Let $G$ be a group with center $C$. Let $\phi: G/C\to G$ be a homomorphism with $\phi(gC)\in gC,\forall g\in G$. Prove that $G\cong C\times(G/C)$. Let $G$ be a group and let $C$ denote the center of $G$. Suppose there exists a group homomorphism $\phi: G/C \longrightarrow G$ with the property that $\phi(gC) \in gC$ for all $g \in G$. Prove that $G \cong C \times (G/C)$. 
The idea I had in mind was to use the following Theorem: If $G$ is a group with normal subgroups $H$ and $K$ such that $HK = G$ and $H \cap K = \{e\}$, then $G \cong H \times K$. 
Here, $C$ is a normal subgroup of $G$, as the center of a group is always a normal subgroup of that group. 
But, although I know that $G/C$ has a group structure, how can I say that $G/C$ is a normal subgroup of $G$ ? 
Further, I believe it's true that $C \cap G/C = \{e\}$ (inherently, since $G/C$ is, by definition, modding out by $C$). 
But, how can I see that $C(G/C) = G$ here ? I believe I'm supposed to use the property given about the group homomorphism, that $\phi(gC) \in gC$ for all $g \in G$ -- but I'm not sure how to use this property in a clever way to conclude the desired product is all of $G$. 
I appreciate your time and help. (=
 A: Sketch of the proof. 
The map $\phi: G/Z(G) \rightarrow G$ is a homomorphism. Hence we can define a map $\theta : G/Z(G) \rightarrow Z(G)$ through $\phi(\overline{g})=g\theta(g)$.One can check that since $\phi$ is a homomorphism, $\theta: G \rightarrow Z(G)$ is also a homomorphism and $\theta(z)=z^{-1}$ for all $z \in Z(G)$, in particular $\theta^2(z)=\theta(\theta(z))=z$ for all $z \in Z(G)$.
Now, define a map $f: G \rightarrow G/Z(G) \times Z(G)$, by $f(g)=(\overline{g}, \theta(g))$. It is easy to see that $f$ is a homomorphism. 
Let us check injectivity of $f$: assume $(\overline{g}, \theta(g))=(\overline{1},1)$. Then $\overline{g}=\overline{1}$, hence $g \in Z(G)$ and $\theta(g)=1=g^{-1}$, so $g=1$.
Finally the surjectivity of $f$: take an arbitrary element $(\overline{g},z) \in G/Z(G) \times Z(G)$. Put $x=z^{-1}g\theta(g)$, then $f(x)=(\overline{x}, \theta(z^{-1}g\theta(g))=(\overline{g},z\theta(g)\theta(g)^{-1})=(\overline{g},z)$.
A: Let $H=im \phi$. H is isomorphic to $G/Z.$
Notice that $H$ has the following properties, which follow from your definition of $\phi$


*

*H and Z intersect trivially at identity.

*Every element of G is the product of some element of H and some element of Z.

*H and Z commute element wise (as Z is the center).


Therefore we can apply the direct product theorem to get your result.
