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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. (=

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    $\begingroup$ 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? $\endgroup$
    – verret
    Jan 19, 2020 at 23:26
  • $\begingroup$ 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 $CH = G$ ? $\endgroup$ Jan 20, 2020 at 1:57
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    $\begingroup$ 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 $h\in H$? $\endgroup$
    – verret
    Jan 20, 2020 at 2:24
  • $\begingroup$ 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! (= $\endgroup$ Jan 20, 2020 at 2:25

3 Answers 3

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

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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.

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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.

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