In Aluffi's book "Algebra: Chapter 0", he uses the categorical definition of free group in pp. 71, also see in wiki. This defines the free group up to group isomorphism. This is easy to get through.

Now the problem is, he also use this to define the subgroup generated by a subset in pp. 81. I put it here:

Definition (Subgroup generated by a subset). Let $G$ be a group. If $A\subset G$ is any subset, we have a unique group homomorphism $$\varphi_A: F(A)\to G,$$ by the universal property of free group. The image of this homomorphism is a subgroup of $G$, the subgroup generated by $A$ in $G$, ofter denoted $\langle A\rangle$.

This will naturally lead to a doubt of the well-definedness, i.e., does this definition depend on the choice of the free group $F(A)$ that is unique up to group isomorphism?

I don't know how to prove it to be well-defined. What I know is that if one can prove the well-definedness, then it's easy to prove that the categorical definition of generated subgroup is equivalent to other two definitions which just follows the categorical one in Aluffi's book, see also here.

Can anyone give some clues or hints on the well-definedness? TIA!


1 Answer 1


Let $ F'(A)$ be some other free group which is isomorphic to $F(A)$ via $ \psi$. Let $ \varphi_A'$ be the unique homomorphism $ F'(A)\rightarrow G$. We also get a homomorphism from $ F'(A) \rightarrow G $ by taking $ \varphi_A\circ \psi$. By uniqueness, these are the same homomorphism, so $$\varphi_{A'}(F'(A))=\varphi_A\circ \psi(F'(A))=\varphi_A(F(A))$$ where the last equality follows from the fact that $\psi$ is an isomorphism.

  • $\begingroup$ I guess the more interesting point is that both $\phi$ and $\phi'$ have the same image. $\endgroup$ May 1, 2018 at 3:33
  • $\begingroup$ @KevinCarlson Is this right: $\varphi_A'(F'(A))=\varphi_A\circ\psi(F'(A))=\varphi_A(F(A))$? $\endgroup$
    – Dreamer
    May 1, 2018 at 3:56
  • $\begingroup$ I think what we would like to say more is that the group homomorphism $\varphi_A\circ \psi: F'(A) \rightarrow G$ is exactly the one such that the diagram in the definition of free group $F'(A)$ commutes. $\endgroup$
    – Dreamer
    May 1, 2018 at 4:03
  • $\begingroup$ I have edited my proof slightly to include this equality, which is correct @Q.Huang. I think that this is enough to show that the image of both homomorphisms is the same, which is what you need to show well-defined. $\endgroup$
    – user293794
    May 1, 2018 at 4:09

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