I was reading this proof by Steinberg of Nielsen-Schreier theorem and i had a doubt about proposition 1, that basically is an alternative universal property of free groups. It says:

Let $X$ be a set and $F$ a group equipped with a map $i: X \rightarrow F$. Then $F$ is a free group on $X$ (with respect to the mapping $i$) if and only if given any set $A$ and any map $σ: X \rightarrow S_A$, there is a unique action of $F$ on $A$ such that: $ai(x)=\sigma(x)(a)$

Note that in the paper when he says actions he will always refer to a right action. In the paper he says that the right implication was obvious and i thought so, but later when i tried to check it i came up with nothing. My idea was to use the universal property of free groups and then the inducted homomorphism:

$F$ is free so given the map $\sigma$ there's a unique homomorphism $\varphi: F \rightarrow S_A$ such that $\varphi(i(x))=\sigma(x)$ Then i thought that the unique actions could be the action $*: A \times F \rightarrow A$ with $(a,w) \rightarrow \varphi(w)(a)$. The only problem with that is that it is a left action and not a right action. To define the equivalent right actions that should be $(a,w) \rightarrow \varphi(w^{-1})(a)$ but this way the condition $ai(x)=\sigma(x)(a)$ is not true anymore.

Any ideas to prove this statement?


You just have to use the action $(a,w)\to\varphi(w^{-1})(a)$ for the unique homomorphism $\varphi:F\to S_A$ such that $\varphi(i(x))=\sigma(x)^{\color{red}{-1}}$.

Alternatively, I would guess that actually the author is using the reverse composition operation on $S_A$ (not the usual composition order of functions), so that a homomorphism to $S_A$ gives a right action rather than a left action.

  • $\begingroup$ How can i state that such homomorphism exists and that is unique? Is this equivalent to the "standard" universal proof? $\endgroup$ – Andrea Licata Feb 3 at 1:36
  • $\begingroup$ It is literally just an instance of the usual universal property, for the map $x\mapsto \sigma(x)^{-1}$. $\endgroup$ – Eric Wofsey Feb 3 at 2:26

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