MHB Is an Abelian Group Divisible?

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Here is this week's POTW:

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An abelian group $G$ is called divisible if for every $n\in \Bbb N$ and $g\in G$, there exists $x\in G$ such that $nx = g$. Show that for abelian groups $G$, $G$ is injective if and only if $G$ is divisible.

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No one answered this week's problem. You can read my solution below.

Spoiler

Let $G$ be an abelian group.

Suppose $G$ is injective. Fix $n\in \Bbb N$ and $g\in G$. Let $\phi : \Bbb Z\to \Bbb Z$ be given by $\phi(m) = nm$, and $\psi : \Bbb Z\to G$ be defined by $\psi(m) = mg$. Since $\phi$ is a monomorphism, injectivity of $G$ allows for a lift $\alpha : \Bbb Z \to G$ such that $\alpha(\psi(m)) = \phi(m)$ for all $m\in \Bbb Z$. So if $x = \alpha(1)$, then $nx = \alpha(n) = \alpha(\psi(1)) = \phi(1) = g$. Since $n$ and $g$ were arbitrary, $G$ is divisible.

Conversely, suppose $G$ is divisible. Let $\phi : A \to B$ be a monomorphism of abelian groups and $\psi : A \to G$ be a homomorphism. Without loss of generality, assume $A\subset B$. Let $\mathcal{C}$ be the collection of all pairs $(X,f)$ where $X$ is an abelian group with $A \subset X \subset B$ and $f : X \to G$ is an homomorphic extension of $\phi$. Partially order $\mathcal{C}$ by setting $(X,f) < (X', f')$ if and only if $X\subset X'$ and $f'$ is a homomorphic extension of $f$. Then $\mathcal{C}$ is inductive ordered set, and hence (by Zorn's lemma) $\mathcal{C}$ has a maximal element $(\tilde{X},\tilde{f})$. Suppose $\tilde{X} \neq B$. Then $B\setminus \tilde{X}$ contains some element $x_0$. If $x_0$ has the property that $nx_0\in \tilde{X}$ implies $n = 0$, then $\tilde{f}$ has a homomorphic extension $\tilde{F} : \tilde{X} + \Bbb Z x_0 \to G$ such that $\tilde{F}(x + nx_0) = \tilde{f}(x)$ for all $x\in \tilde{X}$ and $n\in \Bbb Z$. Then $(\tilde{X} + \Bbb Z x_0, \tilde{F})$ is a an element of $\mathcal{C}$ greater than $(\tilde{X}, \tilde{f})$, contradicting maximality of $(\tilde{X}, \tilde{f})$. On the other hand, if there exists an $n\in \Bbb Z$ such that $nx_0\in \tilde{X}$, then we may consider $n_0$, the least positive integer $n$ such that $nx_0\in \tilde{X}$. Since $G$ is divisible, there exists a $g\in G$ such that $n_0 g = \tilde{f}(n_0x_0)$. The mapping $F : \tilde{X} + \Bbb Z x_0 \to G$ given by $F(x + nx_0) = f(x) + ng$ for all $x\in \tilde{X}$ and $n\in \Bbb Z$, is a homomorphic extension of $\tilde{f}$. Yet again, this contradicts maximality of $(\tilde{X}, \tilde{f})$. Therefore $\tilde{X} = B$ and $\tilde{f}\circ \phi = \psi$. Since $\phi$ and $\psi$ were arbitrary, $G$ is injective.