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. (=
Best Answer
In this answer, I tried to address all the OP' questions from the post and comments and realize their idea.
Given a homomorphism $\varphi:G/\mathcal Z(G)\to G,$ with the property $\varphi(g\mathcal Z(G))\in g\mathcal Z(G),$ we can conclude the following:
Since $\varphi$ is injective, $G/\mathcal Z(G)\cong\operatorname{im}\varphi.$
$\operatorname{im}\varphi$ intersects each coset $g\mathcal Z(G)$ at exactly one point.
Now, $$\begin{aligned}&\forall g\in G,\color{blue}{\varphi(g\mathcal Z(G))\in} g\mathcal Z(g)\overset{\substack{\mathcal Z(G)\\\text{normal}\\\text{ in } G}}{=}\color{blue}{\mathcal Z(G)g}\\\Leftrightarrow &\forall g\in G, g\in\mathcal Z(G)\varphi(g\mathcal Z(G))\\\Leftrightarrow &\forall g\in G,\exists z(g)\in\mathcal Z(G),\boxed{g=z(g)\varphi(g\mathcal Z(G))}.(**)\end{aligned}$$
$\operatorname{im}\varphi$ is normal in $G.$
Therefore:
(1) $\operatorname{im}\varphi,\mathcal Z(G)\unlhd G$
(2) $\operatorname{im}\varphi\cap\mathcal Z(G)=\{e_G\}$ (by $(*)$).
(3) $G\overset{(**)}{=}\operatorname{im}\varphi\mathcal Z(G)$
and the claim follows just as the OP had in mind.