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Consider the following (typical) notion of a free group with a basis:

Notion (U)
Fix a group $F$ and a subset $B\subset F$. We say $F$ is a free group with basis $B$ if for any function $f:B\to H$, where $H$ a group, there exists a unique group homomorphism $\tilde f:F\to H$ that extends $f$.

Note above did not require $B$ to generate $F$. It is claimed that $B$ generating $F$ is equivalent to the uniqueness of above homomorphism. Namely, the following notion (G), should be equivalent to (U):

Notion (G)
Fix a group $F$ and a subset $B\subset F$. We say $F$ is a free group with basis $B$ if $B$ generates $F$, and for any function $f:B\to H$, where $H$ is a group, there exists a group homomorphism $\tilde f:F\to H$ that extends $f$.

Now, it is straightforward to show (G) $\implies$ (U), but I am unable to establish (U) $\implies$ (G). This equivalence is remarked in Lyndon and Schupp's text Combinatorial Group Theory, shortly defining free group with a basis.

I looked at several "obvious thing" to try, but perhaps I'm too silly and not seeing how to finish it. For instance, suppose group $F$ and subset $B\subset F$ satisfy (U), then the inclusion function $B\hookrightarrow F$ extends uniquelty to the identity function $1_F$, and somehow we need to show if there exists $z\in F-\langle B\rangle$, there exist a different homomorphism $F\to F$ extending the inclusion.

Or, show the inclusion function $B\hookrightarrow \langle B\rangle$ extends to an injective homomorphism $F\to B$.

Apologies if this is simply trivial and I'm not seeing it.

goblin GONE
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bonsoon
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    The proof I know is by explicitly constructing a group that satisfies these conditions, and is clearly generated by $B$. Then the uniqueness shows every free group on $B$ is generated by $B$. There is a nice direct way to do this, when the subgroup generated by $B$ is normal (so, this works for free abelian groups): let $H$ be the subgroup generated by $B$. Then there are two homomorphisms $G\rightarrow G/H$, given by the trivial map, and the quotient map. In both cases, all elements of $B$ get mapped to the identity. By the uniqueness condition, $G/H$ must be trivial. – Steve D Sep 13 '17 at 05:04
  • @SteveD Right, I've consider the situation where $\langle B \rangle$ is normal, though I was puzzled as to this subgroup is not a priori normal. Certainly true in abelian case as you mentioned. I'll try your recommendation. I am guessing this explicit construction would then involve words over $B$? Thanks for the comment by the way! – bonsoon Sep 13 '17 at 05:23
  • It might be possible to use the fact that every epimorphism in $\mathbf{Grp}$ is surjective to prove this. I'm a bit hazy on the details, and I think this approach is overkill, but nonetheless I'll leave this comment since it may be helpful to someone, somewhere. – goblin GONE Sep 13 '17 at 05:23
  • Yes, the construction I'm suggesting is the usual one: show that all reduced words in elements of $B$ form a group. Then you have maps between the two (this constructed group and your $G$), and uniqueness implies this is an isomorphism. Since the constructed group is clearly generated by $B$, so was the original. – Steve D Sep 13 '17 at 05:26
  • @goblin: that is usually proved using free products, which I think is enough to prove this question directly (by showing $\ast_B\mathbb{Z}$ is a free group). – Steve D Sep 13 '17 at 05:27
  • @SteveD, fair enough. So my approach, even if I can get it to work, would be circular? – goblin GONE Sep 13 '17 at 05:34

1 Answers1

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The ideas being bounced around in the comments seem to me to be translatable into a proof which does not use any special construction.

Let $B$ be a free basis of $F$.

Assuming that $B$ does not generate $F$, I'll argue to a contradiction. From the assumption it follows that the subgroup $G<F$ generated by $B$ is a proper subgroup of $F$.

Consider the inclusion function $i : B \to F$. I will construct two distinct extensions of $i$ to homomorphisms $\tilde i, \tilde i' : F \to F$, which will provide the contradiction.

The extension $\tilde i$ is just the identity function on $F$. Note that $\tilde i$ is surjective.

For the other extension $\tilde i'$, first include $B$ into $G$. This extends uniquely to a homomorphism $F \to G$, and by composing with the inclusion $G \hookrightarrow F$ I obtain a nonsurjective homomorphism $\tilde i' : F \to F$ that extends $i$.

Lee Mosher
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  • Ah yes! Thanks for completing the proof. I didn't see that for some reason. Brilliant! – bonsoon Sep 13 '17 at 13:16
  • Nice! $;!;!$ – goblin GONE Sep 14 '17 at 09:53
  • This is actually a direct proof, written in a "proof by contradiction" style. IMHO it is clearer to give the direct proof then : extending the inclusion $B\to G$ gives a group morphism $F\to G$; when composed with the inclusion $G \to F$ (which is a group morphism), it gives a group morphism $F\to F$ that extends (just check it) the inclusion $B\to F$, so it is $\mathrm{id}_F$ by unicity; this shows that the inclusion $G\to F$ has a section, which means that $G = F$. – Pece Sep 15 '17 at 10:59