin's proof uses only that the inclusion of a subgraph into the full graph is $\pi_1$-injective, and that covering maps are $\pi_1$-injective. The subgraph and covering map to which these facts are applied are very concrete in her paper. – Lee Mosher Jun 21 at 22:17
	See the comment added by Sergio Buschi added to #2. I don't know enough homotopy theory to see why that map is injective on $\pi_1$. Actually, I should have recalled that Eilenberg-Moore had it. I was very familiar with it back when it first came out. But it was kind of overtaken by Beck's thesis and other things and disappeared from my memory. – Michael Barr Jun 22 at 18:23

Andreas Thom · Answer 2 · 2012-06-21 15:33:11Z UTC

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.

Tom Goodwillie · Answer 3 · 2012-06-21 14:41:18Z UTC

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.

No injective groups with more than one element?

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