Why is this element in the center? (Part two) Suppose that $A$ is an abelian group, and that $G$ is a group. Suppose we have a group homomorphism $$f: A \times G \to G,$$ and suppose further that $f(a,f(b,g)) = f(ab,g)$ for all $a,b \in A, g \in G.$  I have seem the claim that this implies that $f(a,1)$ is in the center of $G.$ Why is this true?

Related Why is this element in the center?
This is a left $A$-act on $G$ with the action given by $f$. Denoting $G= (G,\cdot)$ and $A = (A,+)$.

Assuming $G$ is unitary i.e $f$ surjective we would have $g = f(s,h)$
\begin{align*}
f(a,1)\cdot g = f(a,1)\cdot f(s,h) &amp;= f(a+s, 1\cdot h) = f(s+a,h\cdot 1)\\ &amp;= f(s,h)\cdot f(a,1) = g\cdot f(a,1)
\end{align*}

which does not require the 'associativity' of the action only that $f$ is a homomorphism.  

Not sure atm what do if $f$ is not surjective.
This is a left $A$-act on $G$ with the action given by $f$. Denoting $G= (G,\cdot)$ and $A = (A,+)$.

Assuming $G$ is unitary i.e $f$ surjective we would have $g = f(s,h)$
\begin{align*}
f(a,1)\cdot g = f(a,1)\cdot f(s,h) &amp;= f(a+s, 1\cdot h) = f(s+a,h\cdot 1)\\ &amp;= f(s,h)\cdot f(a,1) = g\cdot f(a,1)
\end{align*}

which does not require the 'associativity' of the action only that $f$ is a homomorphism.  

Not sure atm what do if $f$ is not surjective.