I am reading an older paper by Jamali and Mousavi.
On the second page there is the following proposition 2.2
I marked fourplaces in red.
The first one seems like a typo: ".. for every $f$ in $\operatorname{Hom}(G,Z(G))$" makes more sense to me.
The second one ".. is an isomorphism" - why? The map is certainly a bijection, but an isomorphism needs groups as domain and range and $\operatorname{Hom}(G,Z(G))$ is no group. What am I missing?
The third says "..$\operatorname{Hom}(G,Z(G)) \cong \operatorname{Hom}(G/G',Z(G))$.." - again why? There is no group on either side. But even if it is only a bijection: is this obvious?
Fourth mark: what implies this conclusion?
All in all I am certainly missing something essential - perhaps something obvious and/or easy? Can you tell me what it is? Thank you!!
One thing that's going on is this:
If $A$ is an Abelian group, and $G$ is a group, and $G'$ the commutator subgroup, then $Hom(G,A) \cong Hom(G / G', A)$, where this indicates that the natural map $\phi: Hom(G / G', A) \to Hom(G,A)$ (induced by the map $G \to G / G'$) is a bijection.
This is straightforward:
Injectivity of $\phi$ follows because $G \to G / G'$ is surjective.
Surjectivity of $\phi$ follows because if you have any $G \to A$, it must be zero on $G'$, and hence factors through $G / G'$.
I'm not sure about the rest. (What does the subscript $c$ in $Aut_c$ indicate?)