Uniqueness of Extensions and Generators of Groups

Question

Let $G$ be a group, $X$ a non empty subset of $G$ and $\iota: X \rightarrow G$ the inclusion. Consider the following three properties:

(i) for every group $H$ and every map $f:X \rightarrow H$, there is at most one homomorphism $\phi:G \rightarrow H$ such that $f=\phi \circ \iota$;

(ii) for every abelian group $H$ and every map $f:X \rightarrow H$, there is exactly one homomorphism $\phi:G \rightarrow H$ such that $f=\phi \circ \iota$;

(iii) for every abelian group $H$ and every map $f:X \rightarrow H$, there is at most one homomorphism $\phi:G \rightarrow H$ such that $f=\phi \circ \iota$;

(iv) G is abelian and for every abelian group $H$ and every map $f:X \rightarrow H$, there is at most one homomorphism $\phi:G \rightarrow H$ such that $f=\phi \circ \iota$.

Does property one of these properties implies that $X$ generates $G$? I have thought about it, but I have no idea of the answers for now, since I was neither able to prove that the answers are in the affirmative nor I could find counterexamples.

Thank you very much for your attention.

Note This question was motivated by the definition of free (abelian) group, which is as follows. An (abelian) group $G$ is said to be free (abelian) on a non-empty subset $X$ of $G$, if for every (abelian) group $H$ and every map $f:X \rightarrow H$, there is exactly one homomorphism $\phi:G \rightarrow H$ such that $f=\phi \circ \iota$. In that case you can conclude that $X$ actually generates $G$ by the following simple argument. Take $H=\langle X \rangle$, let $f:X \rightarrow H$ be the inclusion and let $\phi:G \rightarrow H$ be the homomorphism satisfying $f= \phi \circ \iota$. Then, if $g:H \rightarrow G$ is the inclusion, $g \circ \phi:G \rightarrow G$ is a homomorphism extending the inclusion $g \circ f$ of $X$ into $G$. Since also the identity map $1_{G}:G \rightarrow G$ is a homomorphism extending $g \circ f$, we must have $g \circ \phi = 1_{G}$, so that $G= \text{Im} (1_{G}) = \text{Im} (g \circ \phi) = \text{Im} (\phi) \subseteq H$.

Answer 1

Yes, property (i) implies $X$ generates $G$, because epimorphisms in $\mathsf{Group}$ are surjective. Your condition is equivalent to asking that the inclusion $i\colon\langle X\rangle\hookrightarrow G$ be an epimorphism in $\mathsf{Group}$. Note that $\iota$ completely determines $i$, so it is equivalent to talk about the inclusion of the set $X$ into $G$ and of the inclusion of the subgroup generated by $X$ into $G$.

Indeed, suppose that $i\colon\langle X\rangle\hookrightarrow G$ is an epimorphism in $\mathsf{Group}$. Let $f\colon X\to H$ be any set map. If this does not extend to a group morphism $\overline{f}\colon\langle X\rangle \to H$, then there are no morphisms $G\to H$ extending $f$ at all. If it does extend to a group morphism $\overline{f}\colon\langle X\rangle\to H$, and $\phi,\psi\colon G\to H$ satisfy that $\phi\circ i = \psi\circ i$, then because $i$ is an epimorphism then $\phi=\psi$. Thus, there is at most one extension of $f$ to a morphism from $G$ to $H$.

Conversely, if the property holds then $i$ is an epimorphism, since an equality $\phi\circ i=\psi\circ i$ imply $\phi=\psi$ by uniqueness.

And since $i\colon\langle X\rangle \to G$ is an epimorphism, and epimorphisms are surjective, we have $\langle X\rangle = G$.

Likewise (iv) implies that $X$ generates $G$, because epimorphisms in $\mathsf{AbGroup}$ are surjective, though the argument there is simpler: you can just take the zero and quotient maps of $G$ onto $G/\langle X\rangle$ to see that unless $\langle X\rangle=G$ then you have two morphisms.

On the other hand, neither (ii) nor (iii) imply that $\langle X\rangle=G$. For example, let $G=A_5\oplus\mathbf{Z}$, and let $X$ consist of a generator of the infinite cyclic factor $\mathbf{Z}$ (or more generally, let $G$ be the direct sum of a free abelian group and a nonabelian group with no nontrivial abelian quotients). Then every set map form $X$ to an abelian group has one and only one extension to $G$, but $X$ does not generate $G$.