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. (=

• Your idea to use the "$H$ and $K$" result is a good one. On the other hand, $G/C$ is not a subset of $G$, so $C \cap G/C$ is not defined. You can still use your general approach though if, instead of using $G/C$, you use a subgroup of $G$ isomorphic to $G/C$. Consider the mapping $\phi$ you are given. What can you prove about it? Is it injective? If yes, what does that tell you? Jan 19, 2020 at 23:26
• I see. I believe the mapping $\phi$ I'm given gives us an isomorphism from $G/C$ to a subgroup of $G$, correct ? If I call this subgroup $H$, how can I see that this is a normal subgroup of $G$ ? And that $CH = G$ ? Jan 20, 2020 at 1:57
• What have you tried? Have you tried proving $H$ is normal directly from the definition? Same for $CH=G$. Take an arbitrary element $g\in G$. Can you see how to write it as $g=ch$ with $c\in C$ and $h\in H$? Jan 20, 2020 at 2:24
• Okay! Will do. I thought there might be some tricks to it. I'll try it directly from the definitions, as you suggested. Thank you! (= Jan 20, 2020 at 2:25

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 $$f(x)=(\overline{x}, \theta(z^{-1}g\theta(g))=(\overline{g},z\theta(g)\theta(g)^{-1})=(\overline{g},z)$$.

Let $$H=im \phi$$. H is isomorphic to $$G/Z.$$ Notice that $$H$$ has the following properties, which follow from your definition of $$\phi$$

1. H and Z intersect trivially at identity.
2. Every element of G is the product of some element of H and some element of Z.
3. H and Z commute element wise (as Z is the center).

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

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:

1. $$\varphi$$ is injective:

indeed, if $$g_1\mathcal Z(G),g_2\mathcal Z(G)\in G/\mathcal Z(G),g_1\mathcal Z(G)\ne g_2\mathcal Z(G),$$ then $$g_1\mathcal Z(G)\cap g_2\mathcal Z(G)=\emptyset$$ and $$\varphi(g_1\mathcal Z(G))\in g_1\mathcal Z(G)$$ and $$\varphi(g_2\mathcal Z(G))\in g_2\mathcal Z(G)$$ must be different as elements of two disjoint sets.

1. Since $$\varphi$$ is injective, $$G/\mathcal Z(G)\cong\operatorname{im}\varphi.$$

2. $$\operatorname{im}\varphi$$ intersects each coset $$g\mathcal Z(G)$$ at exactly one point.

Otherwise, if there were $$g\in G$$ such that $$\operatorname{im}\varphi\cap g\mathcal Z(G)$$ contained two distinct elements $$h_1, h_2,\in g\mathcal Z(G),h_1=\varphi(g_1\mathcal Z(G)), h_2=\varphi(g_2\mathcal Z(G)),$$ then $$g_1\mathcal Z(G),g_2\mathcal Z(G)\in G/\mathcal Z(G)$$ would have been distinct with $$\varphi(g_1\mathcal Z(G))\mathcal Z(G)=h_1\mathcal Z(G)=g\mathcal Z(G)=h_2\mathcal Z(G)=\varphi(g_2\mathcal Z(G))\mathcal Z(G),$$ (because $$h_i\in g\mathcal Z(G)\Leftrightarrow h_i\mathcal Z(G)=g\mathcal Z(G)$$) and we reach a contradiction with 1.

1. $$\operatorname{im}\varphi\cap\mathcal Z(G)=\{e_G\}.(*)$$

From the previous point, having in mind $$\varphi(\mathcal Z(G))=e_G\in\mathcal Z(G)\le G,$$ we conclude that $$\operatorname{im}\varphi\cap\mathcal Z(G)=\{e_G\}.$$

1. 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}

2. $$\operatorname{im}\varphi$$ is normal in $$G.$$

Let $$\varphi(g\mathcal Z(G))\in\operatorname{im}\varphi$$ and $$x\in G$$ be arbitrary. Then, by the previous point: \begin{aligned}x\varphi(g\mathcal Z(G))x^{-1}&=\underbrace{z(x)}_{\in\mathcal Z(G)}\varphi(x\mathcal Z(G))\varphi(g\mathcal Z(G))\varphi(x\mathcal Z(G))^{-1}\underbrace{z(x)^{-1}}_{\in\mathcal Z(G)}\\&\overset{\substack{\varphi \text{ is a}\\\text{homo-}\\\text{ morphism}}}{=}z(x)z(x)^{-1}\varphi(xgx^{-1}\mathcal Z(G))\\&=\varphi(xgx^{-1}\mathcal Z(G))\in\operatorname{im}\varphi\end{aligned}

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.