I have to demonstrate: Let B a group and X a nonempty set. So, $$Z[Aplc(X,G)]=Aplc[X,Z(G)]$$ where $Z$ is the center of the group. And $Aplc(X,G)$ is the set of all the applications from X to G. ($Aplc[X,G]$ is a group with $\circ$)
My little work. Im so lost in this, but I try make a double implication. From an element of $Z[Aplc(X,G)]$ reach an element of $Aplc[X,Z(G)]$.
$\Rightarrow)$ Let $g\in Z[Aplc(X,G)]$, so $g\in Aplc(X,G)$ and $ag=ga$ $\forall a\in Aplc(X,G)$. But I see a problem and this occurs when I try to define the characteristic of $g$ in the center. Well, ik that $g\circ a=a\circ g$ so I get $g(a(x))=a(g(x))$ and then I don't know what to make. What should I do for demonstrate that $g\in Aplc[X,Z(G)]$ :(
$\Leftarrow)$ Well, I start with Let $n\in Aplc[X,Z(G)]$ so, $n:X\to Z(G)$. And I have to show that $n\in Z[Aplc(X,G)]$. For that, I need to see that $an=na$ $\forall a\in Aplc(X,G)]$. Well, i try that with $$an(x)=(a\circ n)(x)=a(n(x))$$ And I don't know how to proceed. What should I do. :'(
Any help would be awesome. I've tried this all this days and nothing works.
PDT: I think I can make this finding an isomorphism between the two sets. Buuut, I could not find it and I discard that idea.
Thanks to all!