I have a proof of the fact that an abelian group $I$ is injective $\iff$ it is divisible.

If $I$ is injective, then applying the definition of injective to the inclusion $n\mathbb{Z} \hookrightarrow \mathbb{Z}$ and the homomorphism $n\mathbb{Z}\rightarrow \mathbb{Z}$ taking $nk\mapsto kd$ shows that $I$ is divisible.

Why? I can't connect this with definition of divisible group.


2 Answers 2


Take the homomorphism $f:n\mathbb Z\to I$ given by $n\mapsto d$.

It extends to $\hat f:\mathbb Z\to I$ such that $\hat f(nk)=f(nk)$.

As $\hat f$ is $\mathbb Z$-linear, you have that $\hat f(nk)=n\hat f(k)$.

When $k=1$, you have $d=f(n)=\hat f(n\cdot 1)=n\hat f(1)$.

Thus the element $\hat f(1)\in I$ multiplies with $n$ to make $d$. Thus $I$ is divisible.

  • 1
    $\begingroup$ This makes use of the inclusion $n\mathbb Z\to \mathbb Z$ that you mentioned. Along with $f$ and $\hat f$ it makes the commutative diagram. $\endgroup$
    – rschwieb
    Commented Mar 12, 2017 at 17:05
  • $\begingroup$ Quick question: How do we know that for any $d\in I$, there will exist homomorphism $f\in$ Hom$(n\mathbb{Z},I)$ where $f:n\mapsto d$? I suspect it has to do with the fact -- I know for finite groups -- that $f:x\mapsto d$ (f\in Hom$(<x:x^n>,G)$) induces a homomorphism if and only if $o(d)\ |\ o(x)$. $\endgroup$
    – JAG131
    Commented May 19 at 23:37

Consider an injective module $I$ and $x\in I$. Suppose $n>0$ and consider

  1. the homomorphism $f\colon\mathbb{Z}\to I$, $f(z)=zx$

  2. the monomorphism $g\colon\mathbb{Z}\to\mathbb{Z}$, $g(z)=nz$

By injectivity, there exists $h\colon\mathbb{Z}\to I$ such that $hg=f$.

In particular, $$ x=f(1)=h(g(1))=h(n)=nh(1) $$

Proving that divisible modules are injective exploits the fact that $\mathbb{Z}$ is a PID. There are domains where divisible modules (with obvious definition) may not be injective.


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