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I am reading an older paper by Jamali and Mousavi.

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On the second page there is the following proposition 2.2

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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!!

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    $\begingroup$ If $A$ is an abelian group, $Hom(G,A)$ can be given a group law by pointwise multiplication of functions: $(f\cdot g)(x)=f(x)g(x)$ for $x\in G$. $\endgroup$ – user120527 May 13 '18 at 16:46
  • $\begingroup$ Thank you. Associativity seems clear to me from your definition. What about unity/identity and inverse elements? If it is obvious, I apologize! $\endgroup$ – Moritz May 13 '18 at 17:00
  • $\begingroup$ the identity is the constant map from $G$ to $A$ equal to the neutral of $A$. The inverse $f^{-1}$ (the notation is maybe not very clear): $f^{-1}(x)=(f(x))^{-1}$. It's just a subgroup of the classical product group $A^G$, identified as a set of functions from $G$ to $A$. $\endgroup$ – user120527 May 13 '18 at 17:04
  • $\begingroup$ Great! Thank you very much. $\endgroup$ – Moritz May 13 '18 at 17:07
  • $\begingroup$ Please ask one question at a time. $\endgroup$ – Shaun May 13 '18 at 18:22
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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?)

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  • $\begingroup$ Short ingenious argument. Thank you and +1 from me. $Aut_c$ is the normal subgroup of $Aut(G)$ that contains all automorphisms that commute with every element of $Inn(G)$ - the inner automorphisms of $G$. $\endgroup$ – Moritz May 13 '18 at 17:05
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    $\begingroup$ @Moritz This is the 'the universal property of the abelianization.' You can find further discussion about this kind of situation here: math.stackexchange.com/questions/40692/… $\endgroup$ – Lorenzo Najt May 13 '18 at 19:40

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