Why is $ \sigma (g) = \phi (g) g $ for some $ \phi \in \mathrm{Hom} (G,Z) $? Let $ G $ be a group. 
Let $ Z $ be the center of $ G $ which is not finite, such that $ G/Z $ is a finite group.
If $ \sigma \in \mathrm{Aut} ( G ) $, then $ \sigma ( Z ) = Z $, hence, we obtain the two following morphisms : $ \mathrm{Aut} ( G ) \to \mathrm{Aut} ( Z ) $ and $  \mathrm{Aut} ( G ) \to \mathrm{Aut} ( G / Z ) $
Question :
Why is :
$ \sigma \in \ker ( \mathrm{Aut} ( G ) \to \mathrm{Aut} ( G/Z )) $ if and only if $ \sigma (g) = \phi (g) g $ for some $ \phi \in \mathrm{Hom} (G,Z) $ ?
Thanks in advance for your help.
 A: By definition, $\mathrm{Aut}(G) \to \mathrm{Aut}(G/Z)$ sends an automorphism $\sigma$ to $\bar{\sigma}$ induced by $\sigma$, such that $\pi\circ \sigma = \bar{\sigma}\circ \pi$, where $\pi$ is the natural projection $G \to G/Z$.
Now, since $\sigma$ is in the kernel of this map, it must mean that $\bar{\sigma}$ is the identity morphism, then $\pi \circ \sigma = \pi$. Which means that $\pi(\sigma(g)) = \pi(g)$ for every $g$. Equivalently, $\pi(\sigma(g)g^{-1}) = 0$. This means that $\sigma(g)g^{-1}$ is in $Z$. Put $\phi: G \to Z$, that sends $g$ to $\sigma(g)g^{-1}$. You have to check that this is a group homomorphism. 
\begin{align}
\sigma(gh)(gh)^{-1} &= \sigma(g)\sigma(h)h^{—1}g^{-1}
\end{align}
But since $Z$ isn't just any subgroup, but the center of $G$, and since $\sigma(h)h^{—1} \in Z$, $\sigma(h)h^{—1}g^{-1} = g^{-1}\sigma(h)h^{—1}$.
You get that \begin{align}\sigma(g)\sigma(h)h^{—1}g^{-1} &= \sigma(g)g^{-1}\sigma(h)h^{—1}\\
&= \phi(g)\phi(h)
\end{align}
So $\phi \in \mathrm{Hom}(G,Z)$
By definition of $\phi$, $\sigma(g) = \phi(g)g$.
Notice that I didn't use the fact $G/Z$ is finite.
