Let $S$ be any set and let $f:S \rightarrow F$ denote the free group on $S$. By definition, this means that for any group $X$ and any function $g:S \rightarrow X$ there exists a unique homomorphism $h:F \rightarrow X$ such that $h \circ f = g$. How can I show that $f(S)$ generates $F$? By "generate $F$" I mean that the intersection of all subgroups of $F$ containing $f(S)$ equals $F$.


This is a standard property that can be derived from the universal property of $F$.

Let $H$ be the subgroup generated by $f(S)$ in $F$. Then consider the embedding $\iota\colon S\hookrightarrow H$ induced by $f$. By the universal property, there exists a unique homomoprhism $\varphi\colon F\to H$ such that $\varphi\circ f =\iota$. If we compose $\iota$ with the inclusion $j\colon H\to F$, then we obtain a map $j\circ\iota\colon S\to F$, and so there is a unique homomorphism $\psi\colon F\to F$ such that $\psi\circ f = j\circ\iota$. Of course, $\psi=\mathrm{id}_F$ works. On the other hand, so does $j\circ\varphi$, since $$(j\circ\varphi)\circ f = j\circ(\varphi\circ f) = j\circ \iota.$$ Therefore, by the uniqueness clause of the universal property, $j\circ\varphi=\mathrm{id}_F$. Since $j\circ\varphi$ is a bijection, it follows that $j$ is onto. Since $j$ is the inclusion $H\to F$ and is onto, then $H=F$, as desired.

Note. The above argument works in many settings! It shows that the free semigroup on $S$ is generated by the (image of) $S$, same for free rings, relatively free groups, free lattices, etc. In fact, it works in any variety of algebras (in the sense of universal algebra).

  • $\begingroup$ By the "induced embedding" $\iota$ do you just mean the function defined by $\iota(s) = f(s)$ for every $s \in S$? $\endgroup$ – ItsNotObvious May 15 '12 at 18:41
  • $\begingroup$ @IsNotObvious: Yes; formally (as, "in the sense of category theory"), $\iota$ and $f$ are not equal because they have different codomain. $\endgroup$ – Arturo Magidin May 15 '12 at 18:44
  • $\begingroup$ Ok, I'm with you up until the penultimate paragraph where you conclude that $j$ is surjective; isn't $j \circ \phi = id_F$ enough to conclude this ? i.e., a function is surjective if and only if it has a right inverse. $\endgroup$ – ItsNotObvious May 15 '12 at 19:59
  • $\begingroup$ @ItsNotObvious: Same difference; if a composition is bijective, then the first map is one-to-one and the second map is onto. So you can conclude it either from the fact that $\varphi$ is a right inverse to $j$, or from the fact that $j$ is the last function in a composition that is a bijective function. Either way, $j$ is surjective, and since it was one-to-one to begin with, it's a bijection and we are done. Note, however, that "surjective" and "right inverse" are not equivalent in the category of groups (though right invertible does imply surjective): e.g., $\mathbb{Z}\to\mathbb{Z}_2$. $\endgroup$ – Arturo Magidin May 15 '12 at 20:00
  • $\begingroup$ I see; it was, in fact, obvious. Thanks. $\endgroup$ – ItsNotObvious May 15 '12 at 20:02

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